Chapter 6: Random Variables and the Normal Distribution 6 ...jga001/chapter 6 Alford.pdf · Chapter...

Chapter 6 Random Variables and the Normal Distribution

61 Discrete Random Variables

62 Binomial Probability Distribution

63 Continuous Random Variables and the Normal Probability Distribution

64 Standard Normal Distribution

65 Applications of the Normal Distribution


Objectives

By the end of this section I will be

able tohellip

1) Identify random variables

2) Explain what a discrete probability distribution is and construct probability distribution tables and graphs

3) Calculate the mean variance and standard deviation of a discrete random variable

Random Variables

A variable whose values are determined by chance

Chance in the definition of a random variable is crucial

Example 62 - Notation for random variables

Suppose our experiment is to toss a single fair

die and we are interested in the number

rolled We define our random variable X to be

the outcome of a single die roll

a Why is the variable X a random variable

b What are the possible values that the

random variable X can take

c What is the notation used for rolling a 5

d Use random variable notation to express the probability of rolling a 5

Example 62 continued

Solution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Types of Random Variables

Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process

Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example


(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4


06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas



Objectives


able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example


5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands


ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur


bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA


bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P


bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN


bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP


256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP


bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment


1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)


4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula


(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)


n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)


4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example


Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)




ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable


Objectives


able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean


standard deviations of the mean



FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it


Objectives


able tohellip

1) Find areas under the standard normal curve given a Z-value

2) Find the standard normal Z-value given an area

The Standard Normal (Z) Distribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

When the normal curve is plotted we plot Z on the horizontal axis

The Empirical Rule For Standard Normal Distribution

0 -3 -2 -1 1 2 3

Find the area that lies under the standard normal curve between Z=-1 and Z=2

Example

ANSWER

034+034+0135=0815

Example

Find the area that lies under the standard normal curve that lies above Z=3

Example

ANSWER

Since 100-997=03=0003

we take half of 0003 to get

00015

Example


In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z

Table C in Appendix on pages T-9 and T-10

1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1

2 It is on two pages with one page for negative z-scores and the other page for positive z-scores

3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score

Table C Standard Normal Distribution

Table continues to -00 in the appendix

Table continues to 34 in the appendix

4 When working with a graph avoid confusion

between z-scores and areas

z Score

Distance along horizontal scale of the standard

normal distribution refer to the leftmost

column and top row of Table C

Area

Region under the curve refer to the values in

the body of Table C

5 The part of the z-score denoting hundredths is

found across the top


Steps for finding areas under the standard normal curve

Table 67

Page 307

Do problems 18 and 20

Example

ANSWER

(18)

Example

Use Table C for Z=200

ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)

Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be

1-09830=0017

Example

Example

ANSWER

28(c)

Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be

1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)

Use Table C to find that the area to the left of Z=128 is 08997


We want the area to the between

Z=128 and Z=196 which will be

09750-08997=00753

Example

Example

ANSWER

38(c)

Use Table C to find that the area to the left of Z=-201 is 00222



Z=-201 and Z=237 which will be

09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example

Note this answer is the Z value in the

table that corresponds to

1 - 05120 = 04880

Summary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

Summary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal

Distribution

Objectives


able tohellip

1) Compute probabilities for a given value of any normal random variable

2) Find the appropriate value of any normal random variable given an area or probability

3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)

Standardizing a Normal Random Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Normal Distribution Step 1

Determine the random variable X the mean μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values


Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X

Step4

Find the area under the standard normal curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use

Do problem 12 again with calculator

Calculator Use

Do problems 68 again with calculator

Or use 10^99 as larger value and -10^99

as the smaller value

Finding Normal Data Values for Specified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X


Step 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest


Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use

Do problem 18 and 22 again with calculator

Calculator Use


Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values

Edmundscom reported that the average

amount that people were paying for a 2007

Toyota Camry XLE was $23400 Let X = price

and assume that price follows a normal

distribution with μ = $23400 and σ = $1000

Find the prices that separate the middle

95 of 2007 Toyota Camry XLE prices from

the bottom 25 and the top 25


Solution

Step 1


Let X = price μ = $23400 and σ = $1000

The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649

FIGURE 649


Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196


Solution

Step 3

Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360

The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

Page 324

Example

Solution

Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find

Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that

that is the 5th percentile (only 5 of his games did he have a lower score)

Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution

The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual

Example

Summary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

Summary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ


Objectives


able tohellip

1) Identify random variables

2) Explain what a discrete probability distribution is and construct probability distribution tables and graphs

3) Calculate the mean variance and standard deviation of a discrete random variable

Random Variables














Solution











Example

ANSWER

(a) continuous

(b) discrete

Example





Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Random Variables














Solution











Example

ANSWER

(a) continuous

(b) discrete

Example





Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary













Solution











Example

ANSWER

(a) continuous

(b) discrete

Example





Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Solution











Example

ANSWER

(a) continuous

(b) discrete

Example





Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary








Example

ANSWER

(a) continuous

(b) discrete

Example





Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





Example

ANSWER

(a) continuous

(b) discrete

Example





Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

(a) continuous

(b) discrete

Example





Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

(a) discrete

(b) continuous

(c) continuous

Example










Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary











Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






Example



That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




That is ΣP(X) = 1




Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Example


Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Example





06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example



25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP




Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





Denoted as E(X )





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





X P X



22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




22

2

Definition Formulas

X P X

X P X

2 2 2

2 2


X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001





Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

Page 270

Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example


Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

Page 270

Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

ANSWER


850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

ANSWER




x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

ANSWER


750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Summary





Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Summary







Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Objectives


able tohellip





Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Factorial symbol


0 = 1

1 = 1



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

Example

504012345677 2

2412344 1


Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Calculator

7 MATH PRB 4

which is


7

Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Combinations








Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





Identify r and n



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Example





Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






Example

Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Combination Formula




n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C


Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C




ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

9605982552C

Example






Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary







Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA




aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





aAAaAA






NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





NGGG GNGG GGNG GGGN



256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




256

27

4

3

4

3

4

3

4

1


256

27

4

3

4

3

4

1

4

3



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



256

27

4

3

4

1

4

3

4

3


256

27

4

1

4

3

4

3

4

3




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1


3

P

Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Binomial Experiment


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Binomial Experiment

Four Requirements





Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Binomial Experiment










where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




where

XnX

Xn ppCXP )1()(

XnX

nCXn


Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Let q=1-p

XnX

Xn qpCXP )(


Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Probability Formula


(n ndash x )x


trials



(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




(n ndash x )x


trials




4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


2ND VARS

Abinompdf(4 75 3)


n p x








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary








Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X



Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


20 coin tosses



Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 284

Example


Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Four Requirements





ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER



Example

229)5840)(50(np

ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER


Example

Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 285

Example


Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Four Requirements





ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER


Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER



Example

)3()2()1()0()3( XPXPXPXPXP



Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




Example



)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER


Example

213)110)(120(np

ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER


Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811





2score-2 z

Sample Population

Z score Formulas

s

xxz x

z



981433

21320z

Example

Summary



Summary




Objectives


able tohellip

















Variables)





above an interval







FIGURE 619










The Empirical Rule



properties apply
















Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

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Solution

Example

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)30(1)30( XPXP

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Example

Solution

Example

14640)2010( XP

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Solution

Find so that


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Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






2score-2 z

Sample Population

Z score Formulas

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Page 295

Example

ANSWER

Example

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Since 100-997=03=0003


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Table 67

Page 307


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ANSWER

(18)

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(20)

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Page 307


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26(c)


1-09830=0017

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Example

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28(c)


1-02451=07549

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Example

Page 308


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Example

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32(c)





09750-08997=00753

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38(c)





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Example

Do problem 42

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Example

Do problem 52

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Do problem 58

Example



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Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


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Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

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598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


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050)( 1XXP

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Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Sample Population

Z score Formulas

s

xxz x

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21320z

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Page 295

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Since 100-997=03=0003


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Table 67

Page 307


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ANSWER

(18)

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(20)

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26(c)


1-09830=0017

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Example

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28(c)


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Page 308


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32(c)





09750-08997=00753

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38(c)





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Example

Do problem 42

Example

Page 308

Example

Do problem 52

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Example

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Example



1 - 05120 = 04880

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Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

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Example

Calculator Use

Calculator Use


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Step 1





Step 2





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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

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1X

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Gives the value

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050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



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Page 295

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Table 67

Page 307


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ANSWER

(18)

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(20)

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26(c)


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Example

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28(c)


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32(c)





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38(c)





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Example

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Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

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Calculator Use

Calculator Use


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Step 1





Step 2





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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


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Example

ANSWER

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Since 100-997=03=0003


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Table 67

Page 307


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(20)

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26(c)


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28(c)


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32(c)





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Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

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Calculator Use

Calculator Use


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Step 1





Step 2





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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

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)30(1)30( XPXP

Page 324

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Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


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Solution

Gives the value

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points 84141X

Page 324

Example

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Example

ANSWER

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Summary



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the body of Table C





Table 67

Page 307


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ANSWER

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(20)

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26(c)


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28(c)


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32(c)





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Example

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Page 323

Example

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Solutions

6) 01587

8) 00062

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Calculator Use

Calculator Use


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Step 2





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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

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)30(1)30( XPXP

Page 324

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Solution

Example

14640)2010( XP

Page 324

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Solution

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Gives the value

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points 84141X

Page 324

Example

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(20)

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26(c)


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Page 323

Example

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Solutions

6) 01587

8) 00062

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Example

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Step 2





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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

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)30(1)30( XPXP

Page 324

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Solution

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Table 67

Page 307


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(20)

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26(c)


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32(c)





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Page 323

Example

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Solutions

6) 01587

8) 00062

10) 08413

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Example

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Step 2





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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

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)30(1)30( XPXP

Page 324

Example

Solution

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Page 324

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Table 67

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Page 323

Example

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Solutions

6) 01587

8) 00062

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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Example

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Table 67

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ANSWER

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26(c)


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Page 323

Example

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Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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Example

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Page 324

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Table 67

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(20)

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26(c)


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28(c)


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Example

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Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

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Step 2





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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


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28(c)


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Solutions

6) 01587

8) 00062

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

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X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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Example

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The Empirical Rule



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26(c)


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28(c)


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Solutions

6) 01587

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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Example

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26(c)


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28(c)


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32(c)





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Solutions

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

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Example

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Table 67

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26(c)


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28(c)


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32(c)





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Example

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Solutions

6) 01587

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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Example

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(20)

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26(c)


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28(c)


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32(c)





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Example

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Solutions

6) 01587

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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(20)

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26(c)


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28(c)


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32(c)





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Example

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Solutions

6) 01587

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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(20)

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26(c)


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28(c)


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32(c)





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Example

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Solutions

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8) 00062

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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1X

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(20)

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28(c)


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Example

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Solutions

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8) 00062

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Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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Solutions

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8) 00062

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Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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Solutions

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Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


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32(c)





09750-08997=00753

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Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

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Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


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Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

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)30(1)30( XPXP

Page 324

Example

Solution

Example

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Page 324

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Solution

Find so that


Example

050)( 1XXP

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Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




0 -3 -2 -1 1 2 3


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ANSWER

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Since 100-997=03=0003


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Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

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Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

Example

Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




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ANSWER

034+034+0135=0815

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ANSWER

Since 100-997=03=0003


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Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 296

Example

ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




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Example

ANSWER

034+034+0135=0815

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ANSWER

Since 100-997=03=0003


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Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

Example

Summary



Summary





Objectives


able tohellip





mean μ = 0 and




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Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


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Example












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Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

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Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Summary



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Objectives


able tohellip





mean μ = 0 and




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Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


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Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

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Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Summary





Objectives


able tohellip





mean μ = 0 and




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Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


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Example












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Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


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Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Objectives


able tohellip





mean μ = 0 and




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Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


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Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




mean μ = 0 and




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Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


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Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



0 -3 -2 -1 1 2 3


Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Example

ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

034+034+0135=0815

Example


Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Example

ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

Since 100-997=03=0003


00015

Example












z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary













z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary










z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary






z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary





z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




z Score




Area


the body of Table C





Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Table 67

Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 307


Example

ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

(18)

Example


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

(20)

Example


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 307


Example

Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

26(c)


1-09830=0017

Example

Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

28(c)


1-02451=07549

Example

Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 308


Example

Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 308


Example

Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

32(c)





09750-08997=00753

Example

Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


ANSWER

38(c)





09911-00222=09689

Example

Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 308

Example

Do problem 42

Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 308

Example

Do problem 52

Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example

Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 308

Example

Do problem 58

Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Example



1 - 05120 = 04880

Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Summary




Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Summary




Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Distribution

Objectives


able tohellip






xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




xZ







Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary








Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary




Step4



Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 323

Example

Do problems 681012

Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Solutions

6) 01587

8) 00062

10) 08413

12) 08400

Example

Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Calculator Use

Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Calculator Use


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Calculator Use





Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Step 1





Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Step 2





Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Step 3



Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 323

Example

Do problems 182022

Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Solutions

18) X=572

20) X=5355 and X=8645

22) X=442 and X=958

Example

Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Calculator Use


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Calculator Use












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary












Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Solution

Step 1


Let X = price μ = $23400 and σ = $1000


FIGURE 649


Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Solution

Step 2






Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary



Solution

Step 3


X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440

X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 324

Example

Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Solution


Example

)30(XP

Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Solution

Example

04013

598701

)30(1)30( XPXP

Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 324

Example

Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Solution

Example

14640)2010( XP

Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 324

Example

Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Solution

Find so that


Example

050)( 1XXP

1X

Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Solution

Gives the value

Example

050)( 1XXP

points 84141X

Page 324

Example

Solution


Example

Summary



Summary


Page 324

Example

Solution


Example

Summary



Summary


Solution


Example

Summary



Summary


Summary



Summary


Summary


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