COMPLETE BUSINESS STATISTICS

4-1

COMPLETE COMPLETE BUSINESS BUSINESS

STATISTICSSTATISTICSbyby

AMIR D. ACZELAMIR D. ACZEL

&&

JAYAVEL SOUNDERPANDIANJAYAVEL SOUNDERPANDIAN

66thth edition (SIE) edition (SIE)

4-2

Chapter 4 Chapter 4

The Normal DistributionThe Normal Distribution

4-3

Using StatisticsProperties of the Normal DistributionThe TemplateThe Standard Normal DistributionThe Transformation of Normal Random

VariablesThe Inverse TransformationThe Normal Approximation of Binomial

Distributions

The Normal DistributionThe Normal Distribution44

4-4

Identify when a random variable will be normally distributed

Use the properties of normal distributionsExplain the significance of the standard normal distributionUse normal distribution tables to compute probabilitiesTransform a normal distribution into a standard normal

distributionConvert a binomial distribution into an approximate normal

distributionUse spreadsheet templates to solve normal distribution

problems

LEARNING OBJECTIVESLEARNING OBJECTIVES44

After studying this chapter, you should be able to:After studying this chapter, you should be able to:

4-5

As n increases, the binomial distribution approaches a ...

n = 6 n = 14n = 10

Normal Probability Density Function:

6543210

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0.0

x

P(x

)

Binomial Distribution: n=6, p=.5

109876543210

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0.2

0.1

0.0

x

P(x

)


14131211109876543210

0.3

0.2

0.1

0.0

x

P(x

)


50-5

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0.1

0.0

x

f(x)

Normal Distribution: = 0, = 1

4-1 Introduction4-1 Introduction

...14159265.3 and ...7182818.2 where

for 22

2

22

1)(

e

x

x

exf

4-6

The normal probability density function:

50-5

0.4

0.3

0.2

0.1

0.0

x

f(x)


The Normal Probability DistributionThe Normal Probability Distribution

f x e

x

x

e

( )

. ... . ...

1

2 2

2

2 2

2 7182818 314159265

for

where and

4-7

• The normal is a family ofBell-shaped and symmetric distributions. because the

distribution is symmetric, one-half (.50 or 50%) lies on either side of the mean.

Each is characterized by a different pair of mean, , and variance, . That is: [X~N()].

Each is asymptotic to the horizontal axis.The area under any normal probability density

function within k of is the same for any normal distribution, regardless of the mean and variance.

4-2 Properties of the Normal 4-2 Properties of the Normal DistributionDistribution

4-8

• If several independent random variables are normally distributed then their sum will also be normally distributed.

• The mean of the sum will be the sum of all the individual means.

• The variance of the sum will be the sum of all the individual variances (by virtue of the independence).

4-2 Properties of the Normal 4-2 Properties of the Normal Distribution (continued)Distribution (continued)

4-9

• If X1, X2, …, Xn are independent normal random variable, then their sum S will also be normally distributed with

• E(S) = E(X1) + E(X2) + … + E(Xn)

• V(S) = V(X1) + V(X2) + … + V(Xn)

• Note: It is the variances that can be added above and not the standard deviations.


4-10

Example 4.1: Let X1, X2, and X3 be independent random variables that are normally distributed with means and variances as shown.

4-2 Properties of the Normal 4-2 Properties of the Normal Distribution – Example 4-1Distribution – Example 4-1

Mean Variance

X1 10 1

X2 20 2

X3 30 3

Let S = X1 + X2 + X3. Then E(S) = 10 + 20 + 30 = 60 and V(S) = 1 + 2 + 3 = 6. The standard deviation of S is = 2.45.

6

4-11

• If X1, X2, …, Xn are independent normal random variable, then the random variable Q defined as Q = a1X1 + a2X2 + … + anXn + b will also be normally distributed with

• E(Q) = a1E(X1) + a2E(X2) + … + anE(Xn) + b• V(Q) = a1

2 V(X1) + a22 V(X2) + … + an

2 V(Xn)• Note: It is the variances that can be added above

and not the standard deviations.


4-12

Example 4.3: Let X1 , X2 , X3 and X4 be independent random variables that are normally distributed with means and variances as shown. Find the mean and variance of Q = X1 - 2X2 + 3X2 - 4X4 + 5

4-2 Properties of the Normal 4-2 Properties of the Normal Distribution – Example 4-3Distribution – Example 4-3

Mean Variance

X1 12 4

X2 -5 2

X3 8 5

X4 10 1

E(Q) = 12 – 2(-5) + 3(8) – 4(10) + 5 = 11

V(Q) = 4 + (-2)2(2) + 32(5) + (-4)2(1) = 73

SD(Q) = 544.873

4-13Computing the Mean, Variance and Standard Computing the Mean, Variance and Standard Deviation for the Sum of Independent Random Deviation for the Sum of Independent Random Variables Using the TemplateVariables Using the Template

4-14

All of these are normal probability density functions, though each has a different mean and variance.

Z~N(0,1)

50-5

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0.0

z

f(z)

Normal Distribution: =0, =1

W~N(40,1) X~N(30,25)

454035

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w

f(w)


6050403020100

0.2

0.1

0.0

x

f(x)


Y~N(50,9)

65554535

0.2

0.1

0.0

y

f(y)


50

Consider:

P(39 W 41)P(25 X 35)P(47 Y 53)P(-1 Z 1)

The probability in each case is an area under a normal probability density function.

Normal Probability DistributionsNormal Probability Distributions

4-15

4-3 Computing Normal Probabilities 4-3 Computing Normal Probabilities Using the TemplateUsing the Template

4-16

The standard normal random variable, Z, is the normal random variable with mean = 0 and standard deviation = 1: Z~N(0,12).

543210- 1- 2- 3- 4- 5

0 . 4

0 . 3

0 . 2

0 . 1

0 . 0

Z

f(z )

Standard Normal Distribution

= 0

=1{

4-4 The Standard Normal 4-4 The Standard Normal DistributionDistribution

4-17

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41771.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44411.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45451.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46331.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47061.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.47672.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48172.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48572.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.48902.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.49162.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.49642.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49742.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.49812.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.49863.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

543210-1-2-3-4-5

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Z

f( z)


1.56{

Standard Normal Probabilities

Look in row labeled 1.5 and column labeled .06 to find P(0 z 1.56) = 0.4406

Finding Probabilities of the Standard Finding Probabilities of the Standard Normal Distribution: Normal Distribution: P(0 < Z < 1.56)P(0 < Z < 1.56)

4-18

To find P(Z<-2.47):Find table area for 2.47

P(0 < Z < 2.47) = .4932

P(Z < -2.47) = .5 - P(0 < Z < 2.47) = .5 - .4932 = 0.0068

543210-1-2-3-4-5

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Z

f(z)


Table area for 2.47P(0 < Z < 2.47) = 0.4932

Area to the left of -2.47P(Z < -2.47) = .5 - 0.4932= 0.0068

Finding Probabilities of the Standard Finding Probabilities of the Standard Normal Distribution: Normal Distribution: P(Z < -2.47)P(Z < -2.47)

z ... .06 .07 .08. . . .. . . .. . . .2.3 ... 0.4909 0.4911 0.49132.4 ... 0.4931 0.4932 0.49342.5 ... 0.4948 0.4949 0.4951...

4-19

Finding Probabilities of the Standard Finding Probabilities of the Standard Normal Distribution: Normal Distribution: P(1< Z < 2)P(1< Z < 2)

z .00 ... . . . . . .0.9 0.3159 ...1.0 0.3413 ...1.1 0.3643 ... . . . . . .1.9 0.4713 ...2.0 0.4772 ...2.1 0.4821 ... . . . . . .

To find P(1 Z 2):1. Find table area for 2.00

F(2) = P(Z 2.00) = .5 + .4772 =.9772

2. Find table area for 1.00

F(1) = P(Z 1.00) = .5 + .3413 = .8413

3. P(1 Z 2.00) = P(Z 2.00) - P(Z 1.00)

= .9772 - .8413 = 0.1359

543210-1-2-3-4-5

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Z

f(z)


Area between 1 and 2P(1 Z 2) = .9772 - .8413 = 0.1359

4-20

Finding Values of the Standard Normal Finding Values of the Standard Normal Random Variable: Random Variable: P(0 < Z < z) = 0.40P(0 < Z < z) = 0.40

To find z such that P(0 Z z) = .40:

1. Find a probability as close as possible to .40 in the table of standard normal probabilities.

2. Then determine the value of z from the corresponding row and column.

P(0 Z 1.28) .40

Also, since P(Z 0) = .50

P(Z 1.28) .90543210-1-2-3-4-5

0.4

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0.0

Z

f(z)


Area = .40 (.3997)

Z = 1.28

Area to the left of 0 = .50P(z 0) = .50

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4-21

z .04 .05 .06 .07 .08 .09. . . . . . . . . . . . . .. . . . . . .2.4 ... 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 ... 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 ... 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964. . . . . . .. . . . . . .. . . . . . .

To have .99 in the center of the distribution, there should be (1/2)(1-.99) = (1/2)(.01) = .005 in each tail of the distribution, and (1/2)(.99) = .495 in each half of the .99 interval. That is:

P(0 Z z.005) = .495

Look to the table of standard normal probabilities to find that:

z.005 z.005

P(-.2575 Z ) = .99

To have .99 in the center of the distribution, there should be (1/2)(1-.99) = (1/2)(.01) = .005 in each tail of the distribution, and (1/2)(.99) = .495 in each half of the .99 interval. That is:

P(0 Z z.005) = .495

Look to the table of standard normal probabilities to find that:

z.005 z.005

P(-.2575 Z ) = .99

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)

-z.005 z.005

Area in right tail = .005Area in left tail = .005

Area in center right = .495

Area in center left = .495

2.575-2.575

Total area in center = .99

99% Interval around the Mean99% Interval around the Mean

4-22

The area within k of the mean is the same for all normal random variables. So an area under any normal distribution is equivalent to an area under the standard normal. In this example: P(40 X P(-1 Z sinceand

The area within k of the mean is the same for all normal random variables. So an area under any normal distribution is equivalent to an area under the standard normal. In this example: P(40 X P(-1 Z sinceand

1009080706050403020100

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0.00

X

f(x)


=10{

543210-1-2-3-4-5

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Z

f(z)


1.0

{

Transformation

(2) Division by x)

The transformation of X to Z:The transformation of X to Z:

The inverse transformation of Z to X:The inverse transformation of Z to X:

4-5 The Transformation of Normal 4-5 The Transformation of Normal Random VariablesRandom Variables

(1) Subtraction: (X - x)

ZX x

x

X x Z x

4-23

Example 4-9Example 4-9

X~N(160,302)


X~N(160,302)


X~N(127,222)


X~N(127,222)

Using the Normal TransformationUsing the Normal Transformation

P X

PX

P Z

P Z

( )

.. . .

100 180100 180

100 160

30

180 160

302 6667

0 4772 0 2475 0 7247

P X

PX

P Z

P Z

( )

.. . .

150150

150 127

22

1 0450 5 0 3520 0 8520

4-24


X~N(383,122)


X~N(383,122)

440390340

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0.00

X

f( X)


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Z

f(z)


Equivalent areas

Using the Normal Transformation - Using the Normal Transformation - Example 4-11Example 4-11

Template solutionTemplate solution

P X

PX

P Z

P Z

( )

. .

. . .

394 399

394 399

394 383

12

399 383

12

0 9166 1 333

0 4088 0 3203 0 0885

4-25

The transformation of X to Z:The transformation of X to Z: The inverse transformation of Z to X:The inverse transformation of Z to X:

The transformation of X to Z, where a and b are numbers::The transformation of X to Z, where a and b are numbers::

The Transformation of Normal The Transformation of Normal Random VariablesRandom Variables

P X a P Za

P X b P Zb

P a X b Pa

Zb

( )

( )

( )

ZX x

x

X

xZ

x

4-26

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Z

f(z)

S tand ard N o rm al D is trib utio n• The probability that a normal random variable will be within 1 standard deviation from its mean (on either side) is 0.6826, or approximately 0.68.

• The probability that a normal random variable will be within 2 standard deviations from its mean is 0.9544, or approximately 0.95.

• The probability that a normal random variable will be within 3 standard deviation from its mean is 0.9974.

Normal Probabilities (Empirical Rule)Normal Probabilities (Empirical Rule)

4-27

z .07 .08 .09 . . . . . . . . . . . . . . .1.1 . . . 0.3790 0.3810 0.38301.2 . . . 0.3980 0.3997 0.40151.3 . . . 0.4147 0.4162 0.4177 . . . . . . . . . . . . . . .

The area within k of the mean is the same for all normal random variables. To find a probability associated with any interval of values for any normal random variable, all that is needed is to express the interval in terms of numbers of standard deviations from the mean. That is the purpose of the standard normal transformation. If X~N(50,102),

That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 = + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal distribution.

The area within k of the mean is the same for all normal random variables. To find a probability associated with any interval of values for any normal random variable, all that is needed is to express the interval in terms of numbers of standard deviations from the mean. That is the purpose of the standard normal transformation. If X~N(50,102),

That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 = + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal distribution.

Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28) 0.10x = + z = 124 + (1.28)(12) = 139.36


18013080

0.04

0.03

0.02

0.01

0.00

X

f(x)


4-6 The Inverse Transformation4-6 The Inverse Transformation

0.01

139.36

P X Px

P Z P Z( ) ( )

70

70 70 50

102

4-28



Template Solution for Example 4-12Template Solution for Example 4-12

4-29

Example 4-13 X~N(5.7,0.52)P(X > x)=0.01 and P(Z > 2.33) 0.01x = + z = 5.7 + (2.33)(0.5) = 6.865

Example 4-13 X~N(5.7,0.52)P(X > x)=0.01 and P(Z > 2.33) 0.01x = + z = 5.7 + (2.33)(0.5) = 6.865

Example 4-14 X~N(2450,4002)P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)P(1666 < X < 3234) = 0.95

Example 4-14 X~N(2450,4002)P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)P(1666 < X < 3234) = 0.95

z .05 .06 .07 . . . . . . . . . . . . . . .1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . .

4000300020001000

0.0015

0.0010

0.0005

0.0000

X

f(x)

Normal Distribution: = 2450 = 400

4000300020001000

0.0015

0.0010

0.0005

0.0000

543210-1-2-3-4-5

Z

.4750.4750

.0250.0250

-1.96 1.96

The Inverse Transformation (Continued)The Inverse Transformation (Continued)

z .02 .03 .04 . . . . . . . . . . . . . . .2.2 . . . 0.4868 0.4871 0.48752.3 . . . 0.4898 0.4901 0.49042.4 . . . 0.4922 0.4925 0.4927 . . . . . . . . . . . . . . .

8.27.26.25.24.23.2

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

8.27.26.25.24.23.2

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

X

f(x)

Normal Distribution: = 5.7 = 0.5

543210-1-2-3-4-5

z Z.01 = 2.33

Area = 0.49

Area = 0.01

X.01 = +z = 5.7 + (2.33)(0.5) = 6.865

4-30

4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)


.

.

.

.

.

.

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)

Standard Norm al D istribution

1. Draw pictures of the normal distribution in question and of the standard normal distribution.


Finding Values of a Normal Random Finding Values of a Normal Random Variable, Given a ProbabilityVariable, Given a Probability

4-31



2. Shade the area corresponding to the desired probability.



4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)


.

.

.

.

.

.

.4750.4750

.9500

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


.4750.4750

.9500

4-32

z .05 .06 .07 . . . . . . . . . . . . . . .1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . .

3. From the table of the standard normal distribution, find the z value or values.







4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)


.

.

.

.

.

.

.4750.4750

.9500

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


.4750.4750

.9500

-1.96 1.96

4-33

4. Use the transformation from z to x to get value(s) of the original random variable.

4. Use the transformation from z to x to get value(s) of the original random variable.

x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)

x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)


z .05 .06 .07 . . . . . . . . . . . . . . .1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . .







4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)


.

.

.

.

.

.

.4750.4750

.9500

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


.4750.4750

.9500

-1.96 1.96

4-34

1050

0.3

0.2

0.1

0.0

X

f(x)

Normal Distribution: = 3.5, = 1.323

76543210

0.3

0.2

0.1

0.0

X

P(x

)

Binomial Distribution: n = 7, p = 0.50

The normal distribution with = 3.5 and = 1.323 is a close approximation to the binomial with n = 7 and p = 0.50.

The normal distribution with = 3.5 and = 1.323 is a close approximation to the binomial with n = 7 and p = 0.50.

P(x<4.5) = 0.7749

MTB > cdf 4.5;SUBC> normal 3.5 1.323.Cumulative Distribution Function

Normal with mean = 3.50000 and standard deviation = 1.32300

x P( X <= x) 4.5000 0.7751

MTB > cdf 4.5;SUBC> normal 3.5 1.323.Cumulative Distribution Function

Normal with mean = 3.50000 and standard deviation = 1.32300

x P( X <= x) 4.5000 0.7751

MTB > cdf 4;SUBC> binomial 7,.5.Cumulative Distribution Function

Binomial with n = 7 and p = 0.500000

x P( X <= x) 4.00 0.7734

MTB > cdf 4;SUBC> binomial 7,.5.Cumulative Distribution Function

Binomial with n = 7 and p = 0.500000

x P( X <= x) 4.00 0.7734

P( x 4) = 0.7734


4-35

4-7 The Normal Approximation of Binomial 4-7 The Normal Approximation of Binomial DistributionDistribution

1050

0.3

0.2

0.1

0.0

X

f(x)

Normal Distribution: = 5.5, = 1.6583

11109876543210

0.2

0.1

0.0

X

P(x

)

Binomial Distribution: n = 11, p = 0.50

The normal distribution with = 5.5 and = 1.6583 is a closer approximation to the binomial with n = 11 and p = 0.50.

The normal distribution with = 5.5 and = 1.6583 is a closer approximation to the binomial with n = 11 and p = 0.50.

P(x < 4.5) = 0.2732

P(x 4) = 0.2744

4-36

or:

NOTE: If p is either small (close to 0) or large (close to 1), use the Poisson approximation.

NOTE: If p is either small (close to 0) or large (close to 1), use the Poisson approximation.

Approximating a Binomial Probability Approximating a Binomial Probability Using the Normal Distribution Using the Normal Distribution

P a X b Pa np

np pZ

b npnp p

( ) ( ) ( )

1 1

for large (n 50) and not too close to 0 or 1.00n p

P a X b Pa np

np pZ

b npnp p

( ) .( )

.( )

0 51

0 51

for moderately large (20 n < 50).n

4-37

Using the Template for Normal Approximation Using the Template for Normal Approximation of the Binomial Distributionof the Binomial Distribution

Date post:	03-Jan-2016
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