Statistics-The Normal Probability Distribution 1

8/3/2019 Statistics-The Normal Probability Distribution 1

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The Normal Probability

DistributionChapter


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TO LIST THE CHARACTERISTICS OF THENORMAL DISTRIBUTION.

TO DEFINE AND CALCULATE ZVALUES.

TO DETERMINE PROBABILITIESASSOCIATED WITH THE STANDARD

NORMAL DISTRIBUTION.

TO USE THE NORMAL DISTRIBUTION TO

APPROXIMATE THE BINOMIAL

DISTRIBUTION.

THIS CHAPTERS GOALS


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The normal curve is bell-shapedand has a singlepeak at the exact center of the distribution.

The arithmetic mean, median, and mode of the

distribution are equal and located at the peak.Half the area under the curve is above this

center point, and the other half is below it.

The normal probability distribution issymmetricalabout its mean.

It is asymptotic - the curve gets closer and closer

to the x-axis but never actually touches it.

CHARACTERISTICS OF A NORMAL

PROBABILITY DISTRIBUTION


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CHARACTERISTICS OF A NORMAL DISTRIBUTION

Theoretically, curve

extends to - infinity

Theoretically, curve

extends to + infinityMean, median, and

mode are equal

Tail Tail

Normal curve is symmetrical

- two halves identical -


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Normal Distributions with Equal Means

but Different Standard Deviations.

Q!

W!W!3.9W = 5.0


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Normal Probability Distributions with Different

Means and Standard Deviations.

Q= 5, W = 3Q = 9, W= 6Q = 14, W = 10


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A normal distribution with a mean of 0 and astandard deviation of 1 is called the standard

normal distribution.

zvalue: The distance between a selected value,designatedX, and the population mean Q,divided by the population standard deviation, W

The z-value is the number of standard deviations

Xis from the mean.

THE STANDARD NORMAL

PROBABILITY DISTRIBUTION

Z

X! QW


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The monthly incomes of recent MBA graduatesin a large corporation are normally distributed

with a mean of $2,000 and a standard deviation

of $200. What is the zvalue for an incomeXof

$2,200? $1,700?

For X = $2,200 and since z= (X - QWthenz= (2,200 - 2,000)/200 = 1.

A zvalue of 1 indicates that the value of $2,200 is1 standard deviation above the mean of $2,000.

EXAMPLE


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For X = $1,700 and sincez

= (X - QWthenz= (1,700 - 2,000)/200 = -1.5.A zvalue of -1.5 indicates that the value of

$2,200 is 1.5 standard deviation below the mean

of $2,000.

How might a corporation use this type of

information?

EXAMPLE (continued)


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About

68percentof the area under the normalcurve is within plus one and minus one standard

deviation of the mean. This can be written as Q 1W.

About 95percentof the area under the normal

curve is within plus and minus two standard

deviations of the mean, written Q 2W.

Practically all (99.74 percent) of the area underthe normal curve is within three standard

deviations of the mean, written Q 3W.

AREAS UNDER THE NORMAL CURVE


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Q QW QW QWQWQWQW

Between:

1. 68.26%

2. 95.44%

3. 99.97%


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A typical need is to determine the probability of

a z-value being greater than or less than some

value.

Tabular Lookup (Appendix D, page 474)

EXCELFunction =NORMSDIST(z)

P(z)=?


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The daily water usage per person in Toledo,

Ohio is normally distributed with a mean of 20

gallons and a standard deviation of 5 gallons.

About 68% of the daily water usage per person

in Toledo lies between what two values?

EXAMPLE


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The daily water usage per person in Toledo (X),Ohio is normally distributed with a mean of 20

gallons and a standard deviation of 5 gallons.

What is the probability that a person selected at

random will use less than 20 gallons per day?

What is the probability that a person selected at

random will use more than 20 gallons per day?

EXAMPLE


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What percent uses between 20 and 24 gallons? The zvalue associated with X = 20 is z= 0 and

with X = 24, z= (24 - 20)/5 = 0.8 P(20

< X < 24) = P(0 < z< 0.8) = 0.2881=28.81%

What percent uses between 16 and 20 gallons?

EXAMPLE (continued)


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0.8

P(0 < z < 0.8)

= 0.2881


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What is the probability that a person selected atrandom uses more than 28 gallons?

EXAMPLE (continued)


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P(z > 1.6) =

0.5 - 0.4452 =

0.0048

Area =

z


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EXAMPLE (continued)

What percent uses between 18 and 26 gallons?


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Area =

0.1554 Area =

0.3849

Total area =

0.1554 + 0.3849 =

0.5403

z


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How many gallons or more do the top 10% ofthe population use?

LetXbe the least amount. Then we need to find

Ysuch that P(X u Y) = 0.1 To find the

corresponding zvalue look in the body of the

table for (0.5 - 0.1) = 0.4. The corresponding z

value is 1.28 Thus we have (Y- 20)/5 = 1.28,

from which Y= 26.4. That is, 10% of thepopulation will be using at least 26.4 gallons

daily.

EXAMPLE (continued)


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1.28

0.4

0.1

z

(Y- 20)/5 = 1.28

Y= 26.4


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A professor has determined that the final

averages in his statistics course is normally

distributed with a mean of 72 and a standard

deviation of 5. He decides to assign his grades

for his current course such that the top 15% ofthe students receive an A. What is the lowest

average a student must receive to earn an A?

EXAMPLE


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1.04

0.15

0.35

z

(Y- 72)/5 = 1.04

Y= 77.2


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The amount of tip the waiters in an exclusive

restaurant receive per shift is normally

distributed with a mean of $80 and a standard

deviation of $10. A waiter feels he has provided

poorservice if his total tip for the shift is lessthan $65. Based on his theory, what is the

probability that he has providedpoorservice?

EXAMPLE


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z- 1.5

Area =

Area =

0.5 - 0.4332 =0.0668


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Using the normal distribution (a continuousdistribution) as a substitute for a binomial

distribution (a discrete distribution) for large

values ofn

seems reasonable because asn

increases, a binomial distribution gets closer and

closer to a normal distribution.

When to use the normal approximation?

The normal probability distribution is generally

deemed a good approximation to the binomial

probability distribution when np and n(1 -p) are

both greater than 5.

THE NORMAL APPROXIMATION TO

THE BINOMIAL


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Binomial Distribution with n = 3 andp = 0.5.

0 1 2

0.3

0.4

0.5

P(r)

r0.25


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Binomial Distribution with n = 5 andp = 0.5.P(r)

r


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Binomial Distribution with n = 20 andp = 0.5.P(r)

r

Observe

the

Normal

shape.


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Recall for the binomial experiment: There are only two mutually exclusive outcomes

(success or failure) on each trial.

A binomial distribution results from countingthe number of successes.

Each trial is independent.

The probabilityp is fixed from trial to trial, andthe number of trials n is also fixed.

THE NORMAL APPROXIMATION

(continued)


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The value 0.5subtracted or added, depending onthe problem, to a selected value when a binomial

probability distribution, which is a discrete

probability distribution, is being approximatedby a continuous probability distribution--the

normal distribution.

The basic concept is that a slice of the normal

curve from x-0.5 to x+0.5 is approximately equal

to P(x).

CONTINUITY CORRECTION

FACTOR


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A recent study by a marketing research firm

showed that 15% of the homes had a video

recorder for recording TV programs. A sample

of 200 homes is obtained. (Let X be the number

of homes).

Of the 200 homes sampled how many would you

expect to have video recorders?

Q= np (0.15)(200) = 30 & n(1 - p) = 170What is the variance?

W2 = np(1 - p) = (30)(1- 0.15) = 25.5

EXAMPLE


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What is the standard deviation?

W = (25.5) = 5.0498.What is the probability that less than 40 homes

in the sample have video recorders?

We need P(X < 40) = P(X e39). So, using thenormal approximation, P(X e39.5)

} P[ze(39.5 - 30)/5.0498] = P(ze 1.8812)}P(ze 1.88) = 0.5 + 0.4699 = 0.9699Why did I use 39.5 ? ...

How would you calculate P(X=39) ?

EXAMPLE (continued)


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1.88

0.50.4699

P(z e 1.88) =

0.5 + 0.4699 =0.9699

z


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EXAMPLE (continued)What is the probability that more than 24 homes

in the sample have video recorders?


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-1.09

0.50.3621

P(z u -1.09) =

0.5 + 0.3621 =0.8621.

z


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EXAMPLE (continued)

What is the probability that exactly 40 homes inthe sample have video recorders?


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1.88

2.08

0.4699

0.4812

P(1.88 e z e2.08)= 0.4812 - 0.4699

= 0.0113

z

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