Normal Distribution[1]

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1The Normal Probabil i ty

Distribution

Chapter


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2TO LIST THE CHARACTERISTICS OF THE

NORMAL DISTRIBUTION.

TO DEFINE AND CALCULATE ZVALUES.

TO DETERMINE PROBABILITIES

ASSOCIATED WITH THE STANDARD

NORMAL DISTRIBUTION.

TO USE THE NORMAL DISTRIBUTION TO

APPROXIMATE THE BINOMIAL

DISTRIBUTION.

THIS CHAPTERS GOALS


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3 The normal curve isbell-shapedand has a single

peak 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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4CHARACTERISTICS 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 -

0.5 0.5


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5Normal Distributions with Equal Meansbut Different Standard Deviations.

m = 20

s = 3.1s = 3.9s = 5.0


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6Normal Probability Distributions with DifferentMeans and Standard Deviations.

m = 5, s = 3m = 9, s = 6m = 14, s = 10


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7

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

normal distribution.

z value:The distance between a selected value,designated X, and the population mean m,divided by the population standard deviation, s.

The z-value is the number of standard deviations

Xis from the mean.

THE STANDARD NORMAL

PROBABILITY DISTRIBUTION

Z

X=

- m

s


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8 The monthly incomes of recent MBA graduates

in a large corporation are normally distributedwith a mean of $2,000 and a standard deviation

of $200. What is the zvalue for an income Xof

$2,200? $1,700? For X = $2,200 and since z= (X - m)/s, thenz= (2,200 - 2,000)/200 = 1.

Azvalue of 1 indicates that the value of $2,200 is1 standard deviationabovethe mean of $2,000.

EXAMPLE


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9 For X = $1,700 and since z= (X - m)/s, thenz= (1,700 - 2,000)/200 = -1.5.

Azvalue of -1.5 indicates that the value of

$2,200 is 1.5 standard deviation belowthe mean

of $2,000.

How might a corporation use this type of

information?

EXAMPLE (cont inued)


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10About 68 percentof the area under the normal

curve is within plus one and minus one standarddeviation of the mean. This can be written as m 1s.

About 95 percentof the area under the normalcurve is within plus and minus two standard

deviations of the mean, written m 2s. Practically all (99.74 percent)of the area under

the normal curve is within three standard

deviations of the mean, written m 3s.

AREAS UNDER THE NORMAL CURVE


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11

m m+1s m+2s m+3s-1s-2s+3s

Between:

1. 68.26%

2. 95.44%

3. 99.97%


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12A typical need is to determine the probability of

a z-value being greater than or less than somevalue.

Tabular Lookup (Appendix D, page 474)

EXCEL Function =NORMSDIST(z)

P(z)=?


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13 The daily water usage per person in Toledo,

Ohio is normally distributed with a mean of 20gallons 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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14 The daily water usage per person in Toledo (X),

Ohio is normally distributed with a mean of 20gallons and a standard deviation of 5 gallons.

What is the probability that a person selected at

random will use less than20 gallons per day?

What is the probability that a person selected at

random will use more than20 gallons per day?

EXAMPLE


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15What percent uses between 20 and 24 gallons?

The zvalue associated with X = 20 isz= 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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16

0.8

P(0 < z < 0.8)

= 0.2881


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17What is the probability that a person selected at

random uses more than 28 gallons?



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18

P(z > 1.6) =

0.5 - 0.4452 =

0.0048

Area =0.4452

1.6 z


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19EXAMPLE (cont inued)What percent uses between 18 and 26 gallons?


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20

1.2.4

Area =

0.1554 Area =

0.3849

Total area =

0.1554 + 0.3849 =

0.5403

z


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21How many gallons or more do the top 10% of

the population use?

Let Xbe the least amount. Then we need to find

Ysuch that P(X Y) = 0.1 To find thecorresponding zvalue look in the body of thetable 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 usingat least26.4 gallons

daily.



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22

1.28

0.4

0.1

z

(Y- 20)/5 = 1.28

Y= 26.4


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23A professor has determined that the final

averages in his statistics course is normallydistributed 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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24

1.04

0.15

0.35

z

(Y- 72)/5 = 1.04

Y= 77.2


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25 The amount of tip the waiters in an exclusive

restaurant receive per shift is normallydistributed 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 provided poorservice?

EXAMPLE


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26

z- 1.5

Area =0.4332

Area =

0.5 - 0.4332 =

0.0668

THE NORMAL APPROXIMATION TO


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27

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

distribution (a discrete distribution) for large

values ofnseems reasonable because as n

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 npand n(1 - p)are

both greater than 5.

THE NORMAL APPROXIMATION TO

THE BINOMIAL

Bi i l Di t ib ti ith 3 d 0 5


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28Binomial Distribution with n= 3 and p= 0.5.

0 1 2

0.3

0.4

0.5

P(r)

r

0.25



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29Binomial Distribution with n= 5 and p= 0.5.P(r)

r



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30Binomial Distribution with n= 20 and p= 0.5.P(r)

r

Observe

the

Normal

shape.

THE NORMAL APPROXIMATION


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31

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 probability pis fixed from trial to trial, andthe number of trials nis also fixed.

THE NORMAL APPROXIMATION

(continued)


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32

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 approximated

by 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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33A recent study by a marketing research firm

showed that 15% of the homes had a videorecorder 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?

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

s2 = np(1 - p)= (30)(1- 0.15) = 25.5

EXAMPLE


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34What is the standard deviation?

s = (25.5) = 5.0498.What is the probability that less than40 homes

in the sample have video recorders?

We need P(X < 40) = P(X 39). So, using thenormal approximation, P(X 39.5) P[z (39.5 - 30)/5.0498] = P(z 1.8812) P(z 1.88) = 0.5 + 0.4699 = 0.9699

Why did I use 39.5 ? ...

How would you calculate P(X=39) ?



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35

1.88

0.5 0.4699

P(z 1.88) =0.5 + 0.4699 =0.9699

z


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36EXAMPLE (cont inued)What is the probability that more than24 homes

in the sample have video recorders?


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37

-1.09

0.50.3621

P(z -1.09) =0.5 + 0.3621 =0.8621.

z


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38EXAMPLE (continued)What is the probability that exactly40 homes in

the sample have video recorders?


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39

1.88

2.08

0.4699

0.4812

P(1.88 z 2.08)= 0.4812 - 0.4699

= 0.0113

z

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Normal Distribution[1]