The Normal Probability Distribution

1

The Normal Probability Distribution

The Normal Probability DistributionChapter

2TO LIST THE CHARACTERISTICS OF THE

NORMAL DISTRIBUTION.TO DEFINE AND CALCULATE Z VALUES.TO DETERMINE PROBABILITIES

ASSOCIATED WITH THE STANDARD NORMAL DISTRIBUTION.

TO USE THE NORMAL DISTRIBUTION TO APPROXIMATE THE BINOMIAL DISTRIBUTION.

TO LIST THE CHARACTERISTICS OF THE NORMAL DISTRIBUTION.

TO DEFINE AND CALCULATE Z VALUES.TO DETERMINE PROBABILITIES

ASSOCIATED WITH THE STANDARD NORMAL DISTRIBUTION.

TO USE THE NORMAL DISTRIBUTION TO APPROXIMATE THE BINOMIAL DISTRIBUTION.

THIS CHAPTER’S GOALSTHIS CHAPTER’S GOALS

3The normal curve is bell-shaped and 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 is

symmetrical about its mean.It is asymptotic - the curve gets closer and closer

to the x-axis but never actually touches it.

The normal curve is bell-shaped and 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 is symmetrical about 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

CHARACTERISTICS OF A NORMAL PROBABILITY DISTRIBUTION

4CHARACTERISTICS OF A NORMAL DISTRIBUTIONCHARACTERISTICS OF A NORMAL DISTRIBUTION

Theoretically, curveextends to - infinity

Theoretically, curve extends to + infinityMean, median, and

mode are equal

Tail Tail

Normal curve is symmetrical - two halves identical -

Normal Distributions with Equal Means but Different Standard Deviations.

Normal Distributions with Equal Means but Different Standard Deviations.

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313.9 = 5.0

313.9 = 5.0

6Normal Probability Distributions with Different Means and Standard Deviations.

Normal Probability Distributions with Different Means and Standard Deviations.

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

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

7A normal distribution with a mean of 0 and a standard

deviation of 1 is called the standard normal distribution.

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

The z-value is the number of standard deviations X is from the mean.

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

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

The z-value is the number of standard deviations X is from the mean.

THE STANDARD NORMAL PROBABILITY DISTRIBUTIONTHE STANDARD NORMAL

PROBABILITY DISTRIBUTION

Z X

8The monthly incomes of recent MBA graduates

in a large corporation are normally distributed with a mean of $2,000 and a standard deviation of $200. What is the z value for an income X of $2,200? $1,700?

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

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

The monthly incomes of recent MBA graduates in a large corporation are normally distributed with a mean of $2,000 and a standard deviation of $200. What is the z value for an income X of $2,200? $1,700?

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

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

EXAMPLE EXAMPLE

9For X = $1,700 and since z = (X - then

z = (1,700 - 2,000)/200 = -1.5.A z value 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?

For X = $1,700 and since z = (X - then z = (1,700 - 2,000)/200 = -1.5.

A z value 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) EXAMPLE (continued)

1About 68 percent of the area under the normal

curve is within plus one and minus one standard deviation of the mean. This can be written as ± 1.

About 95 percent of the area under the normal curve is within plus and minus two standard deviations of the mean, written ± 2.

Practically all (99.74 percent) of the area under the normal curve is within three standard deviations of the mean, written ± 3.

About 68 percent of the area under the normal curve is within plus one and minus one standard deviation of the mean. This can be written as ± 1.

About 95 percent of the area under the normal curve is within plus and minus two standard deviations of the mean, written ± 2.

Practically all (99.74 percent) of the area under the normal curve is within three standard deviations of the mean, written ± 3.

AREAS UNDER THE NORMAL CURVEAREAS UNDER THE NORMAL CURVE

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1 2 3123

Between:

1. 68.26%

2. 95.44%

3. 99.97%

Between:

1. 68.26%

2. 95.44%

3. 99.97%

12A 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)EXCEL Function =NORMSDIST(z)

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)EXCEL Function =NORMSDIST(z)

P(z)=? P(z)=?

13The 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?

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 EXAMPLE

14The 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?

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 EXAMPLE

1What percent uses between 20 and 24 gallons?The z value 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?

What percent uses between 20 and 24 gallons?The z value 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?


16

0.80.8

P(0 < z < 0.8) = 0.2881

P(0 < z < 0.8) = 0.2881

17What is the probability that a person selected at

random uses more than 28 gallons?

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


18

P(z > 1.6) =0.5 - 0.4452 =0.0048

P(z > 1.6) =0.5 - 0.4452 =0.0048

Area = 442

Area = 442

1616z

19 EXAMPLE (continued) EXAMPLE (continued)

What percent uses between 18 and 26 gallons?What percent uses between 18 and 26 gallons?

2

1244

Area = 0.1554

Area =0.3849

Area =0.3849

Total area =0.1554 + 0.3849 =0.5403

Total area =0.1554 + 0.3849 =0.5403

z

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

the population use?Let X be the least amount. Then we need to find

Y such that P(X Y) = 0.1 To find the corresponding z value 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 the population will be using at least 26.4 gallons daily.

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

Let X be the least amount. Then we need to find Y such that P(X Y) = 0.1 To find the corresponding z value 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 the population will be using at least 26.4 gallons daily.


22

1.281.28

0.40.4

0.10.1

z

(Y - 20)/5 = 1.28Y= 26.4

23A 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% of the students receive an A. What is the lowest average a student must receive to earn an A?

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% of the students receive an A. What is the lowest average a student must receive to earn an A?

EXAMPLE EXAMPLE

24

1.041.04

0.150.15

0.350.35

z

(Y - 72)/5 = 1.04

Y = 77.2

2The 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 poor service if his total tip for the shift is less than $65. Based on his theory, what is the probability that he has provided poor service?

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 poor service if his total tip for the shift is less than $65. Based on his theory, what is the probability that he has provided poor service?

EXAMPLE EXAMPLE

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z- 1.5- 1.5

Area = 4332

Area = 4332

Area =0.5 - 0.4332 =0.0668

Area =0.5 - 0.4332 =0.0668

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Using the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seems 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 np and n(1 - p) are both greater than 5.

Using the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seems 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 np and n(1 - p) are both greater than 5.

THE NORMAL APPROXIMATION TO THE BINOMIAL

THE NORMAL APPROXIMATION TO THE BINOMIAL

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

0 1 2

0.3

0.4

0.5P(r)

r0.25


P(r)

r


P(r)

r

Observe the Normal shape.

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 counting

the number of successes.Each trial is independent.The probability p is fixed from trial to trial, and

the number of trials n is also fixed.

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

(success or failure) on each trial.A binomial distribution results from counting

the number of successes.Each trial is independent.The probability p is fixed from trial to trial, and

the number of trials n is also fixed.

THE NORMAL APPROXIMATION (continued)

THE NORMAL APPROXIMATION (continued)

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The value 0.5 subtracted or added, depending on the 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).

The value 0.5 subtracted or added, depending on the 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 FACTORCONTINUITY CORRECTION FACTOR

33A 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?

= np (0.15)(200) = 30 & n(1 - p) = 170What is the variance?2 = np(1 - p) = (30)(1- 0.15) = 25.5

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?

= np (0.15)(200) = 30 & n(1 - p) = 170What is the variance?2 = np(1 - p) = (30)(1- 0.15) = 25.5

EXAMPLE EXAMPLE

34What is the standard deviation? = (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 39). So, using the

normal 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) ?

What is the standard deviation? = (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 39). So, using the

normal 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) ?


3

1.881.88

0.50.5 0.46990.4699

P(z 1.88) = 0.5 + 0.4699 = 0.9699

P(z 1.88) = 0.5 + 0.4699 = 0.9699

z


What is the probability that more than 24 homes in the sample have video recorders?

What is the probability that more than 24 homes in the sample have video recorders?

37

-1.09-1.09

0.50.50.36210.3621

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

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

z


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

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

39

1.881.88

2.082.08

0.4699

0.4812

P(1.88 z 2.08)= 0.4812 - 0.4699 = 0.0113

P(1.88 z 2.08)= 0.4812 - 0.4699 = 0.0113

z

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The Normal Probability Distribution