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- The Normal Probability

Distribution

Chapter

THIS CHAPTERS GOALS

- 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.

CHARACTERISTICS OF A NORMAL PROBABILITY

DISTRIBUTION

- 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 DISTRIBUTION

Normal curve is symmetrical - two halves

identical -

Tail

Tail

0.5

0.5

Theoretically, curve extends to - infinity

Theoretically, curve extends to infinity

Mean, median, and mode are equal

Normal Distributions with Equal Means but

Different Standard Deviations.

s 3.1 s 3.9 s 5.0

m 20

Normal Probability Distributions with Different

Means and Standard Deviations.

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

THE STANDARD NORMAL PROBABILITY DISTRIBUTION

- 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 m, divided

by the population standard deviation, s. - The z-value is the number of standard deviations

X is from the mean.

EXAMPLE

- 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 - m)/s, 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 (continued)

- For X 1,700 and since z (X - m)/s, 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?

AREAS UNDER THE NORMAL CURVE

- 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

m 1s. - About 95 percent of the area under the normal

curve 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.

Between 1. 68.26 2. 95.44 3. 99.97

m

m1s

m2s

m3s

m-1s

m-2s

m3s

P(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)

EXAMPLE

- 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

- 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 (continued)

- 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 lt X lt 24) P(0 lt z lt 0.8)

0.288128.81 - What percent uses between 16 and 20 gallons?

P(0 lt z lt 0.8) 0.2881

0.8

EXAMPLE (continued)

- What is the probability that a person selected at

random uses more than 28 gallons?

P(z gt 1.6) 0.5 - 0.4452 0.0048

Area 0.4452

z

1.6

EXAMPLE (continued)

- What percent uses between 18 and 26 gallons?

Total area 0.1554 0.3849 0.5403

Area 0.1554

Area 0.3849

z

- .4

1.2

EXAMPLE (continued)

- 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 ³ 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.

(Y - 20)/5 1.28

Y 26.4

0.4

0.1

z

1.28

EXAMPLE

- 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?

(Y - 72)/5 1.04

Y 77.2

0.35

0.15

z

1.04

EXAMPLE

- 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?

Area 0.5 - 0.4332 0.0668

Area 0.4332

z

- 1.5

THE NORMAL APPROXIMATION TO THE BINOMIAL

- 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.

Binomial Distribution with n 3 and p 0.5.

P(r)

0.5

0.4

0.3

0.25

r

0

1

2

Binomial Distribution with n 5 and p 0.5.

P(r)

r

Binomial Distribution with n 20 and p 0.5.

P(r)

Observe the Normal shape.

r

THE NORMAL APPROXIMATION (continued)

- 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.

CONTINUITY CORRECTION FACTOR

- 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 x0.5 is approximately equal

to P(x).

EXAMPLE

- 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? - m np (0.15)(200) 30 n(1 - p) 170
- What is the variance?
- s2 np(1 - p) (30)(1- 0.15) 25.5

EXAMPLE (continued)

- What is the standard deviation?
- s Ö(25.5) 5.0498.
- What is the probability that less than 40 homes

in the sample have video recorders? - We need P(X lt 40) P(X 39). So, using the

normal approximation, P(X 39.5) - Pz (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(X39) ?

P(z 1.88) 0.5 0.4699 0.9699

0.5

0.4699

z

1.88

EXAMPLE (continued)

- What is the probability that more than 24 homes

in the sample have video recorders?

P(z ³ -1.09) 0.5 0.3621 0.8621.

0.5

0.3621

z

-1.09

EXAMPLE (continued)

- What is the probability that exactly 40 homes in

the sample have video recorders?

P(1.88 z 2.08) 0.4812 - 0.4699 0.0113

1.88

2.08

0.4699

0.4812

z