1. 7- 1
Chapter Seven
The Normal Probability Distribution
GOALS
When you have completed this chapter, you will be able to:
ONE
List the characteristics of the normal probability distribution.
TWO
Define and calculate z values.
THREE
Determine the probability an observation will lie between two
points using the standard normal distribution.
FOUR
Determine the probability an observation will be above or
below a given value using the standard normal distribution.

2. 7- 2
Chapter Seven continued
The Normal Probability Distribution
GOALS
When you have completed this chapter, you will be able to:
FIVE
Compare two or more observations that are on different
probability distributions.
SIX
Use the normal distribution to approximate the binomial
probability distribution.

3. 7- 3
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.
Thus half the area under the curve is above the
mean and half is below it.

4. 7- 4
Characteristics of a Normal
Probability Distribution
• The normal probability distribution is
symmetrical about its mean.
The normal probability distribution is
asymptotic. That is the curve gets closer and
closer to the X-axis but never actually
touches it.

5. 7- 5
r a l i t r b u i o n : µ = 0 , σ2 = 1
Characteristics of a Normal Distribution
0 . 4
Normal
curve is
0 . 3 symmetrical
0 . 2
Theoretically,
curve
f ( x
0 . 1
extends to
infinity
. 0
- 5
a
Mean, median, and
x
mode are equal
McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved

6. 7- 6
The Standard Normal Probability
Distribution
• The standard normal distribution is a normal
distribution with a mean of 0 and a standard
deviation of 1.
• It is also called the z distribution.
A z-value is the distance between a selected value,
designated X, and the population mean , , divided by
the population standard deviation, t . The formula is:
X −µ
z =
σ

7. 7- 7
EXAMPLE 1
• The bi-monthly starting salaries of recent MBA
graduates follows the normal distribution with a
mean of $2,000 and a standard deviation of $200.
What is the z-value for a salary of $2,200?
X −µ
z =
σ
$2,200 − 2,000
$
= =2.00
$200

8. 7- 8
EXAMPLE 1 continued
What is the z-value of $1,700.
X −µ
z =
σ
$1,700 − 2,200
$
= =− .50
1
$200
• A z-value of 1 indicates that the value of $2,200 is
one standard deviation above the mean of
$2,000. A z-value of –1.50 indicates that $1,700 is
1.5 standard deviation below the mean of $2000.

9. 7- 9
Areas Under the Normal Curve
• About 68 percent of the area under the normal curve
is within one standard deviation of the mean.
i
• About 95 percent is within two standard deviations
of the mean.
o 22
• Practically all is within three standard deviations of
the mean. t
33

10. 7- 10
Areas Under the Normal Curve
r a l i t r b u i o n : µ = 0 , σ2 = 1
0 . 4
Between:
B11 - 68.26%
0 . 3 22 - 95.44%
33 - 99.74%
0 . 2
f ( x
0 . 1
. 0
µ µ + 2σ
- 5
µ − 2σ
µ + 3σ
x
µ − 3σ µ −1σ µ +1σ
Copyright © 2002 by The McGraw-Hill Companies, Inc.,.All rights reserved
McGraw-Hill/Irwin

11. 7- 11
EXAMPLE 2
The daily water usage per person in New
Providence, New Jersey is normally distributed
with a mean of 20 gallons and a standard
deviation of 5 gallons. About 68 percent of
those living in New Providence will use how
many gallons of water?
• About 68% of the daily water usage will lie
between 15 and 25 gallons.

12. 7- 12
EXAMPLE 3
• What is the probability that a person from New
Providence selected at random will use between
20 and 24 gallons per day?
X −µ 20 − 20
z = = = .00
0
σ 5
X −µ 24 − 20
z = = = .80
0
σ 5

13. 7- 13
Example 3 continued
• The area under a normal curve between a z-value of
0 and a z-value of 0.80 is 0.2881.
• We conclude that 28.81 percent of the residents use
between 20 and 24 gallons of water per day.
• See the following diagram.

14. 7- 14
r a l i
EXAMPLE 3
t r b u i o n : µ = 0 ,
0 . 4
0 . 3
P(0<z<.8)
=.2881
0 . 2
0<x<.8
f ( x
0 . 1
. 0
- 5
-4 -3 -2 -1 x
0 1 2 3 4
Copyright ©2002 by The McGraw-Hill Companies, Inc.,.All rights reserved
Irwin/McGraw-Hill

15. 7- 15
EXAMPLE 3 continued
• What percent of the population use between 18 and
26 gallons per day?
X −µ 18 − 20
z = = = 0.40
−
σ 5
X −µ 26 − 20
z = = = .20
1
σ 5

16. 7- 16
Example 3 continued
• The area associated with a z-value of –0.40 is .
1554.
• The area associated with a z-value of 1.20 is .
3849.
• Adding these areas, the result is .5403.
• We conclude that 54.03 percent of the residents
use between 18 and 26 gallons of water per day.

17. 7- 17
EXAMPLE 4
• Professor Mann has determined that the scores
in his statistics course are approximately
normally distributed with a mean of 72 and a
standard deviation of 5. He announces to the
class that the top 15 percent of the scores will
earn an A. What is the lowest score a student
can earn and still receive an A?

18. 7- 18
Example 4 continued
• To begin let X be the score that separates an A
from a B.
• If 15 percent of the students score more than X,
then 35 percent must score between the mean
of 72 and X.
The z-value associated corresponding to
35 percent is about 1.04.

19. 7- 19
Example 4 continued
• We let z equal 1.04 and solve the standard normal
equation for X. The result is the score that
separates students that earned an A from those
that earned a B.
X −72
1.04 =
5
X =72 + .04(5) =72 +5.2 =77.2
1
Those with a score of 77.2 or more earn an A.

20. 7- 20
The Normal Approximation to the
Binomial
• The normal distribution (a continuous
distribution) yields a good approximation of the
binomial distribution (a discrete distribution) for
large values of n.
The normal probability distribution is generally a
good approximation to the binomial probability
distribution when nn and n(1- n ) are both greater
than 5.

21. 7- 21
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 is fixed from trial to trial, and the
number of trials n is also fixed.

22. 7- 22
Continuity Correction Factor
• The value .5 subtracted or added, depending
on the problem, to a selected value when a
binomial probability distribution (a discrete
probability distribution) is being approximated
by a continuous probability distribution (the
normal distribution).

23. 7- 23
EXAMPLE 5
A recent study by a marketing research firm
showed that 15% of American households owned
a video camera. For a sample of 200 homes, how
many of the homes would you expect to have
video cameras?
µ = nπ = (.15)(200) = 30
This is the mean of a binomial distribution.

24. 7- 24
EXAMPLE 5 continued
• What is the variance?
σ = nπ (1 − π ) = (30)(1−.15) = 25.5
2
What is the standard deviation?
σ = 25.5 = 5.0498

25. 7- 25
Example 5
What is the probability that less than 40 homes in
the sample have video cameras?
• We use the correction factor, so X is 39.5.
• The value of z is 1.88.
X −µ 39.5 − 30.0
z= = = 1.88
σ 5.0498

26. 7- 26
Example 5 continued
• From Appendix D the area between 0 and 1.88 on
the z scale is .4699.
• So the area to the left of 1.88 is .5000 + .4699 = .
9699.
• The likelihood that less than 40 of the 200 homes
have a video camera is about 97%.

27. 7- 27
µ σ2
EXAMPLE 5
r a l i t r b u i o n : = 0 , = 1
0 . 4
P(z<1.88)
0 . 3
=.5000+.4699
=.9699
0 . 2
f ( x
0 . 1
z=1.88
. 0
- 5
0 1 2 3 4 z
Copyright ©2002 by The McGraw-Hill Companies, Inc., . All rights reserved
McGraw-Hill/Irwin