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6-1
Chapter Six
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.
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6-2
Chapter Six
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.
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6-3
Characteristics of a Normal
Probability Distribution
The characteristics are:
• It is bell-shaped and has a single peak at the
exact center of the distribution.
•The arithmetic mean, median, and mode are
equal and located at the peak.
•Half the area under the curve is above the mean
and half is below it.
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6-4
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.
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6-5
r
a
l
i
t r
b
u
i o
n
:
m
=
0
,
s2
=
1
Characteristics of a Normal Distribution
0
. 4
. 3
0
. 2
Theoretically,
curve
extends to
infinity
f ( x
0
Normal
curve is
symmetrical
0
. 1
. 0
- 5
a
Mean, median, and
mode are equal
x
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6-6
The Standard Normal Probability
Distribution
The standard normal distribution has mean
of 0 and a standard deviation of 1. It is
also called the z distribution.
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6-7
The Standard Normal Probability
Distribution
A z-value is the distance between a selected
value, designated X, and the population mean
μ, divided by the population standard deviation,
σ. The formula is:
z
McGraw-Hill/ Irwin
X m
s
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6-8
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?
z 
X m
s
$2,200  $2,000

 2.00
$200
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6-9
EXAMPLE 1
continued
What is the z-value corresponding to $1,700?
z 
X m
s
$1,700  $2,200

 1.50
$200
A z-value of 1 indicates $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.
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6-10
Areas Under the Normal Curve
About 68 percent of the area under the normal curve is
within one standard deviation of the mean.
μ±σ
About 95 percent is within two standard deviations of
the mean.
μ± 2σ
Practically all is within three standard deviations of the
mean. μ ± σ
μ ± 3σ
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6-11
m
s2
Areas Under the Normal Curve
r
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0
. 3
0
. 2
0
. 1
l
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t r
b
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=
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,
=
1
Between:
±1σ - 68.26%
±2σ - 95.44%
±3σ - 99.74%
f ( x
0
a
. 0
m
m  2s
- 5
m  3s
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x
m  1s
m  2s
m  1s
m  3s
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6-12
EXAMPLE 2
The daily water usage per person in New
Providence, New Jersey follows the a normal
distribution 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.
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6-13
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 m
20  20
z 

 0.00
s
5
X m
24  20
z 

 0.80
s
5
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6-14
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.
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6-15
r
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0
. 3
0
. 2
0
. 1
l
i
EXAMPLE 3
t r
b
u
i o
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:
m
=
0
,
P(0<z<.8)
=.2881
f ( x
0
a
0<x<.8
. 0
- 5
-4
-3 -2 -1
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x
0
1
2
3
4
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6-16
EXAMPLE 3 continued
What percent of the population use
between 18 and 26 gallons per day?
z 
z 
X m
s
X m
s
McGraw-Hill/ Irwin
18  20

 0.40
5
26  20

 1.20
5
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6-17
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.
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6-18
EXAMPLE 4
Professor Mann reports the scores on Test
1 in his statistics course follow a normal
distribution 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?
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6-19
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. See Appendix D.
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6-20
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  1.04 (5)  72  5.2  77 .2
Those with a score of 77.2 or more earn an A.
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6-21
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.
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6-22
The Normal Approximation to the
Binomial
The normal probability distribution is a good
approximation to the binomial probability
distribution when n π and n(1- π ) are both
greater than 5.
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6-23
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.
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6-24
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 used to
approximate a continuous probability
distribution (the normal distribution).
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6-25
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?
m  n  (.15)(200)  30
This is the mean of a binomial distribution.
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6-26
EXAMPLE 5 continued
Calculate the variance?
s  n (1   )  (30)(1.15)  255
.
2
Calculate the standard deviation?
s  25.5  5.0498
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6-27
Example 5 continued
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  m 39 .5  30 .0
z

 1.88
s
5.0498
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6-28
Example 5 continued
From Appendix D the area between 0 and 1.88 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%.
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6-29
r
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EXAMPLE 5
0
,
s2
=
1
. 4
. 3
0
. 2
0
. 1
f ( x
0
P(z<1.88)
=.5000+.4699
=.9699
z=1.88
. 0
- 5
0
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2
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z
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