7.3
CHAPTER 7
The Standard Normal Distribution (Z)
The standard normal deviation is a normal distribution
with a mean of 0 and a standard deviation of 1.
The Normal Distribution
34.13% 34.13%
2.28%
µ − 3σ
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1
e
−( X − µ )2
2σ 2
µ−σ
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µ
µ+σ
µ + 2σ
2.28%
µ + 3σ
About 68%
About 95%
About 99.7%
2
Procedure
σ 2π
1. Draw the picture
2. Shade the area desired
All normally distributed variables can be transformed into
the standard normally distributed variable by using the
formula for the standard score.
3. Find the correct figure
4. Follow the direction to get the desired area
value − mean
X −µ
or z =
z=
σ
standard deviation
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µ − 2σ
13.59%
Finding Areas under the Standard Normal Distribution
Curve
The formula for the standard normal distribution is
y=
13.59%
3
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1
Case 1: Between 0 and z value
• Look up the z value in the table to get the area
Case 2: In any tail
• Look up the z value to get the area
• Subtract the area from 0.5000
0
+z
-z
0
-z
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Case 3: Between two z values on the same side of the mean:
• Look up both z values to get the areas
0
0
6
Case 4: Between two z values on opposite sides of the mean
• Look up both z value to get the areas
• Add the areas
z1 z2
-z
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+z
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• Subtract the smaller area from the larger area
-z1 -z2 0
0
7
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0
+z
8
2
Case 5: To the left of any z value, where z is greater than the mean
Case 6: To the right of any z value, where z is less than the mean
• Look up the z value to get the area
• Look up the z value in the table to get the area
• Add 0.5000 to the area
• Add 0.5000 to the area
0
+z
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-z
9
0
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Example 7.1 (Ex 7-8)
Case 7: In any two tails
• Look up the z value in the table to get the area
Find the area between z = 0 and z = -0.48
Solution:
• Subtract both areas from 0.500
Area for z = 0.48 is 0.1844
• Add the answers
So area between z = 0 and z = -0.48
is 0.1844
-0.48
0
Example 7.2 (Ex 7-10)
Find the area to the right of z = 1.09
Solution:
Area for z = 1.09 is 0.3621
So area = 0.5 - 0.3621
-z
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0
= 0.1379
+z
11
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0
+1.09
12
3
Example 7.3 (Ex 7-16)
Find the area between z =-0.87 and z = -0.21
Example 7.5 (Ex 7-20)
Find the area to the left of z = 1.31
Solution:
Area between z =0 and z = 0.87 is 0.3708
Area between z =0 and z = 0.21 is 0.0832
So, area between z =-0.87 and z = -0.21 is
0.3708 – 0.832 = 0.2246
Solution:
Area between z =0 and z = 1.31 is 0.4049
So, area to the left of z =1.31 is
0.5 + 0.4049 = 0.9049
Example 7.4 (Ex 7-18)
Find the area between z = 0.24 and z = -1.12
Solution:
Area between z =0 and z = 0.24 is 0.0948
Area between z =0 and z = -1.12 is 0.3686
So,area between z =0.24 and z = -1.12 is
0.0948 + 0.3686 = 0.4634
-1.12
0
+0.24
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Example 7.7 (Ex 7-24)
-1.92
-2.15
0
0
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Solution:
Area between z = 0 and z = 1.69 is 0.4545
So P (0 < z < 1.69) = 0.4545
0
1.69
Example 7.9 (Ex7-28)
Find probabilities for P (-1.23 < z < 0),
using the standard normal distribution
+1.62
Solution:
Area between z = 0 and z = 1.23 is 0.3907
So P (-1.23 < z < 0) = 0.3907
Thus, the area to the left of z = -2.15 and to the right of z = 1.62 is
0.0158 + 0.0526 = 0.0684
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Example 7.6 (Ex 7-22)
Find the area to the right of z = -1.92
Solution:
Area between z =0 and z = 1.92 is 0.4726
So, area to the right of z = -1.92 is
0.5 + 0.4726 = 0.9726
Example 7.8 (Ex 7-26)
Find probabilities for P (0 < z < 1.69),
using the standard normal distribution
Find the area to the left of z = -2.15 and to the right of z = 1.62
Solution:
Area between z =0 and z = 2.15 is
0.4842
So, area to the left of z = -2.15 is
0.5 – 0.4842 = 0.0158
Area between z =0 and z = 1.62 is
0.4474
So, area to the right of z = 1.62 is
0.5 - 0.4474 = 0.0526
+1.31
0
-0.87 -0.21 0
15
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-1.23
0
16
4
Example 7.10 (Ex 7-36)
Find probabilities for P (1.32 < z < 1.51),
using the standard normal distribution
7.4
Example 7.12 (Ex.7-56)
Solution:
Area between z = 0 and z = 1.32 is 0.4066
Area between z = 0 and z = 1.51 is 0.4345
So P (1.32 < z < 1.51) = 0.4345 – 0.4066 = 0.0279
0
1.32 1.51
Example 7.11 (Ex 7-42)
Find the z value for the given area.
Solution:
The area between 0 and z is
0.5-0.0239 = 0.4761
Thus, from the table, z = 1.98
Application of the Normal Distribution
The Speedmaster IV automobile gets an average 22.0
miles per gallon in the city. The standard deviation is 3
miles per gallon. Assume the variable is normally
distributed. Find the probability that on any given day the
car will get more than 26 miles per gallon when driven in
the city.
0.0239
0
z
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Solution:
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Example 7.13 (Ex 7-60)
Step 1: Draw the figure and represent
the area, as shown on the right.
The average age of CEOs is 56 years. Assume the variables
Step 2: Find the z value for 26
will be in the following range.
normally distributed. If the standard deviation is four years,
find the probability that the age of a randomly selected CEO
26 − 22
z=
3
Z = 1.33
18
22
a. Between 53 and 59 years old
b. Between 58 and 63 years old
26
c. Between 50 and 55 years old
Step 3: The area between z = 0 and z = 1.333 is 0.4082
Hence, the probability is 0.5 – 0.4082 = 0.0918 or 9.18%
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5
Solution:
b.
a. Find the two z values
X −µ
53 − 56
3
=−
4
4
X − µ 59 − 56 3
=
=
z2 =
σ
4
4
z1 =
σ
=
X −µ
58 − 56 1
=
4
2
X − µ 63 − 56 7
=
=
z2 =
σ
4
4
z1 =
σ
=
Area between z = 0 and z = 0.5 is 0.1915
56
58 63
Area between z = 0 and z = 1.75 is 0.4599
53
56
So, total area is 0.4599 – 0.1915 = 0.2684
59
The area between z = 0 and z = -0.75 is 0.2734
c.
The area between z = 0 and z = 0.75 is 0.2734
So, the total area is 0.5468 or 54.68%
X −µ
50 − 56
3
=−
4
2
1
X − µ 55 − 56
z2 =
=
=−
σ
4
4
Area between z = 0 and z = -1.5 is 0.4332
z1 =
σ
=
50 55 56
Area between z = 0 and z = -0.25 is 0.0987
So, total area is 0.4332 – 0.0987 = 0.3345
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Example 7.14 (Ex 7-70)
7.5
In order to qualify for letter sorting, applicants are given a speedreading test. The scores are normally distributed, with a mean of
80 and a standard deviation of 8. If only the top 15% of the
applicants are selected, find the cutoff score.
The top 15% is in the right tail of the normal
curve. The corresponding z score is found
using area = 0.50 – 0.15 = 0.35.
Thus, z = 1.04
1.04 =
15%
X −µ
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The Central Limit Theorem
A sampling distribution of sample means is a
distribution obtained by using the means computed from
random samples of a specific size taken from a
population
Solution:
z=
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Sampling error is the difference between the sample
measure and the corresponding population measure due
to the fact that the sample is not a perfect representation
of the population.
σ
X − 80
8
80
x
X = 88.32
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6
The Central Limit Theorem
Properties of the Distribution of Sample Means
1.
2.
As the sample size n increases, the shape of the
The mean of the sample means will be the same
as the population mean.
distribution of the sample means taken with replacement
from a population with mean µ and standard deviation σ
The standard deviation of the sample means will
be smaller than the standard deviation of the
population, and it will be equal to the population
standard deviation divided by the square root of
the sample size.
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will approach a normal distribution. As previously shown,
this distribution will have a mean µ and a standard
deviation
25
Summary of Formulas and Their Uses
z =
z =
X − µ
σ
X − µ
σ n
σ
n
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Example 7.15 (Ex 7-104)
The mean grade point average of the engineering majors at
a large university is 3.23, with a standard deviation of 0.72.
In a class of 48 students, find the probability that the mean
grade point average of the student is less than 3.15.
Used to gain information about an
individual data value when the variable is
normally distributed.
Solution:
Used to gain information when applying
the central limit theorem about a sample
mean when the variable is normally
distributed or when the sample size is 30
or more.
z=
X − µ 3.15 − 3.23
=
= 0.77
n 0.72 48
σ
P(z < 0.77) = 0.5 – 0.2794
= 0.2206
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3.15
3.23
28
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Example 7.16 (Ex 7-114)
Solution:
Assume that the mean systolic blood pressure of normal
adults is 120 millimeters of mercury (mm Hg) and the
standard deviation is 5.6. Assume the variable is normally
distributed.
a.
b.
b. If a sample of 30 adults is randomly selected, find the
probability that the sample mean will be between 120
and 121.8 mm Hg.
121 . 8 − 120
= 1 .76
5 .6
30
120
121.8
c. Sample means are less variable than
individual data.
29
7.1 (*Ex 7-48)
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Review Exercise
7.2 (Ex 7-78)
A small publisher wishes to publish self-improvement books. After
a survey of the market, the publisher finds that the average cost of
the type of the book that she wishes to publish is $12.80. If she
wants to price her books to sell in the middle 70% range, what
should the maximum and minimum prices of the books be? The
standard deviation is $0.83 and the variable is normally distributed.
Find two z values so that 40% of the middle area is bounded
by them.
Solution:
Area between 0 and z is 0.2
Solution:
X −µ
z=
σ
From the table, z = 0.52
Thus, the two values are 0.52
and –0.52
z=
P(0 < z <1.76) = 0.4608
c. Why is the answer to Part a so much smaller than the
answer to Part b?
Review Exercise
121.8 − 120
= 0.32
5. 6
P(0 < z < 0.32) = 0.1255
a. If an individual is selected, find the probability that the
individual’s pressure will be between 120 and 121.8
mm Hg.
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z=
-0.52
X − 12.8
0.83
X = 13.66
1.04 =
0.52
and
− 1.04 =
X − 12.8
0.83
X = 11.94
∴ The minimum price is $11.94 and the maximum price is $13.66
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8
Review Exercise
7.3 (Ex 7-140)
Solution:
The average diastolic blood pressure of a certain age group
of people is 85mm Hg. The standard deviation is 6. If an
individual from this age group is selected, find the probability
that his or her pressure will be the following. Assume the
variable is normally distributed
a.
90 − 85
= 0.83
6
area between z = 0 and z = 0.83
is 0.2967
z=
P(z > 0.83) = 0.5 – 0.2967
85
= 0.2033
a. Greater than 90
b. Below 80
b.
c. Between 85 and 95
d. Between 88 and 92
80 − 85
= −0.83
6
area between z = 0 and z = -0.83
is 0.2967
z=
P(z < -0.83) = 0.5 – 0.2967
= 0.2033
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Solution:
95 − 85
= 1.67
c. z =
6
80
85
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Review Exercise
7.4 (Ex 7-144)
The average weight of an airline passenger’s suitcase is 45
pounds. The standard deviation is 2 pounds. If 15% of the
suitcases are overweight, find the maximum weight allowed
by the airline. Assume the variable is normally distributed.
area between z = 0 and z = 1.67
is 0.4525
P(0 < z <1.67) = 0.4525
90
85
95
Solution:
92 − 85
= 1.17
6
88 − 85
z=
= 0. 5
6
d. z =
The 15% overweight suitcases are in the right tail, the
corresponding z score for the area is 1.04
area = 0.3790
area = 0.1915
z=
X −µ
P( 0.5 < z < 1.17)
= 0.3790 - 0.1915
85
88 92
σ
1.04 =
X − 4.5
2
X = 47.08
= 0.1875
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