2. Outline
6
The Normal Distribution
6-1 Normal Distributions
6-2 Applications of the Normal Distribution
6-3 The Central Limit Theorem
6-4 The Normal Approximation to the Binomial
Distribution
3. Objectives
6
The Normal Distribution
1 Identify distributions as symmetric or skewed.
2 Identify the properties of a normal distribution.
3 Find the area under the standard normal distribution,
given various z values.
4 Find probabilities for a normally distributed variable
by transforming it into a standard normal variable.
5 Find specific data values for given percentages, using
the standard normal distribution.
4. Objectives
6
The Normal Distribution
6 Use the central limit theorem to solve problems
involving sample means for large samples.
7 Use the normal approximation to compute
probabilities for a binomial variable.
5. 6.1 Normal Distributions
• Many continuous variables have distributions that
are bell-shaped and are called approximately
normally distributed variables.
• The theoretical curve, called the bell curve or the
Gaussian distribution, can be used to study many
variables that are not normally distributed but are
approximately normal.
Bluman, Chapter 6 5
6. Normal Distributions
2 2
( ) (2 )
2
X
e
y
The mathematical equation for the normal
distribution is:
2.718
3.14
population mean
population standard deviation
where
e
Bluman, Chapter 6 6
7. Normal Distributions
• The shape and position of the normal distribution
curve depend on two parameters, the mean and the
standard deviation.
• Each normally distributed variable has its own
normal distribution curve, which depends on the
values of the variable’s mean and standard
deviation.
Bluman, Chapter 6 7
9. Normal Distribution Properties
• The normal distribution curve is bell-shaped.
• The mean, median, and mode are equal and located
at the center of the distribution.
• The normal distribution curve is unimodal (i.e., it
has only one mode).
• The curve is symmetrical about the mean, which is
equivalent to saying that its shape is the same on
both sides of a vertical line passing through the
center.
Bluman, Chapter 6 9
10. Normal Distribution Properties
• The curve is continuous—i.e., there are no gaps or
holes. For each value of X, there is a corresponding
value of Y.
• The curve never touches the x-axis. Theoretically, no
matter how far in either direction the curve extends,
it never meets the x-axis—but it gets increasingly
closer.
Bluman, Chapter 6 10
11. Normal Distribution Properties
• The total area under the normal distribution curve is
equal to 1.00 or 100%.
• The area under the normal curve that lies within
– one standard deviation of the mean is
approximately 0.68 (68%).
– two standard deviations of the mean is
approximately 0.95 (95%).
– three standard deviations of the mean is
approximately 0.997 ( 99.7%).
Bluman, Chapter 6 11
13. Standard Normal Distribution
• Since each normally distributed variable has its own
mean and standard deviation, the shape and location
of these curves will vary. In practical applications,
one would have to have a table of areas under the
curve for each variable. To simplify this, statisticians
use the standard normal distribution.
• The standard normal distribution is a normal
distribution with a mean of 0 and a standard
deviation of 1.
Bluman, Chapter 6 13
14. z value (Standard Value)
The z value is the number of standard deviations that a
particular X value is away from the mean. The formula
for finding the z value is:
value mean
standard deviation
z
X
z
Bluman, Chapter 6 14
15. Area under the Standard Normal
Distribution Curve
1. To the left of any z value:
Look up the z value in the table and use the area
given.
Bluman, Chapter 6 15
16. Area under the Standard Normal
Distribution Curve
2. To the right of any z value:
Look up the z value and subtract the area from 1.
Bluman, Chapter 6 16
17. Area under the Standard Normal
Distribution Curve
3. Between two z values:
Look up both z values and subtract the
corresponding areas.
Bluman, Chapter 6 17
19. Example 6-1: Area under the Curve
Find the area to the left of z = 2.06.
The value in the 2.0 row and the 0.06 column of Table E
is 0.9803. The area is 0.9803.
Bluman, Chapter 6 19
21. Example 6-2: Area under the Curve
Find the area to the right of z = –1.19.
The value in the –1.1 row and the .09 column of Table E
is 0.1170. The area is 1 – 0.1170 = 0.8830.
Bluman, Chapter 6 21
23. Example 6-3: Area under the Curve
Find the area between z = +1.68 and z = –1.37.
The values for z = +1.68 is 0.9535 and for
z = –1.37 is 0.0853. The area is 0.9535 – 0.0853 = 0.8682.
Bluman, Chapter 6 23
25. Example 6-4: Probability
a. Find the probability: P(0 < z < 2.32)
The values for z = 2.32 is 0.9898 and for z = 0 is 0.5000.
The probability is 0.9898 – 0.5000 = 0.4898.
Bluman, Chapter 6 25
26. 0.75
0
Find the area under the normal distribution curve.
Between z = 0 and z = 0.75
area = 0.2734
Exercise #7
27. 0 0.79 1.28
0.3997 – 0.2852 = 0.1145
Find the area under the normal distribution curve.
between z = 0.79 and z = 1.28. The area is
found by looking up the values 0.79 and
1.28 in table E and subtracting the areas
as shown in Block 3 of the
Procedure Table.
Exercise #15
28. 0 2.83
Find probabilities, using the standard normal distribution
P(z > 2.83).The area is found by looking up z = 2.83 in
Table E then subtracting the area from 0.5
as shown in Block 2 of the
Procedure Table.
0.5 – 0.4977 = 0.0023
Exercise #31
29. Find the z value that corresponds to the given area.
0.8962 – 0.5 = 0.3962
0
z
0.8962
Exercise #45
30. Find the z value that corresponds to the given area.
Using Table E, find the area 0.3962 and
read the correct z value [corresponding
to this area] to get 1.26. Finally,
because the z value lies to the left
of 0, z = –1.26.
0
z
0.8962
32. Example 6-5: Probability
Find the z value such that the area under the standard
normal distribution curve between 0 and the z value is
0.2123.
Add 0.5000 to 0.2123 to get the cumulative area of
0.7123. Then look for that value inside Table E.
Bluman, Chapter 6 32
33. Example 6-5: Probability
The z value is 0.56.
Add .5000 to .2123 to get the cumulative area of .7123.
Then look for that value inside Table E.
Bluman, Chapter 6 33
34. 6.2 Applications of the Normal
Distributions
• The standard normal distribution curve can be used
to solve a wide variety of practical problems. The
only requirement is that the variable be normally or
approximately normally distributed.
• For all the problems presented in this chapter, you
can assume that the variable is normally or
approximately normally distributed.
Bluman, Chapter 6 34
35. Applications of the Normal
Distributions
• To solve problems by using the standard normal
distribution, transform the original variable to a
standard normal distribution variable by using the z
value formula.
• This formula transforms the values of the variable
into standard units or z values. Once the variable is
transformed, then the Procedure Table (Sec. 6.1) and
Table E in Appendix C can be used to solve
problems.
Bluman, Chapter 6 35
37. Example 6-6: Summer Spending
A survey found that women spend on average $146.21 on
beauty products during the summer months.
Assume the standard deviation is $29.44.
Find the percentage of women who spend less than $160.00.
Assume the variable is normally distributed.
Bluman, Chapter 6 37
38. Example 6-6: Summer Spending
Step 1: Draw the normal distribution curve.
Bluman, Chapter 6 38
39. Example 6-6: Summer Spending
Step 2: Find the z value corresponding to $160.00.
Table E gives us an area of .6808.
68% of women spend less than $160.
160.00 146.21
0.47
29.44
X
z
Step 3: Find the area to the left of z = 0.47.
Bluman, Chapter 6 39
41. Each month, an American household generates an average of 28
pounds of newspaper for garbage or recycling. Assume the
standard deviation is 2 pounds. If a household is selected at
random, find the probability of its generating between 27 and 31
pounds per month. Assume the variable is approximately
normally distributed.
Step 1: Draw the normal distribution curve.
Example 6-7a: Newspaper Recycling
Bluman, Chapter 6 41
42. Example 6-7a: Newspaper Recycling
Step 2: Find z values corresponding to 27 and 31.
Table E gives us an area of 0.9332 – 0.3085 = 0.6247. The
probability is 62%.
27 28
0.5
2
z
Step 3: Find the area between z = -0.5 and z = 1.5.
31 28
1.5
2
z
Bluman, Chapter 6 42
44. Americans consume an average of 1.64 cups of coffee per day.
Assume the variable is approximately normally distributed with
a standard deviation of 0.24 cup.
If 500 individuals are selected, approximately how many will
drink less than 1 cup of coffee per day?
Example 6-8: Coffee Consumption
Bluman, Chapter 6 44
45. Step 1: Draw the normal distribution curve.
Example 6-8: Coffee Consumption
Bluman, Chapter 6 45
46. Step 4: To find how many people drank less than 1 cup of
coffee, multiply the sample size 500 by 0.0038 to get
1.9.
Since we are asking about people, round the answer to
2 people. Hence, approximately 2 people will drink less
than 1 cup of coffee a day.
Example 6-8: Coffee Consumption
Step 2: Find the z value for 1.
1 1.64
2.67
0.24
z
Step 3: Find the area to the left of z = –2.67. It is 0.0038.
Bluman, Chapter 6 46
48. To qualify for a police academy, candidates must score in the
top 10% on a general abilities test. The test has a mean of 200
and a standard deviation of 20. Find the lowest possible score to
qualify. Assume the test scores are normally distributed.
Step 1: Draw the normal distribution curve.
Example 6-9: Police Academy
Bluman, Chapter 6 48
49. The cutoff, the lowest possible score to qualify, is 226.
Example 6-8: Police Academy
Step 2: Subtract 1 – 0.1000 to find area to the left, 0.9000. Look
for the closest value to that in Table E.
200 1.28 20 225.60
X z
Step 3: Find X.
Bluman, Chapter 6 49
51. For a medical study, a researcher wishes to select people in the
middle 60% of the population based on blood pressure. If the
mean systolic blood pressure is 120 and the standard deviation
is 8, find the upper and lower readings that would qualify people
to participate in the study.
Step 1: Draw the normal distribution curve.
Example 6-10: Systolic Blood Pressure
Bluman, Chapter 6 51
52. Area to the left of the positive z: 0.5000 + 0.3000 = 0.8000.
Using Table E, z 0.84.
Area to the left of the negative z: 0.5000 – 0.3000 = 0.2000.
Using Table E, z –0.84.
The middle 60% of readings are between 113 and 127.
Example 6-10: Systolic Blood Pressure
X = 120 + 0.84(8) = 126.72
X = 120 – 0.84(8) = 113.28
Bluman, Chapter 6 52
53. Normal Distributions
• A normally shaped or bell-shaped distribution is
only one of many shapes that a distribution can
assume; however, it is very important since many
statistical methods require that the distribution of
values (shown in subsequent chapters) be normally
or approximately normally shaped.
• There are a number of ways statisticians check for
normality. We will focus on three of them.
Bluman, Chapter 6 53
54. Checking for Normality
• Histogram
• Pearson’s Index PI of Skewness
• Outliers
• Other Tests
– Normal Quantile Plot
– Chi-Square Goodness-of-Fit Test
– Kolmogorov-Smikirov Test
– Lilliefors Test
Bluman, Chapter 6 54
56. A survey of 18 high-technology firms showed the number of
days’ inventory they had on hand. Determine if the data are
approximately normally distributed.
5 29 34 44 45 63 68 74 74
81 88 91 97 98 113 118 151 158
Method 1: Construct a Histogram.
The histogram is approximately bell-shaped.
Example 6-11: Technology Inventories
Bluman, Chapter 6 56
57. Method 2: Check for Skewness.
The PI is not greater than 1 or less than –1, so it can be concluded
that the distribution is not significantly skewed.
Method 3: Check for Outliers.
Five-Number Summary: 5 - 45 - 77.5 - 98 - 158
Q1 – 1.5(IQR) = 45 – 1.5(53) = –34.5
Q3 + 1.5(IQR) = 98 + 1.5(53) = 177.5
No data below –34.5 or above 177.5, so no outliers.
Example 6-11: Technology Inventories
3 79.5 77.5
3( )
PI 0.148
40.5
X MD
s
79.5, 77.5, 40.5
X MD s
Bluman, Chapter 6 57
58. A survey of 18 high-technology firms showed the number of
days’ inventory they had on hand. Determine if the data are
approximately normally distributed.
5 29 34 44 45 63 68 74 74
81 88 91 97 98 113 118 151 158
Conclusion:
• The histogram is approximately bell-shaped.
• The data are not significantly skewed.
• There are no outliers.
Thus, it can be concluded that the distribution is approximately
normally distributed.
Example 6-11: Technology Inventories
Bluman, Chapter 6 58
60. a. Greater than 700,000.
b. Between 500,000 and 600,000.
The average daily jail population in the United States
is 618,319. If the distribution is normal and the
standard deviation is 50,200, find the
probability that on a randomly selected
day, the jail population is…
Section 6-4 Exercise #3
61. a. Greater than 700,000
P(z > 1.63) = 0.5 –0.4484
= 0.0516 or 5.16%
z =
700,000–618,319
50,200
= 1.63
0 1.63
z =
X –
62. b. Between 500,000 and 600,000.
z =
500,000–618,319
50,200
= –2.36
area = 0.4909
z =
600,000–618,319
50,200 = –0.36
area = 0.1406
z =
X –
63. area = 0.4909 area = 0.1406
P( – 2.36 < z < –0.36)
= 0.3503 or 35.03%
= 0.4909–0.1406
– 2 .36 – 0.36
b. Between 500,000 and 600,000.
64. The average credit card debt for college seniors is
$3262. If the debt is normally distributed with a
standard deviation of $1100, find these probabilities.
a. That the senior owes at least $1000
b. That the senior owes more than $4000
c. That the senior owes between
$3000 and $4000
Section 6-4 Exercise #11
65. z = X –
a. That the senior owes at least $1000
z =
1000 – 3262
1100
= –2.06
area = 0.4803
= 0.9803 or 98.03%
P(z • – 2.06) = 0.5 + 0.4803
0.9803 or 98.03%
– 2.06 0
66. z = X –
b. That the senior owes more than $4000
z =
4000 – 3262
1100 = 0.67
= 0.2514 or 25.14%
P(z > 0.67) = 0.5 – 0.2486
area = 0.2486
0.2514 or 25.14%
0.67
0
67. z = X –
c. That the senior owes between $3000 and $4000.
z =
3000 – 3262
1100
= – 0.24
area = 0.0948
= 0.3434 or 34.34%
P( –0.24 < z <0.67) = 0.0948 + 0.2486
0.3434 or 34.34%
0.67
– 0.24 0
68. An advertising company plans to market a product to
low-income families. A study states that for a p
$2
ar
4,
ticu
596
lar
area, the average income per family is and the
standard devia $6256
tion is . If the company plans to
target the bottom of the families based on income,
find the cut off income. Assume the variable is normally
distribu
18%
ted.
Section 6-4 Exercise #27
69. The bottom 18% means that 32% of the area is between
z and 0. The corresponding z score will be .
– 0.92
$18,840.48 $24,596
0.32
0.18
X = –0.92(6256) + 24,596
= $18,840.48
71. The average price of a pound of sliced bacon is $2.02.
Assume the standard deviation is $0.08. If a random
sample of 40 one-pound packages is selected, find the
probability the the mean of the sample will
be less than $2.00.
z = X –
n
= –1.58
=
2.00–2.02
0.08
40
P(z < –1.58) = 0.5 –0.4429
area = 0.4429
= 0.0571or5.71%
Section 6-5 Exercise #13
72. $2.00 $2.02
The average price of a pound of sliced bacon is $2.02.
Assume the standard deviation is $0.08. If a random
sample of 40 one-pound packages is selected, find the
probability the the mean of the sample will
be less than $2.00.
0.0571or 5.71%
73. The average time it takes a group of adults to complete
a certain achievement test is 46.2 minutes. The
standard deviation is 8 minutes. Assume the variable
is normally distributed.
Section 6-5 Exercise #21
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
a. Find the probability that a randomly
selected adult will complete the test
in less than 43 minutes.
b. Find the probability that, if
50 randomly selected adults take
the test, the mean time it takes the
group to complete the test will be
less than 43 minutes.
74. c. Does it seem reasonable that an adult
would finish the test in less than
43 minutes? Explain.
d. Does it seem reasonable that the
mean of 50 adults could be less
than 43 minutes? Explain.
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
75. a. Find the probability that a randomly
selected adult will complete the test in
less than 43 minutes.
z = X –
= 43 – 46.2
8
= –0.4
area = 0.1554
P(z < –0.4) = 0.5 –0.1554 = 0.3446or34.46%
43 46.2
0.3446or34.46%
76. b. Find the probability that, if 50 randomly
selected adults take the test, the mean
time it takes the group to complete the
test will be less than 43 minutes.
z = 43 – 46.2
8
50
= – 2.83
area = 0.4977
P(z < – 2.83) = 0.5 –0.4977 = 0.0023or0.23%
43 46.2
0.0023or0.23%
77. c. Does it seem reasonable that an
adult would finish the test in less
than 43 minutes? Explain.
Yes, since it is within one
standard deviation of the mean.
d. Does it seem reasonable that the
mean of 50 adults could be less
than 43 minutes? Explain.
It is very unlikely, since the
probability would be less than 1%.
78. The average cholesterol of a certain brand of eggs is
215 milligrams, and the standard deviation is 15
milligrams. Assume the variable is normally
distributed.
a. If a single egg is selected, find the
probability that the cholesterol
content will be greater than
220 milligrams.
b. If a sample of 25 eggs is selected,
find the probability that the mean
of the sample will be larger than
220 milligrams.
Section 6-5 Exercise #23
79. z = X –
a. If a single egg is selected, find the
probability that the cholesterol content
will be greater than milligr
220 ams.
=
220–215
15
area = 0.1293
= 0.33
P(z > 0.33) = 0.5 –0.1293= 0.3707or37.07%
220
215
0.3707or37.07%
80. z =
X –
n
=
220– 215
15
25
= 1.67
area = 0.4525
b. If a sample of eggs is selected, find the probability that
the mean of the sample will be larger than milligr
25
220 ams.
215 220
P(z > 1.67) = 0.5 –0.4525
= 0.0475or4.75%
81. Section 6-6
The Normal Approximation to
The Binomial Distribution
Chapter 6
The Normal Distribution
82. Two out of five adult smokers acquired the habit by age
14. If 400 smokers are randomly selected, find the
probability that 170 or more acquired the habit by
age 14.
p =
2
5 = 400(0.4)
= 0.4 = 160
= (400)(0.4)(0.6) = 9.8
z =
169.5 – 160
9.8
area = 0.3340
= 0.97
Section 6-6 Exercise #5
84. The percentage of Americans 25 years or older who
have at least some college education is 50.9%. In a
random sample of 300 Americans 25 years and older,
what is the probability that more than 175 have at least
some college education?
= 300(0.509)
= (300)(0.509)(0.491)
= 152.7
= 8.66
z =
175.5 – 152.7
8.66
area = 0.4957
= 2.63
P(X > 175.5) = 0.5 –0.4957 = 0.0043
Section 6-6 Exercise #7
86. Women comprise 83.3% of all elementary school
teachers. In a random sample of 300 elementary
school teachers, what is the probability that more than
50 are men?
= 300(0.167)
= (300)(0.167)(0.833)
= 50.1
= 6.46
z =
50.5 – 50.1
6.46
area = 0.0239
= 0.06
P(X > 50.5) = 0.5 –0.0239 = 0.4761
Section 6-6 Exercise #11
