Chapter 6
The Normal Distribution
Section 6-3
The Standard Normal
Distribution
Chapter 6
The Normal Distribution
Section 6-3
Exercise #7
Find the area under the normal distribution curve.
Between z = 0 and z = 0.75
0
area = 0.2734
0.75
Chapter 6
The Normal Distribution
Section 6-3
Exercise #15
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.
0 0.79 1.28
0.3997 – 0.2852 = 0.1145
Chapter 6
The Normal Distribution
Section 6-3
Exercise #31
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
0.5 – 0.4977 = 0.0023
2.83
Chapter 6
The Normal Distribution
Section 6-3
Exercise #45
Find the z value that corresponds to the given area.
0.8962
z
0
0.8962 – 0.5 = 0.3962
Find the z value that corresponds to the given area.
0.8962
z
0
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.
Chapter 6
The Normal Distribution
Section 6 - 4
Exercise #3
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…
a. Greater than 700,000.
b. Between 500,000 and 600,000.
a. Greater than 700,000
700,000 – 618,319
= 1.63
z=
50,200
P(z > 1.63) = 0.5 –0.4484
= 0.0516 or 5.16%
0
1.63
z=
X –

b. Between 500,000 and 600,000.
500,000 – 618,319
= – 2.36
z =
50,200
area = 0.4909
600,000 – 618,319
z=
= – 0.36
50,200
area = 0.1406
z=
X –

b. Between 500,000 and 600,000.
area = 0.4909
area = 0.1406
P( – 2.36 < z < – 0.36) = 0.4909 – 0.1406
= 0.3503 or 35.03%
– 2 .36 – 0.36
Chapter 6
The Normal Distribution
Section 6-4
Exercise #11
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
a. That the senior owes at least $1000
–
X
z=

1000 – 3262
= – 2.06
z=
1100
area = 0.4803
P(z • – 2.06) = 0.5 + 0.4803
= 0.9803 or 98.03%
a. That the senior owes at least $1000
0.9803 or 98.03%
– 2.06
0
b. That the senior owes more than $4000
–
X
z=

4000 – 3262
z=
= 0.67
1100
area = 0.2486
P(z > 0.67) = 0.5 – 0.2486
= 0.2514 or 25.14%
b. That the senior owes more than $4000
0.2514 or 25.14%
0
0.67
c. That the senior owes between $3000 and $4000.
–
X
z=

3000 – 3262
= – 0.24
z=
1100
area = 0.0948
P( – 0.24 < z < 0.67) = 0.0948 + 0.2486
= 0.3434 or 34.34%
c. That the senior owes between $3000 and $4000.
0.3434 or 34.34%
– 0.24
0
0.67
Chapter 6
The Normal Distribution
Section 6-4
Exercise #27
An advertising company plans to market a product to
low-income families. A study states that for a particular
area, the average income per family is $24,596 and the
standard deviation is $6256. If the company plans to
target the bottom 18% of the families
based on income, find the cutoff
income. Assume the variable is
normally distributed.
The bottom 18% means that 32% of the area is between
z and 0. The corresponding z score will be – 0.92 .
X = – 0.92(6256) + 24,596
= $18,840.48
0.18
$18,840.48
0.32
$24,596
Chapter 6
The Normal Distribution
Section 6-5
Exercise #13
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.
–
2.00 – 2.02
X
z=
=
= – 1.58

0.08
n
40
area = 0.4429
P(z < –1.58) = 0.5 – 0.4429
= 0.0571or 5.71%
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.
0.0571 or 5.71%
$2.00
$2.02
Chapter 6
The Normal Distribution
Section 6-5
Exercise #21
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.
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.
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
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.
a. Find the probability that a randomly
selected adult will complete the test in
less than 43 minutes.
–
– 46.2
X
43
z=  =
= – 0.4
8
area = 0.1554
P(z < –0.4) = 0.5 –0.1554
= 0.3446or 34.46%
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
0.3446or 34.46%
43
46.2
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
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 = – 2.83
8
50
area = 0.4977
P(z < – 2.83) = 0.5 –0.4977
= 0.0023or 0.23%
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
0.0023or 0.23%
43
46.2
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
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.
Average time = 46.2 minutes, Standard deviation = 8
minutes, variable is normally distributed.
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%.
Chapter 6
The Normal Distribution
Section 6-5
Exercise #23
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.
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.
If a single egg is selected, find the
probability that the cholesterol content
will be greater than 220 milligrams.
220– 215
–
X
= 0.33
z=
=
15

area = 0.1293
P(z > 0.33) = 0.5 – 0.1293
= 0.3707or 37.07%
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.
0.3707or 37.07%
215
220
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.
If a sample of 25 eggs is selected, find the
probability that the mean of the sample will
be larger than 220 milligrams.
z=
X– 

=
220 – 215
n
area = 0.4525
15
25
= 1.67
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.
P(z > 1.67) = 0.5 –0.4525
= 0.0475or 4.75%
215
220
Chapter 6
The Normal Distribution
Section 6-6
Exercise #5
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.
2
p = 5 = 0.4
 = 400(0.4) = 160
 = (400)(0.4)(0.6) = 9.8
169.5 – 160
= 0.97
z=
9.8
area = 0.3340
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(X > 169.5) = 0.5 – 0.3340= 0.1660
160
169.5
Chapter 6
The Normal Distribution
Section 6-6
Exercise #7
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) = 152.7
 = (300)(0.509)(0.491) = 8.66
175.5 – 152.7
= 2.63 area = 0.4957
z=
8.66
P(X > 175.5) = 0.5 – 0.4957 = 0.0043
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?
P(X > 175.5) = 0.5 –0.4957 = 0.0043
152.7
175.5
Chapter 6
The Normal Distribution
Section 6-6
Exercise #11
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) = 50.1
 = (300)(0.167)(0.833) = 6.46
50.5 – 50.1
= 0.06 area = 0.0239
z=
6.46
P(X > 50.5) = 0.5 – 0.0239 = 0.4761
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?
P(X > 50.5) = 0.5 – 0.0239 = 0.4761
50.1
50.5