CCHS
Chapter 5, Section 5.3 Practice
Name__________________________________________________ Date_________ Period_________
Use this formula to find the z-score for each problem:
X 
z

z = Value – Mean
Standard deviation
1. According to a survey conducted by television advertisers, the average adult American watches an
average of 6.98 hours of television per day. The data is normally distributed with a standard deviation
of 3.80 hours. Find the probability that a randomly selected person watches more than 8.00 hours of
television in a day.
2. Insurance companies have determined that US males between the ages of 16 and 24, drive an average of
10,718 miles each year with a standard deviation of 3763 miles. Assume the data is normally
distributed. For a randomly selected male in that age group, find the probability that he drives less than
12,000 miles per year.
3. A company that manufactures sleeping bags says that the average amount of down in an adult sleeping
bag is 32 oz. with a standard deviation of 0.9 oz. What is the probability that a bag chosen at random
has more than 33.2 oz. of down?
Mambou
2/16/2016
CCHS
4. The heights of six-year old girls are normally distributed with a mean of 117.80 cm and a standard
deviation of 5.52 cm. Find the probability that a randomly selected six-year girl has a height between
117.80 cm and 120.56 cm.
5. The average heating bill for a residential area is $123 for the month of November with a standard
deviation of $8. If the amounts of the heating bills are normally distributed, find the probability that the
average bill for a randomly selected resident is more than $125.
6. An IQ test has a mean of 100 with a standard deviation of 15. What is the probability that a randomly
selected adult has an IQ between 85 and 115?
7. The average adult spends 5 hours per week on a home computer, with a standard deviation of 1 hour.
What percent of adults spend more than 6 hours per week on their home computer?
Mambou
2/16/2016
CCHS
Chapter 5, Section 5.3 Practice
Name_______Keys_____________________ Date__February 16, 2016 Period_________
Use this formula to find the z-score for each problem:
X 
z

z = Value – Mean
Standard deviation
1. According to a survey conducted by television advertisers, the average adult American watches an
average of 6.98 hours of television per day. The data is normally distributed with a standard
deviation of 3.80 hours. Find the probability that a randomly selected person watches more than
8.00 hours of television in a day.
8.00  6.98
or
3.8
P( z  0.268)  normalcdf (0.268, 10)  0.3943 or 39.43%
  6.98 hrs   3.80 hrs x  8 hrs z 
z  0.268
2. Insurance companies have determined that US males between the ages of 16 and 24, drive an
average of 10,718 miles each year with a standard deviation of 3763 miles. Assume the data is
normally distributed. For a randomly selected male in that age group, find the probability that he
drives less than 12,000 miles per year.
12,000  10,718
or z  0.341
3763
P( z  0.341)  normalcdf (10, 0.341)  0.6334 or 63.34%
  10718 miles   3763 miles x  12,000 miles z 
3. A company that manufactures sleeping bags says that the average amount of down in an adult
sleeping bag is 32 oz. with a standard deviation of 0.9 oz. What is the probability that a bag
chosen at random has more than 33.2 oz. of down?
33.2  32
or z  1.333
0.9
P( z  1.333)  normalcdf (1.33, 10)  0.0918 or 9.18%
  32oz   0.9oz x  33.2oz z 
4. The heights of six-year old girls are normally distributed with a mean of 117.80 cm and a
standard deviation of 5.52 cm. Find the probability that a randomly selected six-year girl has a
height between 117.80 cm and 120.56 cm.
117.80  117.80
120.56  117.80
z
5.52
5.52
or 0  z  0.5 P(0  z  0.5)  normalcdf (0, 0.5)  0.1915 or 19.15%
  117.80 cm   5.52 cm 117.80  x  120.56
Mambou
2/16/2016
CCHS
5. The average heating bill for a residential area is $123 for the month of November with a standard
deviation of $8. If the amounts of the heating bills are normally distributed, find the probability
that the average bill for a randomly selected resident is more than $125.
125  123
or z  0.25
8
P( z  0.25)  normalcdf (0.25, 10)  0.4013 or 40.13%
  $123   $8 x  $125 z 
6. An IQ test has a mean of 100 with a standard deviation of 15. What is the probability that a
randomly selected adult has an IQ between 85 and 115?
85  100
115  100
z
15
15
or  1  z  1 P(1  z  1)  normalcdf (1, 1)  0.6827 or
  100   15 85  x  115
68.27%
7. The average adult spends 5 hours per week on a home computer, with a standard deviation of 1
hour. What percent of adults spend more than 6 hours per week on their home computer?
65
or z  1
1
P( z  1)  normalcdf (1, 10)  0.1587 or 15.87%
  5hrs   1hr x  6hrs z 
Mambou
2/16/2016