Test Review
The management at Ohio National Bank does not want its customers to wait in line for service for too long. The
manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes
for service. Assume that the waiting time for all customers at this branch have a normal distribution with a
mean of 8 minutes and a standard deviation of 2 minutes.
Find the probability that a randomly selected customer will have to wait for less than 3 minutes.


𝑃(𝑥 < 3)

𝑧=

𝑃 𝑥 < 3 = 𝑃 𝑧 < −2.5 = .0062
𝑥−𝜇
𝜎
=
3−8
2
= −2.5
What percentage of the customers have to wait for 10 to 13 minutes?


𝑃 10 ≤ 𝑥 ≤ 13

𝑧=

𝑃 10 ≤ 𝑥 ≤ 13 = 𝑃 𝑧 ≤ 2.5 − 𝑃 𝑧 ≤ 1 = .9938 − .8413 = .1525
15.25% of customers have to wait for 10 to 13 minutes.

𝑥−𝜇
𝜎
=
10−8
2
= 1 and 𝑧 =
𝑥−𝜇
𝜎
=
13−8
2
= 2.5
Is it possible that a customer may have to wait longer than 16 minutes for service? Explain.


𝑃 𝑥 > 16

𝑧=

𝑃 𝑥 > 16 = 1 − 𝑃 𝑧 < 4 = 1 − 1 = 0 (𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑦)
It is possible that a customer may have to wait longer than 16 minutes for service, but very unlikely.

𝑥−𝜇
𝜎
=
16−8
2
=4
A city is planning to build a hydroelectric power plant. A local newspaper found that
53% of the voters in this city favor the construction of the plant. Assume that this result
holds true for the population of all voters in this city.
What is the probability that more than 50% of the voters in a random sample of 200
voters selected from this city will favor the construction of the plant?


𝑃 𝑝 > .5

𝑧=

𝜎𝑝 =

𝑧=

𝑃 𝑝 > .5 = 1 − 𝑃 𝑧 < −.85 = 1 − .1977 = .8023
𝑝−𝑝
𝜎𝑝
𝑝𝑞
𝑛
𝑝−𝑝
𝜎𝑝
=
=
.53∗.47
200
.5−.53
.0353
= .0353
= −.85
Testing has shown that a new washing machine has a length of life that is normally
distributed with mean equal to 5.1 years and a standard deviation equal to 1.54 years. If
the washers are guaranteed for 2 years, what percentage will fail before the end of the
guarantee time?
𝑃 𝑥<2
𝑧=
𝑥−𝜇
𝜎
=
2−5.1
1.54
= −2.01
 𝑃 𝑥 < 2 = 𝑃 𝑧 < −2.01 = .0222
 2.22% of washers will fail before the end of the guarantee time.
The Toyota Prius hybrid car is estimated to get an average of 50 miles per gallon (mpg) of gas. However, the
gas mileage varies from car to car due to a variety of conditions, driving styles, and other factors and has been
reported to be as high as 70 mpg. Suppose that the distribution of miles per gallon for Toyota Prius hybrid card
has a mean of 50 mpg and a standard deviation of 5.9 mpg. Find the probability that the average miles per
gallon for 38 randomly selected Prius hybrid cars is
 Between 48 and 51
 𝑃(48 ≤ 𝑥 ≤ 51)
 𝑧=
𝑥−𝜇
 𝜎𝑥 =
 𝑧=
𝜎𝑥
𝜎
𝑛
𝑥−𝜇
𝜎𝑥
=
=
5.9
= .9571
38
48−50
.9571
= −2.09 and 𝑧 =
𝑥−𝜇
𝜎𝑥
=
51−50
.9571
= 1.04
 𝑃 48 ≤ 𝑥 ≤ 51 = 𝑃 𝑧 ≤ 1.04 − 𝑃 𝑧 ≤ −2.09 = .8508 − .0183 = .8325
 Less than 53
 𝑃(𝑥 ≤ 53)
 𝑧=
𝑥−𝜇
𝜎𝑥
=
53−50
.9571
= 3.13
 𝑃 𝑥 ≤ 53 = 𝑃 𝑧 ≤ 3.13 = .9991
 Greater than the population mean by 2.5 or more
 𝑃(𝑥 ≥ 52.5)
 𝑧=
𝑥−𝜇
𝜎𝑥
=
52.5−50
.9571
= 2.61
 𝑃 𝑥 ≥ 52.5 = 1 − 𝑃 𝑧 < 2.61 = 1 − .9955 = .0045
According to an article on Yahoo.com, the average salary of actuaries in the U.S. is
$98,620 a year. Suppose that currently the distribution of annual salaries of all actuaries
in the U.S. is approximately normal with a mean of $98,620 and a standard deviation of
$18,000. How much would an actuary have to be paid in order to be in the highest-paid
10% of all actuaries?
 Area under the curve to left of x is .9
 z = 1.28
 𝑥 = 𝜇 + 𝑧𝜎 = 98620 + 1.28 ∗ 18000 = $121,660
According to an estimate, the average price of homes in Martha’s Vineyard, Massachusetts,
was $650,000 in 2011. Suppose that the current population distribution of home prices in
Martha’s Vineyard has a mean of $650,000 and a standard deviation of $140,000, but the shape
of this distribution is unknown. Let 𝑥 be the average price of a random sample of certain
homes selected from Martha’s Vineyard.
 Calculate the mean and standard deviation of the sampling distribution of 𝑥 and describe
its shape for a sample size of 100 homes.
 𝜇𝑥 = 𝜇 = 650,000
 𝜎𝑥 =
𝜎
𝑛
=
140000
100
= 14000
 𝑛 ≥ 30, so normal distribution
 Find the probability that the average price of a random sample of 100 homes selected from
Martha’s Vineyard is within $24,000 of the population mean.
 𝑃 626000 ≤ 𝑥 ≤ 674000
 𝑧=
𝑥−𝜇
𝜎𝑥
=
626000−65000
14000
= −1.71 and 𝑧 =
𝑥 −𝜇
𝜎𝑥
=
674000−65000
14000
= 1.71
 𝑃 626000 ≤ 𝑥 ≤ 674000 = .9564 − .0436 = .9128
 Find the probability that the average price of a random sample of 100 homes selected from
Martha’s Vineyard is not within $24,000 of the population mean.
 1 − 𝑃 626000 ≤ 𝑥 ≤ 674000
 1 − .9128 = .0872
According to a Time/Magazine ABT SRBI poll conducted by telephone, 73% of adults
age 18 years and older said they are in favor of raising taxes on those with annual
incomes of $1 million or more to help cut the federal deficit. Assume that this percentage
is true for the current population of all American adults age 18 years and older. Let 𝑝 be
the proportion of American adults age 18 years and older in a random sample of 900
who will hold the above opinion.
 Find the mean and standard deviation of the sampling distribution of 𝑝 and describe its
shape.
 𝜇𝑝 = 𝑝 = .73
 𝜎𝑝 =
𝑝𝑞
𝑛
=
.73∗.27
900
= .0148
 np = 657 (>5) and nq = 333 (>5) so normal distribution
 Find the probability that the sample proportion is less than .76
 𝑃 𝑝 < .76
 𝑧=
.76−.73
.0148
= 2.03
 𝑃 𝑝 < .76 = 𝑃 𝑧 < 2.03 = .9788
 What is the probability that the sample proportion is greater than the population
proportion by .03 or more?
 𝑃 𝑝 > .76 = 1 − 𝑃 𝑝 < .76 = 1 − .9788 = .0212