amounts. Use this information to answer the following questions.
Probability Calculations
P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339
P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834
Percentiles Calculations
1st percentile = 1,912,245, 5th percentile = 1,961,388
95th percentile = 2,198,612, 99th percentile = 2,247,755
59. What is the probability that this company will sell between 2.0 and 2.15 million cars next year?
Wendy’s fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday, the demand for
hamburgers is normally distributed with a mean of 450 and standard deviation of 80 and the demand for chicken
sandwiches is normally distributed with a mean of 120 and standard deviation of 30. Use this information to answer the
following questions.
60. Why is the independence assumption in Question 74 probably not realistic? Using a more realistic assumption, do you
think the probability in Question 74 would increase or decrease?
The time it takes a technician to fix a computer problem is exponentially distributed with a mean of 15 minutes.
61. What is the probability density function for the time it takes a technician to fix a computer problem?
The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and
a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase
amounts. Use this information to answer the following questions.
Probability Calculations
P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339
P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834
Percentiles Calculations
1st percentile = 1,912,245, 5th percentile = 1,961,388
95th percentile = 2,198,612, 99th percentile = 2,247,755
62. What number of cars, equidistant from the mean, such that 98% of car sales are between these values?
The service manager for a new appliances store reviewed sales records of the past 20 sales of new microwaves to
determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate
that the probability any one of their new microwaves needs a warranty repair in the first 90 days is 0.05. The manager
assumes that calls for warranty repair are independent of one another and is interested in predicting the number of
warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new microwaves sold.
63. What is the standard deviation of the number of the new microwaves sold that will require a warranty repair in the first
90 days?
The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and
a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase
amounts. Use this information to answer the following questions.
Probability Calculations
P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339
P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834