Monthly Archives: January, 2019

Exam P Practice Problem 107 – wait time at a busy restaurant

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Both 107-A and 107-B use the following probability density function.

    \displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{1}{450} \ (30-x) &\ \ \ \ \ \ 0 < x < 30 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.

Problem 107-A

The wait time (in minutes) for a table at a busy restaurant on the weekend is distributed according to the density function f(x) given above.

A customer plans to dine in this restaurant on two different weekends.

Determine the expected value of the longest wait of these two visits to the restaurant.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 0 \bold . \bold 0

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 1 \bold . \bold 0

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 2 \bold . \bold 8

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 3 \bold . \bold 5

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 4 \bold . \bold 0

\text{ }

\text{ }

\text{ }

\text{ }

Problem 107-B

The wait time (in minutes) for a table at a busy restaurant on the weekend is distributed according to the density function f(x) given above.

A customer plans to dine in this restaurant on two different weekends.

Determine the expected value of the shortest wait of these two visits to the restaurant.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 4 \bold . \bold 5

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 6 \bold . \bold 0

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 7 \bold . \bold 0

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 8 \bold . \bold 6

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 0 \bold . \bold 0

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

dan ma statistical actuarial

daniel ma statistical actuarial

\copyright 2019 – Dan Ma

Posted in: Practice Problems | Tagged:

Exam P Practice Problem 106 – average height of students

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Problem 106-A

Heights of male students in a large university follow a normal distribution with mean 69 inches and standard deviation 2.8 inches.

Four male students from this university are randomly selected.

Determine the probability that the average height of the selected students is between 5 feet 7 inches and 5 feet 11 inches.

Note that one feet = 12 inches.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 1 \bold 0 \bold 4

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 5 \bold 2 \bold 2 \bold 2

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 4 \bold 7 \bold 2

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 8 \bold 9 \bold 6

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 9 \bold 0 \bold 7 \bold 5

The answers are based on this normal table from SOA.

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Problem 106-B

Heights of female students in a large university follow a normal distribution with mean 65 inches and standard deviation 2.2 inches.

Sixteen female students are randomly selected.

Determine the probability that the average height of the selected students is greater than 5 feet 6 inches.

Note that one feet = 12 inches.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 0 \bold 3 \bold 4 \bold 4

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 7 \bold 8 \bold 2

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 3 \bold 2 \bold 6 \bold 4

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5 \bold 7 \bold 2

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 7 \bold 3 \bold 6

The answers are based on this normal table from SOA.

\text{ }

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Answers

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

\copyright 2019 – Dan Ma

Posted in: Normal Distribution, Practice Problems | Tagged:

Exam P Practice Problem 105 – testing electronic devices

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Problem 105-A

The length of operation (in years) for an electronic device follows an exponential distribution with mean 4. Ten such devices are being observed for one year for a quality control study.

The lengths of operation for these devices are independent.

Determine the probability that no more than three of the devices stop working before the end of the study.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 2 \bold 5 \bold 7

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 1 \bold 3 \bold 2

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 7 \bold 5 \bold 6 \bold 8

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 3 \bold 8 \bold 9

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 5 \bold 6 \bold 0

\text{ }

\text{ }

\text{ }

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Problem 105-B

Twelve patients are randomly selected from a population of patients with history of heart disease to be tracked in a health study. The study begins with an initial assessment of health status. The participants are instructed to return for a follow up visit one year after the initial assessment.

For these patients, the time (in years) from the initial assessment to the next heart attack has an exponential distribution with mean 6.25 years. The times to the next heart attack for these patients are independent.

Determine the probability that ten or more patients experience no heart attack prior to the one-year follow up visit.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 9 \bold 1 \bold 3

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5 \bold 1 \bold 9

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 5 \bold 4 \bold 8 \bold 1

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 4 \bold 5 \bold 5

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 7 \bold 4 \bold 3 \bold 2

\text{ }

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Answers

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

\copyright 2019 – Dan Ma

Posted in: Exponential distribution | Tagged: , ,