Exam P Practice Problem 100 – find the variance of loss in profit

Problem 100-A

The monthly amount of time X (in hours) during which a manufacturing plant is inoperative due to equipment failures or power outage follows approximately a distribution with the following moment generating function.

    \displaystyle M(t)=\biggl( \frac{1}{1-7.5 \ t} \biggr)^2

The amount of loss in profit due to the plant being inoperative is given by Y=12 X + 1.25 X^2.

Determine the variance of the loss in profit.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,927.20}

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,608.20}

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \text{475,693.76}

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \text{583,358.20}

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \text{601,769.56}

\text{ }

\text{ }

\text{ }

Problem 100-B

The weekly amount of time X (in hours) that a manufacturing plant is down (due to maintenance or repairs) has an exponential distribution with mean 8.5 hours.

The cost of the downtime, due to lost production and maintenance and repair costs, is modeled by Y=15+5 X+1.2 X^2.

Determine the variance of the cost of the downtime.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \text{130,928.05}

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \text{149,368.45}

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \text{181,622.05}

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \text{188,637.67}

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \text{195,369.15}

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers can be found in this page.

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Daniel Ma

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