Chapter 16, Problem 35E

(a)

To determine

To find: the reason that standard deviation is so large.

Explanation of Solution

Given:

  $\begin{array}{c}E\left(X\right)=\$150\\ SD\left(X\right)=\$6000\end{array}$

Sometimes the company would have to pay the policy holder big amount in which case the profit reduces. Therefore, there is big variation in the amount of profit.

(b)

To determine

To find: the mean and standard deviation of the annual profit if writes only two of these policies.

Answer to Problem 35E

Mean = 300

Standard deviation = 154.9

Explanation of Solution

Given:

  $\begin{array}{c}E\left(X\right)=\$150\\ SD\left(X\right)=\$6000\end{array}$

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Suppose Y is the annual profit on 2 such polices.

  $Y=2X$

Mean:

  $\begin{array}{c}E\left(Y\right)=2E\left(X\right)\\ =2\left(150\right)\\ =300\end{array}$

Variance:

  $\begin{array}{c}\mathrm{var}\left(Y\right)=\mathrm{var}\left(2X\right)\\ =4\times \mathrm{var}\left(X\right)\\ =4\times 6000\\ =24000\end{array}$

Standard deviation:

  $\begin{array}{c}SD\left(Y\right)=\sqrt{\mathrm{var}\left(Y\right)}\\ =\sqrt{24000}\\ =154.9\end{array}$

Therefore, the mean is 300 and standard deviation is 154.9

(c)

To determine

To find: the mean and standard deviation of profit of annual if it writes 10000 of these policies.

Answer to Problem 35E

Mean = 1500000

Standard deviation = 60000000

Explanation of Solution

Given:

  $\begin{array}{c}E\left(X\right)=\$150\\ SD\left(X\right)=\$6000\end{array}$

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Suppose Z is the annual profit on 10000 policies

  $Z=10000X$

Mean:

  $\begin{array}{c}E\left(Z\right)=10000E\left(X\right)\\ =10000\left(150\right)\\ =1500000\end{array}$

Variance:

  $\begin{array}{c}\mathrm{var}\left(Z\right)=\mathrm{var}\left(10000X\right)\\ =\left(10000\right)\mathrm{var}\left(X\right)\\ ={\left(10000\right)}^2{\left(6000\right)}^2\end{array}$

Standard deviation:

  $\begin{array}{c}SD\left(Y\right)=\sqrt{\mathrm{var}\left(Z\right)}\\ =\sqrt{{\left(10000\right)}^2{\left(6000\right)}^2}\\ =\$60000000\end{array}$

(d)

To determine

To Explain: that the company is likely to be profitable.

Explanation of Solution

Given:

  $\begin{array}{c}E\left(X\right)=\$150\\ SD\left(X\right)=\$6000\end{array}$

The company would be profitable for big number of policies although mean profit increases and standard deviation is near to profit.

(e)

To determine

To find: the assumptions underlie the analysis and thought of circumstances under that assumptions may be violated.

Explanation of Solution

Given:

  $\begin{array}{c}E\left(X\right)=\$150\\ SD\left(X\right)=\$6000\end{array}$

The profit made on policy is not dependent of policy made on another, this is the assumptions underlie the analysis.