Chapter 6.1, Problem 18E

(a)

To determine

To construct: a table for the probability distribution

Answer to Problem 18E

Profit (in $)	-199700	300
Probability	0.0002	0.998

Explanation of Solution

Profit=$300 If house destroyed it would be decreased by $200000

(b)

To determine

To Calculate: the expected value and explain it

Answer to Problem 18E

  ${\mu }_Y=E\left(Y\right)=260$

Explanation of Solution

Formula used:

  $E\left(Y\right)={\displaystyle \sum yP(y)}$

Calculation;

y	$\text{P}\left(y\right)$	$\text{yP}\left(y\right)$	$y^2$	$y^2\text{P}\left(y\right)$
-199700	0.0002	-39.94	39880090000	7976018
300	0.9998	299.94	90000	89982
		260		8066000

  $\begin{array}{c}E\left(Y\right)={\displaystyle \sum yP(y)}\\ =260\end{array}$

Mean is $260 Therefore average 260$ profit will be gained by single insurance policy

(c)

To determine

To Calculate: the standard deviation of Y and explain it

Answer to Problem 18E

  $\sigma =2828.14$

Explanation of Solution

Formula used:

  $\sigma =\sqrt{E\left(Y^2\right)+{\left\{E\left(Y\right)\right\}}^2}$

Calculation:

  $\begin{array}{c}E\left(Y^2\right)=y^{2}P\left(y\right)\\ =8066000\end{array}$

  $\begin{array}{c}\sigma =\sqrt{8066000-{260}^2}\\ =\sqrt{8066000-67600}\\ =\sqrt{7998400}\\ =2828.14\end{array}$

Standard deviation is $2828.14 and mean is $260 therefore the profit of policy on an arbitrary choose varies $2828.14 from the mean $260