Chapter 6.2, Problem 63E

To determine

Mean and standard deviation of W .

Answer to Problem 63E

Mean of W ,

  ${\mu }_W=\$303.35$

Standard deviation of W ,

  ${\sigma }_W=\$6864.29$

Explanation of Solution

Given information:

X : amount earned by Life Insurance Company on a randomly chosen 5 −year term life policy

Such that

Mean,

  ${\mu }_X=\$303.35$

Standard deviation,

  ${\sigma }_X=\$9707.57$

X₁ and X₂ are the insurer’s income from two insurance policies.

Such that

Insurer’s average income on two policies,

  $W=\frac{X_1+X_2}{2}$

Calculations:

If X and Y are independent,

Property mean:

  ${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

Property variance:

  ${\sigma }_{aX+bY}^2=a^2{\mu }_X^2+b^2{\mu }_Y^2$

Since W represents insurer’s average income.

Then

  $W=\frac{X_1+X_2}{2}=0.5X_1+0.5X_2$

Thus,

We have

Mean of W ,

  ${\mu }_W={\mu }_{0.5X_1+0.5X_2}=0.5{\mu }_X+0.5{\mu }_X=0.5\left(303.35\right)+0.5\left(303.35\right)=\$303.35$

Standard deviation of W ,

  ${\sigma }_W={\sigma }_{0.5X_1+0.5X_2}=\sqrt{{0.5}^2{\sigma }_X^2+{0.5}^2{\sigma }_X^2}=\sqrt{0.25{\left(9707.57\right)}^2+0.25{\left(9707.57\right)}^2}\approx \$6864.29$