Question

1. A manufacturer’s annual losses follow a distribution with density function f(x) = 2.5(0.6)2.5/ x 3.5 , x > 0.6 0, otherwise. The manufacturer purchases an insurance policy to cover its annual losses with an annual deductible of 2. Calculate the mean of the manufacturer’s annual losses paid by the insurance policy. (A) 0 (B) 0.05 (C) 0.07 (D) 0.12 (E) 0.16

1. A manufacturer's annual losses follow a distribution with density function 2.5(0.6)2.5 f(x)-350.6 0, otherwise The manufacturer purchases an insurance policy to cover its annual losses with an annual deductible of 2. Calculate the mean of the manufacturer's annual losses paid by the insurance policy (A) 0 (B) 0.05 (C) 0.07 (D) 0.12 E) 0.16

Answer 1

The Expected mean annual loss is given by the function

$E(loss)=\int_{0}^{x}xf(x)dx$

Now observe the values that x might take as per the constriants.

since the standard deductible is 2 so for any value of x there will not be any variation .It will be ocnstant at 2. Hence x =0 for x<2.

Simlilarly since 2 is deducted annual

For x>0

The deductible variable will be x-2

Thus

if We denote X = deductible variable then

X=0, x<=0

X=x-2, x>0

Now the expected annual loss

$f_{X}x=\int_{0}^{x}xf(x)dx$

Or, $f_{X}x=\int_{0}^{2}xf(x)dx+\int_{2}^{\infty}xf(x)dx$

Keeping vlaues we have

$E(loss)=\int_{0}^{2}0*f(x)dx+\int_{2}^{\infty}(x-2)*f(x)dx$

Or, $E(loss)=0+\int_{2}^{\infty}((x-2)*2.5*(0.6)^{2.5}/x^{3.5})dx$

Solving it we get

$E(loss)=0.066 \sim 0.070$

Hence option (C) is correct.

Answer 2

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