Question

(13) Th e probability density function of the loss X is /r(x) = { 0.02 (1-100), 0<x <100 100 0< otherwise The amount paid Y is 70 percent of that portion of the loss that e xceeds a deductible of 20. Determine E(Y). A)12 B)15 C)21。23 E) 25 (14) You are given that X is a normal randou variable such that E(X2) -1 A) 0.25 B) 0.33C) 0.64 (D)0.97 E) Does not exist P(X > 0) = 0.92. Find the variance 0.64 D)0.97 E) Does not exist
the circled answers are wrong can show a good work on those two problems to arrive at correct answers please, will rate more points.

Answer 1

here E(Y)=0.7*(X-20)

therefore E(Y)=$\int_{20}^{100}$ 0.7*(X-20)*f(x) dx =$\int_{20}^{100}$0.7*0.02*(x-20)*(1-x/100) dx

=$\int_{20}^{100}$0.014*(x-x²/100-20+x/5) dx =0.014*(x²/2-x³/300-20x+x²/10)|¹⁰⁰₂₀ =11.95~ 12

option A

14)

P(X>0)=0.92

for above z =-1.405

therefore $\mu$-1.405$\sigma$ =0

$\mu$ =1.405$\sigma$ ..........(1)

E(X²) =$\mu$²+$\sigma$² =1 .......(2)

solving above equations:

$\sigma$² =0.33

option B