Independent solution

How to solve this Mixture Distributions question

Setup

Setup

Measure payments in thousands. Partial damage pays the amount of loss above the 1-thousand deductible, total loss pays 14, and the remaining state pays zero.

P(partial)=0.04,P(total)=0.02P(\text{partial})=0.04,\quad P(\text{total})=0.02
payment=max⁡(X−1,0) for partial damage, in thousands\text{payment}=\max(X-1,0)\ \text{for partial damage, in thousands}

Model

Model

Take expectations within the mutually exclusive partial- and total-loss branches, then weight those conditional payments by their branch probabilities.

E[C]=0.02(14)+0.04∫115(x−1)(0.5003)e−x/2 dxE[C]=0.02(14)+0.04\int_1^{15}(x-1)(0.5003)e^{-x/2}\,dx

Compute

Compute

The total-loss branch contributes 0.02(14)=0.28. The deductible-adjusted partial-loss integral is 1.2049345, contributing 0.04(1.2049345)=0.0481974.

∫115(x−1)(0.5003)e−x/2 dx≈1.2049345\int_1^{15}(x-1)(0.5003)e^{-x/2}\,dx\approx1.2049345
E[C]≈0.28+0.04(1.2049345)=0.3281974E[C]\approx0.28+0.04(1.2049345)=0.3281974

Answer

Answer

The combined expected payment is 0.3281974 thousand, or approximately 328, so choice B is selected.

328(B)\boxed{328\quad\text{(B)}}