Independent solution

How to solve this Continuous Random Variables question

Setup

Setup

Let X denote the original loss and U the unreimbursed amount. Convert the payment condition back to a condition on X.

U=(1−0.60)X=0.40XU=(1-0.60)X=0.40X
U<1.20  ⟺  X<1.200.40=3U<1.20\iff X<\frac{1.20}{0.40}=3

Model

Model

Use the loss density over the transformed interval.

Pr⁡(U<1.20)=∫03142(x2+1)2 dx\Pr(U<1.20)=\int_0^3\frac{1}{42}\left(\frac{x}{2}+1\right)^2\,dx

Compute

Compute

Integrate after recognizing the linear inner function and evaluate at both bounds.

∫142(x2+1)2dx=163(x2+1)3\int\frac{1}{42}\left(\frac{x}{2}+1\right)^2dx=\frac{1}{63}\left(\frac{x}{2}+1\right)^3
Pr⁡(U<1.20)=163[(52)3−1]\Pr(U<1.20)=\frac{1}{63}\left[\left(\frac{5}{2}\right)^3-1\right]
Pr⁡(U<1.20)=1356=0.2321428571…\Pr(U<1.20)=\frac{13}{56}=0.2321428571\ldots

Answer

Answer

Rounding to three decimals gives 0.232.

Pr⁡(U<1.20)≈0.232(E)\boxed{\Pr(U<1.20)\approx0.232\quad\text{(E)}}