Independent solution

How to solve this Mixture Distributions question

Setup

Setup

Let m be the unconditional median. It is positive because the atom at zero contains only 20% of the distribution.

FC(m)=0.20+0.80Φ(m1000400)F_C(m)=0.20+0.80\Phi\left(\frac{m-1000}{400}\right)

Model

Model

Set the unconditional cumulative probability equal to one half and isolate the percentile inside the positive-loss component.

0.20+0.80Φ(m1000400)=0.500.20+0.80\Phi\left(\frac{m-1000}{400}\right)=0.50
Φ(m1000400)=0.300.80=0.375\Phi\left(\frac{m-1000}{400}\right)=\frac{0.30}{0.80}=0.375

Compute

Compute

Convert the conditional percentile to its standard-normal quantile and rescale.

Φ1(0.375)=0.3186394\Phi^{-1}(0.375)=-0.3186394
m=1000+400(0.3186394)=872.5443m=1000+400(-0.3186394)=872.5443

Answer

Answer

The nearest listed median claim amount is 873.

873(C)\boxed{873\quad\text{(C)}}