Independent solution

How to solve this Compound Distributions question

Setup

Setup

A single recruit generates 0, 1, or 2 pensions with probabilities 0.6, 0.1, and 0.3.

P(W=0)=0.6,P(W=1)=0.1,P(W=2)=0.3P(W=0)=0.6,\quad P(W=1)=0.1,\quad P(W=2)=0.3

Model

Model

Compute the first two moments of the three-point count, giving mean 0.7 and variance 0.81 per recruit.

E[W]=0.7E[W]=0.7
E[W2]=1.3E[W^2]=1.3
Var(W)=1.30.72=0.81\operatorname{Var}(W)=1.3-0.7^2=0.81

Compute

Compute

For 100 recruits the total has mean 70 and standard deviation 9. Applying the 90.5 continuity-corrected boundary gives z=2.2778 and probability 0.9886.

E[S]=70,SD(S)=9E[S]=70,\quad\operatorname{SD}(S)=9
P(S90)Φ(90.5709)=Φ(2.2778)=0.9886P(S\le90)\approx\Phi\left(\frac{90.5-70}{9}\right)=\Phi(2.2778)=0.9886

Answer

Answer

The normal approximation rounds to 0.99, corresponding to choice E.

0.99(E)\boxed{0.99\quad\text{(E)}}