Independent solution

How to solve this Joint Distributions question

Setup

Setup

Convert the expected Type I cost into the mean count of Type I claims, then use the row for X=1 to determine its missing cell.

E⁡[X]=345750=0.46\operatorname{E}[X]=\frac{345}{750}=0.46
Pr⁡(X=1)=0.21+0.13+q=0.34+q\Pr(X=1)=0.21+0.13+q=0.34+q
q=0.12q=0.12

Model

Model

The known table entries total 0.70, so normalization gives p+q=0.30 and hence p=0.18. Sum columns to obtain the marginal law of Y.

Pr⁡(Y=0)=0.52,Pr⁡(Y=1)=0.31,Pr⁡(Y=2)=0.17\Pr(Y=0)=0.52,\qquad \Pr(Y=1)=0.31,\qquad \Pr(Y=2)=0.17

Compute

Compute

Calculate the first two moments and apply the variance identity.

E⁡[Y]=0.31+2(0.17)=0.65\operatorname{E}[Y]=0.31+2(0.17)=0.65
E⁡[Y2]=0.31+4(0.17)=0.99\operatorname{E}[Y^2]=0.31+4(0.17)=0.99
Var⁡(Y)=0.99−(0.65)2=0.5675\operatorname{Var}(Y)=0.99-(0.65)^2=0.5675

Answer

Answer

The variance rounds to 0.57.

0.57(B)\boxed{0.57\quad\text{(B)}}