Independent solution

How to solve this Variance of Linear Combinations question

Setup

Setup

Discard the fixed shift when computing variance.

Var(3X+2Y5)=Var(3X+2Y)\operatorname{Var}(3X+2Y-5)=\operatorname{Var}(3X+2Y)

Model

Model

Expand the linear-combination variance and use independence.

Var(3X+2Y)=32Var(X)+22Var(Y)+12Cov(X,Y)\operatorname{Var}(3X+2Y)=3^2\operatorname{Var}(X)+2^2\operatorname{Var}(Y)+12\operatorname{Cov}(X,Y)
Cov(X,Y)=0\operatorname{Cov}(X,Y)=0

Compute

Compute

Insert the two component variances.

Var(Z)=9(3)+4(4)=27+16=43\operatorname{Var}(Z)=9(3)+4(4)=27+16=43

Answer

Answer

The score variance is 43.

43(D)\boxed{43\quad\text{(D)}}