Linear Combination of Normal Random Variables

Question

An exercise from my textbook: $X$ is normally distributed with mean $100$ and variance $6400$. $Y$ is normally distributed with mean $350$ and variance $10000$. Calculate the probability that $12X - 3Y$ is positive. X and Y are independent. The solution in the back of the book says $12X-3Y$ is normally distributed with mean $12(100)-3(350)=150$ and variance $12(6400)+3(10000)=106800$. I'm pretty sure this is wrong, as iirc $Var(aX+bY)=a^2Var(X)+b^2Var(Y)$ and thus $Var(12X-3Y)=1011600$. But I wanted to check. Thanks.

Answer 1

You're correct, the solution in the book is wrong. $$ \text{Var}(12X - 3Y) = 12^2 \text{Var}(X) + 3^2 \text{Var}(Y) = 144 \cdot 6400 + 9 \cdot 10000 = 1011600. $$