Consider $X_1$~normal(4,3) , $X_2$~normal(5,7), and $X_3$~normal(6,4), where $X_1$, $X_2$, and $X_3$ are independent. Obtain variance of $W = 2X_1 - X_2 + 3X_3 + 3.$
Attempt :
Since we are given $\sigma_1$ = 3, $\sigma_2$ = 7, and $\sigma_3$ = 4, we need to to do $\sigma^2_W = (2^2)(3^2) + (-1^2)(7^2) + (3^2)(4^2) = 229$
However, solution is 55. The solution does not square $\sigma_i$ values; I thought that since we're after the variance of W, we must also square standard deviation of each random variable; Why is this?
No, in general, if $X$ follows a normal distribution, then in shorthand it is written as $$X\sim N(\mu,\sigma^2).$$ Thus, for example, the variance of $X_1$ is 3 not $3^2$. Or as you have written it $$\sigma_1^2 = 3$$ not $\sigma_1 = 3$.