Basic stats question If $X\sim N(3,3)$ and $Y\sim N(-0.5,0.25)$ are independent, what is the probability that $W=X+2Y$ is negative?
I attempted to find $X+2Y$ which I found to be $N(2,10)$, not sure where to go from here, or if I am even on the right path?
 A: If $\operatorname{var}(X)=3$ and $\operatorname{var}(Y)= 0.25$ and they are independent, then
$$
\operatorname{var}(X+2Y) = \operatorname{var}(X) + 4\operatorname{var}(Y) = 3 + 4\times0.25 = 4.
$$
If looks as if you found $3^2+4\times0.25,$ in effect treating $3$ as the standard deviation rather than the variance and $0.25$ as the variance, not the standard deviation.
Once you have $X+2Y \sim N(2,4)$, you can say
$$
\Pr(X+2Y<0) = \Pr\left( \frac{X+2Y - 2}{\sqrt 4} < \frac{0-2}{\sqrt 4} \right) = \Pr( Z < -1) = \cdots.
$$
A: Hint:
$$P(W<0)=P\left(\frac{W-\mu}{\sigma}<\frac{-\mu}{\sigma}\right)=\Phi\left(\frac{-2}{\sqrt{4}}\right)=\Phi(-1)$$
A: Hint: Since $X$ and $Y $ are independent $\mathbb E(W)=\mathbb E(X+2Y)=\mathbb E(X)+2\cdot E(Y)$ and 
$\operatorname{Var}(W)=\operatorname{Var}(X+2Y)=\operatorname{Var}(X)+\operatorname{Var}(2Y)=\operatorname{Var}(X)+4\cdot \operatorname{Var}(Y)$ Thus 
$$P(W<0)=\Phi\left(\frac{0-\mathbb E(W)}{\sqrt{\operatorname{Var}(W)}} \right)$$
$\Phi(z)$ is the cdf of the standard normal distribution.
