Calculate a probability involving drawings from bivariate normal variables with Xi and Yi i.i.d

Question

There's a question which has been troubling me along with my earlier post.

To be honest, I'm not entirely sure on how to proceed. All I know is that if X~N(mu,sigma^2) then P(X < A) = P(Z< (A-mu)/sigma)

I just need help on starting this question.

Answer 1

Hint: You are asked to find $P(2\bar{X}+\bar{Y} > 3)$ where $\bar{X}$ and $\bar{Y}$ are individually normal random variables whose means and variances you should be able to work out. More strongly, $\bar{X}$ and $\bar{Y}$ are jointly normal which implies that $2\bar{X}+\bar{Y}$ is also a normal random variable whose mean and variance you should be able to work out. You will need to know various results such as $$\begin{align} E\left[\sum_i a_iZ_i\right] &= \sum_i a_iE[Z_i]\\ \operatorname{var}\left(\sum_i a_iZ_i\right) &= \sum_i a_i^2\operatorname{var}(Z_i) + \sum_i \sum_{j\neq i} a_ia_j \operatorname{cov}(Z_i,Z_j) \end{align}$$