Find the probability that the average of X and Z is greater than Y. Where X, Z, and Y are normal RVs.

Question

Here is the exact statement:

Suppose X,Y , and Z are independent random variables. X is a normal random variable with mean 5 and variance 16, Y is a normal random variable with mean 7 and variance 25, and Z is a normal random variable with mean 10 and variance 9.

Then, Find the probability that the average of X and Z is greater than Y

End problem statement

For this problem, I think I want to figure out the $P\{E[X] + E[Z] > E[Y]\}$ as expected values are averages. However, I am unsure of where to go from here. Should I first find the probability density function for X + Z and Y using the normal pdf and sums of normal random variables? A small hint would be appreciated. This is the follow up question from a part before in which I found the value of $k$ such that $P\{X + Y + Z \le k\}$. However, I feel that this part is unrelated. Is that correct?

I'm sorry for the lack of information, but the past hour of scouring my book and notes for information where to start has not been founded with luck. A nudge in the right direction would be appreciated.

Answer 1

Hint: $$ \Pr \! \left( \frac{X + Z}{2} > Y \right) = \Pr(X + Z - 2 Y > 0). $$

Any linear combination of normal random variables that are jointly normally distributed is also a normal random variable (so its probability density function is completely described by its mean and variance).