Probability of a normal random variable added to a number being greater than another normal random variable, and distribution of average $X$= random height of a male
$Y$= random height of a female
$X$ and $Y$ are independent of each other
For $x$, $\mu=180\text{ cm}$ and $\sigma^2= 16\text{ cm}^2$
For $y$, $\mu=170\text{ cm}$ and $\sigma^2= 25\text{ cm}^2$
$\mu$ is expected value, $\sigma^2$ is variance
Assume one random male and one random female are selected from the population.
a) what is the probability that the male is more than $5$ cm taller than the female?
b) what is the distribution of the average of the two heights?
So to compute the answer I tried to solve for $P(X+5>Y)=P((X+5-Y)>0)$, but for the overall expected value and $\sigma$ value, I don't really understand what procedure to use.  Also, for the distribution of the average of the two heights, I believe you have to use the sample mean, but I don't have $n$, so I am not sure how to compute that either.  Any help would be greatly appreciated, thanks.
 A: The male is more than $5$ cm taller than the female if $X\gt Y+5$.
I would then probably let $W=X-Y$. We want $\Pr(W\gt 5)$.
From a standard result, we have that $W$ has normal distribution, mean  $\mu_X-\mu_Y$, and variance $\sigma_X^2+\sigma_Y^2$.
Now that we know the mean and variance of $W$, we can find $\Pr(W\gt 5)$ in the usual way.
Second question: Let $X$ and $Y$ be independent normally distributed random variables, with means $\mu_X,\mu_Y$ and variances $\sigma_X^2, \sigma_Y^2$. Let $a$ and $b$ be constants. 
Then $aX+bY$ has normal distribution, mean $a\mu_X+b\mu_Y$ and variance $a^2\sigma_X^2+b^2\sigma_Y^2$.
Use this with $a=b=\frac{1}{2}$.
A: The average of the two heights is $(X+Y)/2$.
Its expected value is $\operatorname{E}\left( \frac{X+Y}2 \right)= \frac{180+170}2$.
Its variance is
$$
\operatorname{var}\left(\frac{X+Y}2\right) = \frac 1 4 \operatorname{var}(X+Y) = \frac 1 4 \left(\operatorname{var}(X)+ \operatorname{var}(Y) \right) = \frac 1 4 (16+25).
$$
The average of the two is normally distributed with that expected value and that variance.  (André Nicolas has answered part (a).)
