Distribution of extremal values If an item follows normal distribution, average also follows normal distribution. What about minimum and maximum?
 A: You might also want to read up on the generalized extreme value (GEV) distribution. It turns out that as $n\rightarrow\infty$, the (shifted and scaled) distribution of the maximal value of the sample converges to one of the three special cases of the GEV distribution.
A: You should have a look at the order statistics. 
Here is a very brief overview.
Let $X_{1}, \ldots X_{n}$ be an i.i.d. sample of size $n$ drawn from a population with distribution function $F$ and probability density function $f$. Define $Y_{1}=X_{(1)}, \ldots, Y_{r} = X_{(r)}, \ldots, Y_{n}=X_{(n)}$, where $X_{(r)}$ denotes the $r$th order statistic of the sample $X_{1}, \ldots X_{n}$, i.e., its $r$th smallest value.
It can be shown that the joint probability density function of $Y_{1}, \ldots, Y_{n}$ is
$f_{X_{(1)}, \ldots, X_{(n)}}(y_{1}, \ldots, y_{n}) = n! \prod_{i=1}^{n} f(y_{i})$ if $y_{1} < y_{2} < \ldots < y_{n}$ and $0$ otherwise.
By integrating the previous equation we get
$f_{X_{(r)}}(x) = \frac{n!}{(r - 1)! (n - r)!} f(x) (F(x))^{r-1} (1 - F(x))^{n - r}$
In particular, for the minimum and maximum, we respectively have
$f_{X_{(1)}}(x) = n f(x) (1 - F(x))^{n-1}$
$f_{X_{(n)}}(x) = n f(x) (F(x))^{n - 1}$
A: The sum of Gaussians is Gaussian. That is why the average is normal. The distribution of any non-linear function of (finitely many)  Gaussians need not be Gaussian, and it usually isn't. Such is the case of the maximum function. 
To approximate the maximum of a multivariate Gaussian, Hothorn is a good place to start.
