Would using the mean as pivot speed up quicksort?
Using the mean for your partition does not prevent the $\Omega(n^2)$ worst-case behavior. It occurs when the input list is exponentially increasing. Consider the input:

$1,n^2,n^3,\ldots,n^n$

The mean of this set is (asymptotically) $n^{n-1}$ so you obtain the worst partition possible. This is a bit of a cheat considering that storing the list takes $\Omega(n^2)$ space if the numbers are represented as integers. But if you are sorting floating-point numbers then this scenario is concievable.

However, it is possible to calculate the median of a set (or any other order statistic for that matter) in $O(n)$ time so if you really care about run time guarantees for quick sort you should use that rather than the mean.

However, in all practical scenarios the additional cost of computing the mean/median is so large that picking a random pivot almost always is faster.