Average volume of set of cubes using the mean and variance of its side lengths. First, i tried this question:
The side lengths of a set of squares have an average of 5 and variance of 4. What is their average area?
Let X = The Side Length
From this question , I knew we had to calculate $E(X^2)$.
i.e. $$Var(X) = E(X^2) - E(X)^2$$ and as, $$Var(X) = 4, E(X) = 5$$
The average area of the set of squares is therefore $E(X^2) = 29$.
However, if the question were talking about a set of cubes, how would you work out their average volume? Is it possible from just the information given? (Does it involve $E(X^3)$ by any chance too?)
 A: I agree with your conclusion that the problem would consist of finding $\mathbb{E}(X^3)$.
In general, knowing the mean and variance is not enough to calculate the third moment of a distribution. There are some exceptions, for example the normal distribution.
For some distributions though, you would need to the entire distribution function (or something equivalent, e.g. the characteristic function) to calculate the third moment, even given the first and second moments.
As an example, consider Student's t-distribution for $2 < \nu \le 3$. The third moment doesn't even exist, but the second and first moments do and can be specified.
https://en.wikipedia.org/wiki/Student%27s_t-distribution
Of course, such an example probably is not of much use for this problem, since side lengths are inherently non-negative, and student's t'distribution attains negative values.
Nevertheless, I imagine it would be relatively uncomplicated to make a one-sided version of the Student's t-distribution for $2 <\nu \le 3$ (but don't quote me on this) for which we still have the same problem.
EDIT: Yes, it it possible.
Take $\nu=3$, then
$2\frac{\Gamma(2)}{\sqrt{3\pi} \Gamma(\frac{3}{2})} (1 + \frac{x^2}{3})^{-2}$ for $x\ge 0$
and
0 otherwise
is the density function of a non-negative distribution with finite, hence defined mean and variance, but undefined third moment. Thus the side lengths $X$ would assume physically sensible (i.e. non-negative) values with probability one, as would their variance, but it would still be impossible to make any prediction about the average volume of cubes formed from such sides.
