What you are being asked is as follows. Imagine tossing the die $n$ times. Let $X_i$ be the number obtained on the $i$-th toss, and let
$$\bar{X}=\frac{X_1+X_2+\cdots+X_n}{n}$$
For large $n$, what is the approximate distribution of $\bar{X}$?
First we find the mean $\mu$ and the variance $\sigma^2$ of $X_i$.
The random variable $X_i$ has discrete uniform distribution, taking on values $1$ to $20$ each with probability $\frac{1}{20}$. It follows that
$$\mu=\frac{1+2+\cdots+20}{20}.$$
By a formula you were given earlier in the problem, the sum $1+2+\cdots +20$ is $\frac{(20)(21)}{2}$. It follows that $\mu=\frac{21}{2}$.
For $\sigma^2$, use the fact this is $E(X_i^2)-(E(X_i))^2$. We need to find $E(X_i^2)$. This is
$$\frac{1^2+2^2+\cdots +20^2}{20}.$$
To find the sum of the squares, use the formula given to you for the sum of the first $n$ squares. The sum of the squares is $\frac{(20)(21)(41)}{6}$.
Now you have all the information needed to find $\sigma^2$. Divide the sum of squares by $20$, and subtract $(10.5)^2$.
Finally, the mean of $\bar{X}$ is $\mu$, and the variance of $\bar{X}$ is $\frac{\sigma^2}{n}$. For largish $n$, and it does not have to be terribly large, $\bar{X}$ is approximately normal with mean, variance given in this paragraph. And yes, it is by the Central Limit Theorem that we conclude that $\bar{X}$ has approximately normal distribution.
$\displaystyle \sum_{i=1}^N \left(\frac{x_i-\mu}{\sigma}\right)^2$ has a $\chi_N^2$ distribution (chi-squared with $N$ degrees of freedom) as as the sum of $N$ independent standard normal random variables.
So if $\mu=0$ and $\displaystyle s_2^2=\frac{1}{N}\sum_{i=1}^Nx_i^2$ then $N s_2^2 / \sigma^2$ also has a $\chi_N^2$ distribution.
Best Answer
The distribution of the mean and variance of a normal rv is very well known:
$$\sqrt n \left( \begin{array}{c} \overline X - \mu \\ {S^2} - {\sigma ^2} \end{array} \right) \sim \ \left(\begin{array}{c} \mathcal{N}(0,1) \\ \sigma^2\left(\frac{\sqrt{n}\chi^2_{n-1}}{n-1}-1\right) \end{array} \right)$$