A level Central Limit Theorem question How many times must a fair die be rolled in order for there to be less than 1% chance that the mean of the scores differ from 3.5 by more than 0.1?
The answer is $n≥1936$, but how do you get to this answer?

Find mean and variance for one fair die. Let $X$ be the number of faces showing on one fair die.
$$\mu = E(X) = (1 + 2 + \cdots + 6)/6 = [6(7)/2]/6 = 7/2 = 3.5,$$
where we have used the formula that the sum of the first $n$ integers
is $n(n+1)/2.$ Also,
$$E(X^2) = (1^2 + 2^2 + \cdots + 6^2)/6 = \frac{6(7)(13)/6}{6} = 91/6,$$
where we have used the formula that the sum of the first $n$ squares
is $n(n+1)(2n+1)/6.$ Then
$$ \sigma^2 = V(X) = E(X^2) - \mu^2 = \frac{91}{6} - \frac{49}{4}
\frac{182 - 147}{12} = 35/12.$$
There are, of course, other methods of finding $\mu$ and $\sigma^2.$
Find the mean and variance for the average of $n$ dice. For the average $\bar X_n$ on $n$ rolls of a die, we have
$$\mu_n = E[(X_1 + X_2 + \dots + X_n)/n] = n\mu/n = \mu.$$ 
and, because we assume the $n$ rolls of the die are independent,
$$\sigma^2_n = V[(X_1 + X_2 + \dots + X_n)/n] = n\sigma^2/n^2 = \sigma^2/n.$$
Assume the average is normally distributed. If $n$ is even moderately large
the Central Limit Theorem indicates that $\bar X_n$ has approximately
the normal distribution with mean $\mu_n = 7/2$ and variance 
$\sigma_n^2 = \sigma^2/n = \frac{35}{12n}.$
Thus, upon standardizing (here, dividing by $\sigma_n$), we have
 $$P\{-0.1 < \bar X_n - 7/2 < 0.1\} \approx 
P\{-0.1\sqrt{12n/35} < Z < 0.1\sqrt{12n/35} \},$$
where $Z$ is standard normal.
Because we want this probability to equal $.01 = 1\%,$ we note that
$P\{-2.576 < Z < 2.576\} = 0.1.$
So in order to find the desired $n,$ we set $2.576 = 0.1\sqrt{12n/35},$
solve for $n$ and round up to the nearest integer. This gives the claimed answer $n = 1936$. 
