A level Central Limit Theorem question

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?

Answer 1

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 $n = 1936$.