What is the min number of times a fair die should be rolled to be at least 90% certain the fraction of fives rolled is within .005 of true probability 
What is the min number of times a fair die should be rolled to be at least $90\%$ certain the fraction of fives rolled is within $.005$ of true probability

So I let the random variable $X =$ percent of number of fives rolled, and then I attempted to solve $P(.1627<X<.1717)$. But then I thought I should set that equal to $.9$ because that is the certainty that we're looking for. I am just stumped on how to solve for $n$. Any help would be great.
 A: You can use the normal distribution to solve this:
Let $X$ be a random variable that represents rolling five and let us define it:
$$ X = \begin{cases} 1, & w.p \ \frac{1}{6} \\
0, & w.p \ \frac{5}{6} \end{cases}$$
The expectation for rolling 5 is $\sim 0.1666$, since you are looking for an error margin of $0.005$, you want the next inequality to hold:
$$P(0.161 < \bar{X} < 0.171) \ge 0.9$$
where $\bar{X}$ is the average of rolling a dice.
Now, let us translate it to the normal distribution:
$$P(\frac{-\frac{1}{200}}{\sqrt{\frac{5}{36n}}}< \frac{\bar{X} - \mathbb E[X]}{\sigma\frac{1}{\sqrt n}}< \frac{\frac{1}{200}}{\sqrt{\frac{5}{36n}}}) = P(\frac{\bar{X} - \mathbb E[X]}{\sigma\frac{1}{\sqrt n}}< \frac{\frac{1}{200}}{\sqrt{\frac{5}{36n}}}) - P(\frac{-\frac{1}{200}}{\sqrt{\frac{5}{36n}}}< \frac{\bar{X} - \mathbb E[X]}{\sigma\frac{1}{\sqrt n}}) \ge 0.9$$
Now, you can look at the normal distribution table, and look for a $b\in \mathbb R$ for which $\Phi(b) - \Phi(-b) \ge 0.9$
Turns out this $b$ eqauls $\sim 1.65$.
All you are left to solve is this equation:
$$\frac{\frac{1}{200}}{\sqrt{\frac{5}{36n}}} = 1.65$$
Solving it will result in $n=15125$ which is your final answer.
