We want to know the prbabiltiy $p \in (0,1)$ that a biased die rolls a $6$. We roll the die $n$ independent times $X_i=\begin{cases} 1 & \textrm{if the i-th roll is a 6} \\ 0 & \textrm{else} \\ \end{cases} $ calculate with the clt the approximate min. number of rolls so that $p$ can be estimated with $\bar{X_n}=\frac{1}{n} \sum_{i=1}^n X_i$ and a uncertainty of $0,01$ with a probability of min. $0,9$.
Therefore I have to calculate $P(|\bar X_n-p|\le 0.01)\ge 0.9$
If I use $S=\sum X_i=n\bar X_n$ we have $P(|\frac{\sum X_i}{n}-p|\le 0.01)\ge 0.9$
But I don't know how to continue.
Hint: Using the central limit theorem means using $\ \mathcal{N}\left(\mu_n, \sigma_n^2\right)\ $ to approximate the distribution of $\ \overline{X}_n\ $, where $\ \mu_n\ $ is the mean of $\ \overline{X}_n\ $ and $\ \sigma_n^2\ $ is its variance. What are its mean and variance?
Reply to query from OP: Since \begin{align} P\left(\left|\,\overline{X}_n-p\,\right|\le 0.01\right)&=P\left(-0.01\le\overline{X}_n-p\le 0.01\right)\\ &=P\left(-\frac{0.01}{\sigma_n}\le\frac{\overline{X}_n-p}{\sigma_n}\le \frac{0.01}{\sigma_n}\right)\\ &\approx{\cal N}(0,1)\left(\frac{0.01}{\sigma_n}\right)-{\cal N}(0,1)\left(-\frac{0.01}{\sigma_n}\right)\\ &=2{\cal N}(0,1)\left(0.01\sqrt\frac{n}{p(1-p)}\right)-1\ , \end{align} and $\ {\cal N}(0,1)\ $ is a strictly increasing function, you need to find the (unique) $\ x\ $ such that $\ {\cal N}(0,1)\left(x\right)=\frac{1+0.9}{2}=0.95\ $. The value of $\ n\ $ you want is then the smallest integer such that $\ 0.01\sqrt\frac{n}{p(1-p)}\ge x\ $—that is, such that $\ n\ge 10000\,p(1-p)x^2\ $.