Calculating sample size using Central Limit Theorem How many times do we have to throw a dice to achieve the situation, that the proportion of sixes to all numbers is between $\frac{9}{60}$ and $\frac{11}{60}$ with the probability $\leq\frac{1}{100}$?
This is my solution, that gives a wrong answer.
I know, that number of sixes in multiple dice throws is following the binomic distribution ($n$ is the sample size I need to calculate):
$$
\mu = n \cdot\frac{1}{6}\\
\sigma = \sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}
$$
Based on that, I can normalize the random variable $X$ (expressing number of sixes in $n$ throws):
$$
Z=\frac{\overline{X}-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
$$
Now I can use the Central Limit Theorem:
$$
P\left(
\frac{n\cdot\frac{1}{6}-\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}<
Z<
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
\right)=0.99$$
$$
P\left(Z<
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
\right)=\frac{0.99}{2}=0.495
$$
I can look up this value in the standardized normal distribution table and use it to calculate the value of $n$.
$$
\Phi(-0.012)=0.495
$$
$$
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
=-0.012\\
\frac{n\cdot\frac{1}{6}\cdot\frac{1}{60}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
=-0.012\\
n=0.000595
$$
And that is what seems suspicious. I don't think that i have to throw the dice $0.000595$ times to ensure that the ration of sixes is within a $\frac{1}{60}$ error margin from the mean value with the probability of $\frac{1}{100}$.
Can somebody help me to determine where I am wrong? (Sorry about my English.)
Edit based on Doug M's comment:
I made a mistake in the left tailed distribution, It should be:
$$
P\left(Z<
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
\right)=0.995
$$
With this corrected the result for $n$ seems more reasonable:
$$
\Phi(2.58)=0.995
$$
$$
\frac{n\cdot\frac{1}{6}\cdot{1}{60}}{\sqrt{n\cdot\frac{5}{36}}{}}=2.58\\
n=119815,2
$$
Which means, the optimal sample size is $119816$, which is more than I expected but sounds more reasonable than my previous result.

for a single die
$\mu = \frac 16\\
\sigma^2 = \frac {5}{36}$
the expected number of sixes in $n$ throws
$\mu = \frac 16n\\
\sigma^2 = \frac 5{36}n\\
\sigma = \sqrt{\frac{5n}{36}}$
The average number of sixes in $n$ throws
$\mu = \frac 16\\
\sigma = \frac 1n \sqrt{\frac{5n}{36}}\\
\sigma = \sqrt{\frac{5}{36n}}$
The expected number of sixes is governed by a binomial distribution.  But, for a large number of throws the binomal resembles the normal distribution.
If $99\%$ are inside of a two tailed distribution, $99.5\%$ are to the left of a 1 tailed distribution.
+normsinv(0.995) = 2.575
For how large does $n$ have to be such that $2.57\sigma < \frac {1}{60}$ ?
$60\sqrt 5\cdot 2.57 < 6\sqrt n\\
500\cdot 2.57 < n\\
n>1288$
