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 $2.57\sigma < \frac {1}{60}$ ?
$60\sqrt 5\cdot 2.57 < 6\sqrt n\\ 500\cdot 2.57 < n\\ n>1288$