In a wine competition each participant is required to look at $m$ glasses of (different) wine and choose which one is a Bordeaux. Each participant does so $Q=20$ times and collect 1 point for each correct answer, 0 otherwise. There are participant who do not know about wines, so their probability to get 1 point for each question is $1/m$.
A participant passes if she/he gets 50% of questions right (that is, 10 or more out of 20).
I need to choose $m$ such that the the probability that someone who does not know about wines passes is less or equal than 50% with a 95% confidence interval.
My attempt:
Let $p$ be the probability of success (that is $1/m$) for each question for someone who doesn't know wines.
Facts: the expected value for each of them on the $Q=20$ consecutive questions is $\mu=Qp$ and the variance is $\sigma^2=Qp(1-p)$ (a binomial distribution).
For $n$ participants I would know that their results' mean is normal with mean equal to $\mu$ and variance equal to $\sigma^2/n$, so I would build the 95% confidence interval as $[-2\sigma/\sqrt{n}, 2\sigma/\sqrt{n}]$.
Then I would solve the equation $2\sigma/\sqrt{n} \leq 0.5$ (the upper bound of the CI smaller than the maximum wanted probability) and compute $p$, and then $m$.
However, without the knowledge of $n$ I don't know how to build the standard error $\sigma/\sqrt{n}$ (from which the confidence interval).
$n = 20$ is the number of trials. So for a normal approximation solve for a maximum m the inequality $\frac{20}{m} + 1.96\cdot \sqrt{\frac{20}{m}(1-\frac{1}{m})}< 10$
I was just checking the equation. The probability interval for a normal approximation is
$p \pm z\sqrt{\frac{p(1-p)}{n}}$ where $p = \frac{1}{m}$
$\frac{1}{m} \pm 1.96\cdot \sqrt{\frac{\frac{1}{m}(1-\frac{1}{m})}{20}}$
Converting this to expected values:
$\frac{20}{m} \pm 1.96\cdot \sqrt{\frac{\frac{1}{m}(1-\frac{1}{m})}{20}}\cdot 20$
$= \frac{20}{m} \pm 1.96\cdot \sqrt{\frac{20}{m}(1-\frac{1}{m})}$