$N(p, p q/n)$-distributed Let $Y$ be a $N\left(p, \frac{p q}{n}\right)$-distributed random variable. We want to minimize 
$$ \delta = \delta(n)$$ 
so that the possibility that $p$ covers the interval
$$ J = [Y-\delta, Y+\delta] $$
(for every $p$) is at least $0.95$ ($95\%$).
which means $\frac{p q}{n}$ in the normal distribution $N\left(p, \frac{p q}{n} \right)$?
Does anyone have any advice how I could solve it?
 A: If $X$ has normal distribution with mean $\mu$ and variance $\sigma^2$, then with probability $0.95$,
$$\mu -1.96\sigma \le X \le \mu +1.96\sigma.  \qquad\qquad(\ast)$$
The $1.96$ is (approximately) the value $z$ such that with probability $0.025$, $Z > z$, where $Z$ has standard normal distribution.  Similarly, if we want to have probability of only $0.005$ in each tail, we would use $2.57$ instead of $1.96$. In the old days this kind of information was found in tables.  Now the tables are available online, and the information is also a built in feature  of spreadsheet and other programs.  
In our case, we are assuming that $Y$ is normal with mean $p$ and variance $\frac{pq}{n}$. Because of the standard notation that is used in the question, I expect that $p$ is a probability, that is, $0 \le p \le 1$, and $q=1-p$.
We want a $\delta$ that, as the post says, works for every $p$.  Note that $p(1-p)$ takes on its maximum value at $p=\frac{1}{2}$, so our variance is $\le \frac{1}{4n}$.  Thus from $(\ast)$ we can see that for any $p$, with probability at least $0.95$,
$$p-1.96\frac{1}{2\sqrt{n}} \le Y \le p+1.96\frac{1}{2\sqrt{n}}.$$ 
The above inequalities can be rewritten as 
$$Y-1.96\frac{1}{2\sqrt{n}}\le p \le Y+1.96\frac{1}{2\sqrt{n}}.$$
That means that in the notation of the problem
we can take 
$$\delta=1.96\frac{1}{2\sqrt{n}}.$$
Comment: The graph of $y=x(1-x)$ is flat at $x=1/2$.  What this means is that our variance, which is exactly $\frac{1}{4n}$ when $p=1/2$, does not shrink very much when $p$ is not far from $1/2$, like $p=0.4$.  Thus our estimate for the variance, which is exact when $p=1/2$, is not too far off the truth even when $p$ is some distance from $1/2$.
In the context in which the question arose, $Y$ is almost certainly a sample proportion, so does not have normal distribution.  But if $n$ is large, and $p$ is not too close to $0$ or $1$, the distribution of $Y$ is well-approximated by the normal. 
