Confidence interval of binomial distribution "Suppose that X is $Bin(n,p)$. We want to create a confidence interval $I_p$ for p. We do $n=500$ attempts and get $x=62$. Decide the two side interval when the confidence level is 95%."
What I've done: p is obviously $\frac{62}{500}$. We then know that binomial distribution can be approximated to $N(np, \sqrt{np(1-p)})$. Therefore: we have $N(62, 7.36967)$. I then want to use $2(\frac{s}{\sqrt{n}}\cdot\lambda_{\frac{\alpha}{2}})$ to get the entire span of the interval. When entering my numbers here I do however get the wrong answer so I'm doing something wrong here. I would greatly appreciate it if someone could point me in the right director and/or give some good reading material on this subject. My textbook is very poor. 
 A: Let $\mu = \mathbb{E}[X]$, so that $\mu = np = 500p$, where $p$ is our unknown probability for $X\sim B(500, p)$. Then (appoximately)
$$
X\sim N(\mu, 7.36967^2).
$$
From the critical values for the normal distribution, it follows that
\begin{align*}
0.95 &= \mathbb{P}\left(-1.96 \le \frac{62 - \mu}{7.36967}\le 1.96\right)
\\
&=\mathbb{P}(62 - 1.96\times7.36967\le\mu\le 1.96\times7.36967 + 62)
\\
&=\mathbb{P}(62 - 1.96\times7.36967\le 500p\le 1.96\times7.36967 + 62)
\\
&=\mathbb{P}\left(\frac{62 - 1.96\times7.36967}{500}\le p \le \frac{1.96\times7.36967 + 62}{500}\right).
\\
\end{align*}
The values that bound $p$ inside $\mathbb{P}(\cdot)$ are the boundaries of the 95% confidence interval for $p$. Hence the length of this interval is
$$
2\times\frac{1.96\times7.36967}{500} = 0.0578
$$
to 3 s.f.
I believe the formula you were using is for estimating the true mean $\mu$ using the sample mean $\bar{X}$ when we have i.i.d. random variables $X_1, X_2, \dots, X_n$, hence the presence of $\sqrt{n}$. Also note that I've used $X = 62$ as the observed value and not estimated $p$ via $62/500$, since this is the whole idea of using the confidence interval. Hope this helps.
