For large sample sizes (see discussion below), a confidence interval is constructed from the [[standard normal distribution]]. An approximate $100(1-\alpha)\%$ for the true proportion $p$ is given by $\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p (1 - \hat p)}{n}}$ where $z_{\alpha/2}$ is the [[critical value]] from the standard normal distribution. Note that $p(1-p)$ is the variance for the Bernoulli distribution. A proportion can be thought of as a [[Bernoulli distribution|Bernoulli random variable]] where each measure is classified as $1$ when it is part of the proportion you are calculating and $0$ otherwise. The confidence interval can then be thought of as a confidence interval for the mean of a Bernoulli random variable, whose distribution by the [[Central Limit Theorem]] will be [[normal distribution]] $N \sim (p, \frac{p(1-p)}{n})$. In the case of proportions, the sample size can be considered large enough to invoke the Central Limit Theorem when the following conditions hold. $\begin{align}\hat p - 3 \sqrt{\frac{\hat p (1 - \hat p)}{n}} \ge 0, && \hat p + 3 \sqrt{\frac{\hat p (1 - \hat p)}{n}} \le 1\end{align}$ $3$ is used here because $99.7\%$ of the probability of a standard normal distribution is within $3$ standard deviations of the mean.