We can express the law of large numbers as follows:
\begin{align*}
\lim_{n \rightarrow \infty} \frac{x}{n} = p
\end{align*}
where x is the number of events, n is the number of trials, and p is the probability of the event.
In the situation n \rightarrow \infty , we can consider
\begin{align*}
\mathrm{Bin}(X=x | n, p) &\approx \mathcal{N}(X=x|\mu, \sigma^2) \\
\mu &\approx np \\
\sigma^2 &\approx npq
\end{align*}
where q = 1 - p.
Let \bar{X} = \frac{X}{n} , then n\bar{X} = X
\begin{align*}
\int f_{\bar{X}}(\bar{x}) d\bar{x} &= \int f_{X}(x) dx \\
&= \int f_{X}(n\bar{x}) n d\bar{x} \\
\end{align*}
therefore,
\begin{align*}
f_{\bar{X}}(\bar{x}) &= n f_{X}(n\bar{x}) \\
&= n \mathcal{N}(X = n\bar{x}|\mu, \sigma^2) \\
&= n \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left\{ -\frac{(n \bar{x} - \mu)^2}{2 \sigma^2} \right\} \\
&= n \frac{1}{\sqrt{2 \pi npq}} \exp \left\{ -\frac{(n \bar{x} - np)^2}{2 npq} \right\} \\
&= \frac{1}{\sqrt{2 \pi} \sqrt{\frac{pq}{n}}} \exp \left\{ -\frac{(\bar{x} - p)^2}{2 \frac{pq}{n}} \right\} \\
&= \mathcal{N}(\bar{X}=\bar{x}|\mu_{\bar{X}} = p, \sigma_{\bar{X}}^2 = \frac{pq}{n})
\end{align*}
when n \rightarrow \infty, \sigma_{\bar{X}} \rightarrow 0 and \bar{x} = \frac{x}{n} \rightarrow p.
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