Consider ${\displaystyle I_1,\cdots ,I_n\sim \mathrm{Bernoulli}(p)}$, and ${\displaystyle S_n\sim \mathrm{Binomial}(n,p)}$, so ${\displaystyle S_n\stackrel{\mathrm{d}}{=}{I}_1+\cdots +I_n}$. We know $$\mathbb{P}(S_n=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}=\frac{n!}{k!(n-k)!}{p}^k(1-p{)}^{n-k},0\le k\le n.$$
We want to estimate the Binomial distribution, because factorials are troublesome to work with. The key tool is to use Stirling Approximation $${\mathrm{e}}^{\frac{1}{12n+1}}<\frac{n!}{{\left(\frac{n}{\mathrm{e}}\right)}^n\sqrt{2\pi n}}<{\mathrm{e}}^{\frac{1}{12n}}.$$

1 Entropy Approximation

We can use entropy to approximate Binomial distribution. To be more specific, KL divergence.

2 Normal Approximation

Taylor expansion of ${\displaystyle \mathrm{KL}(f||p)}$ (see as a function about ${\displaystyle f}$) about ${\displaystyle f=p}$ gives $$\mathrm{KL}(f||p)=\frac{(f-p{)}^2}{2p(1-p)}+\frac{2g-1}{6g^2(1-g{)}^2}(f-p{)}^3$$ for some ${\displaystyle g}$ between ${\displaystyle f,p}$.

To obtain a normal approximation, we need the remainder ${\displaystyle R}$ quantity to be small. Notice that $$R=|\frac{n(2g-1)}{6g^2(1-g{)}^2}(f-p{)}^3|\le \frac{n|f-p{|}^3}{6[min(f,p)min(1-f,1-p){]}^2}.$$

• If ${\displaystyle p}$ is away from both ${\displaystyle 0}$ and ${\displaystyle 1}$, and ${\displaystyle n|f-p{|}^3}$ is small, then this upper bound will be small.
• For ${\displaystyle n|f-p{|}^3}$ to be small, we need ${\displaystyle |f-p|}$ to decay faster than ${\displaystyle n^{-\frac{1}{3}}}$ as ${\displaystyle n\to \mathrm{\infty }}$.
• Since ${\displaystyle n|f-p{|}^3=O\left(\frac{1}{\sqrt{n}}\right)\to 0}$, the normal approximation for ${\displaystyle \mathbb{P}(S_n=k)}$ is accurate for large ${\displaystyle n}$.

Good news is that when ${\displaystyle I_1,\cdots ,I_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Bernoulli}(p)}$, by SLLN, ${\displaystyle \frac{S_n}{n}\stackrel{\mathrm{d}}{=}\frac{1}{n}(I_1+\cdots +I_n)\stackrel{\mathrm{a}.\mathrm{s}.}{\to }p}$.
So normal approximation is that $$\mathbb{P}(S_n=k)\sim \frac{1}{\sqrt{2\pi np(1-p)}}{\mathrm{e}}^{-\frac{(k-np{)}^2}{2np(1-p)}}.$$
This is accurate if ${\displaystyle \frac{n|f-p{|}^3}{6[min(f,p)min(1-f,1-p){]}^2}\ll 1}$.
However, in general, normal approximation is less accurate than the entropy approximation.

We can also use CLT:

2.1 Application of CLT for ${\displaystyle {\displaystyle \mathrm{Binomial}(n,p)}}$

Let ${\displaystyle p}$ be the proportion of the population supporting Trump.
${\displaystyle n}$ be the number of people polled u.a.r from the population.
${\displaystyle S_n}$ be the number in the sample who support Trump.
${\displaystyle {\hat{p}}_n=\frac{S_n}{n}}$ be the estimator of ${\displaystyle p}$.
How large should ${\displaystyle n}$ be, s.t. $$\mathbb{P}(p\in [{\hat{p}}_n-\epsilon ,{\hat{p}}_n+\epsilon ])\ge 1-\alpha$$ for given ${\displaystyle \epsilon >0}$ and ${\displaystyle 0<\alpha <1}$?

This is the #ConvidenceLevel . For each trial, ${\displaystyle \hat{p}}$ moves, and this moving interval should cover ${\displaystyle p}$ at least ${\displaystyle 100(1-\alpha )\mathrm{\%}}$ of the time.

Denote $$I_j=\{\begin{array}{rl}& 1,j\text{-th person polled supported Trump},\\ & 0,\text{otherwise}.\end{array}$$
Then ${\displaystyle I_1,\cdots ,I_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Bernoulli}(p),S_n=I_1+\cdots +I_n\sim \mathrm{Binomial}(n,p)}$. By CLT, $$\sqrt{n}\frac{{\hat{p}}_n-p}{\sqrt{p(1-p)}}\stackrel{\mathrm{d}}{\to }Z\sim \mathcal{N}(0,1).$$ so $$\begin{array}{rl}\mathbb{P}({\hat{p}}_n-\epsilon \le p\le {\hat{p}}_n+\epsilon )& =\mathbb{P}(-\epsilon \le {\hat{p}}_n-p\le \epsilon )\\ & =\mathbb{P}(\frac{-\epsilon \sqrt{n}}{\sqrt{p(1-p)}}\le \sqrt{n}\frac{{\hat{p}}_n-p}{\sqrt{p(1-p)}})\\ & \approx \mathrm{\Phi }(z)-\mathrm{\Phi }(-z)=1-2\mathrm{\Phi }(-z),\end{array}$$
so

Since ${\displaystyle p(1-p)\le \frac{1}{4},\mathrm{\forall }p\in [0,1]}$, this yields $$n\ge \frac{1}{4}[-\frac{{\mathrm{\Phi }}^{-1}\left(\alpha /2\right)}{\epsilon }].$$

If ${\displaystyle \epsilon =0.05,\alpha =0.05}$, we have ${\displaystyle n\ge 385}$.
If ${\displaystyle \epsilon =0.01,\alpha =0.05}$, we have ${\displaystyle n\ge 9604}$.