How to prove that the binomial distribution is approximately close to the normal distribution when $np(1-p) \geq 10$ I would like a formal proof for this "rule of thumb." Can you assist me in getting to this solution? I require the insights and creativities of mathematicians. 
We know that if $np(1-p) \geq 10$ the binomial random variable $X$ is approximately normally distributed with mean $np$ and standard deviation $\sqrt{np(1-p)}$ 
 A: I'm a beginner so the following is something that helps my intuition.

By the CLT, if $Y_1, ..., Y_n$ are iid Bernoulli with parameter $p$ then
$$ \sqrt{n} \left( \frac{\bar{Y}_n - p}{\sqrt{pq}} \right) \xrightarrow[]{\text{in dist}} N(0,1) \qquad \text{as $n \rightarrow \infty$}$$
Where $\bar{Y}_n$ is the sample mean and $E[\bar{Y}_n] = \mu_{Y}=p$ and $Var(\bar{Y}_n)  = \frac{\sigma_Y^2}{n} = \frac{pq}{n}$
And then focusing on the LHS of the CLT,
$$ \sqrt{n} \left( \frac{\bar{Y}_n - p}{\sqrt{pq}} \right) =  \sqrt{n} \left( \frac{\frac{Y_1 + ... + Y_n}{n} - p}{\sqrt{pq}} \right)  = \sqrt{n} \left( \frac{Y_1 + ... + Y_n - np}{n\sqrt{pq}} \right) =  \frac{Y_1 + ... + Y_n - np}{\sqrt{npq}}$$
But $Y_1 + ... + Y_n \sim Bin(n,p)$ and so if we let $X = Y_1 + ... + Y_n$ then
$$\frac{X - np}{\sqrt{npq}} \xrightarrow[]{\text{in dist}} N(0,1) \qquad \text{as $n \rightarrow \infty$}$$
And so as $n \rightarrow \infty$,
$$\frac{X - np}{\sqrt{npq}} \approx Z$$
$$X \approx Z\sqrt{npq}+np \sim N(np, npq)$$
So as $n$ gets big, the Binomial RV,
$$X \dot{\sim} N(np, npq)$$
which shows the Normal approximation to the Binomial. I'm not sure if this constitutes a completely rigorous proof but I hope it helps your intuition.
A: I think you can check Curtis Mcmullen's notes on a detailed explanation of normal approximation. I read this about 2 years ago and do not recall all details. Here is the link:
http://www.math.harvard.edu/~ctm/home/text/class/harvard/154/11/html/home/course/course.pdf
I think the proof of DeMoivre–Laplace limit theorem is around page 50. 
A: The Berry–Esseen theorem gives an estimate for the normal approximation of the binomial distribution:
$$\sup_{x\in\mathbb R} \left|P\left(\frac{B(n,p)-np}{\sqrt{np(1-p)}} \le x\right)-\Phi(x)\right| \le \frac{C(p^2+q^2)}{\sqrt{npq}}$$
with $C < 0.4748$. So when $\sqrt{npq} \ge \sqrt{10}$:
$$\sup_{x\in\mathbb R} \left|P\left(\frac{B(n,p)-np}{\sqrt{np(1-p)}} \le x\right)-\Phi(x)\right| \le \frac{C(p^2+q^2)}{\sqrt{10}}$$
Because $p^2+q^2\le 1$:
$$\sup_{x\in\mathbb R} \left|P\left(\frac{B(n,p)-np}{\sqrt{np(1-p)}} \le x\right)-\Phi(x)\right| \le \frac{C}{\sqrt{10}} \le 0.1501$$
