Conditions needed to approximate a Binomial distribution using a Normal distribution? Okay. So, many sources state different conditions for approximating binomial using normal. Variations I've seen are as follows. Assuming that we have Bin(n,p):
1. np and n(1-p) are both large (say, 20 or bigger)
2. large n, p close to 0.5
3. large n, np>9(1-p)
My questions:
1. Are all 3 of these valid? Justify your answer.
2. Is there a set of conditions which can summarise all 3 given above (if they are valid)?
Note: Before any of you say "First, what is your opinion about this?", I personally don't know much about these distributions since I'm studying stats at a basic level. I know quite a few distributions and a bit about pdf's, cdf's, mgf's but not really that much more
 A: The usual conditions are that $n\geq 50$ and both $np$ and $n(1-p)>5$.
Note that if $p$ is close to $0.5$ your third condition is inconsistent with your first condition.
A: The Berry-Esseen theorem says that the uniform CDF error for approximating a sum of $n$ iid variables with 3 moments is bounded by $\frac{C \rho}{\sigma^2 \sqrt{n}}$, where $\rho=E[|X_i-m|^3]$ and $C$ is a universal constant. We do not exactly know the best possible value of $C$ but we do know that $0.4<C<0.5$. For the binomial distribution this gives a bound of 
$$\frac{1}{2} \frac{p^3(1-p)+p(1-p)^3}{p^{3/2} (1-p)^{3/2} n^{1/2}}.$$ 
This bound is not tight, as can be checked by some direct calculation, but it is not bad (it is more pessimistic than the true error by a factor of maybe 4 or so). Cancelling as is possible, you get a bound of 
$$\frac{1}{2} \frac{p^2+q^2}{(npq)^{1/2}}$$
where $q=1-p$. The numerator there is between $1$ and $1/2$, so we can intuitively understand this bound
$$\frac{C}{(npq)^{1/2}}$$
where what I have said so far tells us only that $C$ can be chosen to be smaller than $1/2$, though in actuality for the binomial distribution it can be chosen to be a fair bit smaller than that. 
So if, say, you want a uniform error of at most $0.01$ then it will be enough (by the above analysis) to have $n>\frac{10000}{16 pq}$. In actuality the cutoff is probably more like $n>\frac{100}{pq}$.
