# Normal approximation of the Binomial distribution

So my text books says that if $$X \sim B(n,p)$$ with $$np>5$$ and $$nq>5$$ (where $$q=1-p$$) then $$X$$ can be approximated by a normal distribution of $$X\sim N(\mu,\sigma^2)$$ with $$\mu = E(X) = np$$ and $$\sigma^2 = Var(X) = npq$$

So I understand that if n is very large then $$X$$ will roughly show a normal distribution and that $$X$$ will have $$E(X) = np$$ and $$Var(X) = npq$$ but why must $$E(X)>5$$ and $$nq>5$$. and furthermore, what is $$nq$$ representing?

The CLT says the normal approximation is good for a fixed distribution when $$n$$ is large enough. But when you have another parameter to play with, tweaking that other parameter can slow down the convergence rate (meaning that $$n$$ must get larger to achieve a given error tolerance). In the case of the binomial distribution, there is a sort of complete classification:
• When $$n,np$$ and $$nq$$ are all large, Bin($$n,p$$) behaves like N($$np,npq$$).
• When $$n$$ is large but $$np$$ is not large, Bin($$n,p$$) behaves like Poisson($$np$$) and the normal approximation has a large error.
• When $$n$$ is large but $$nq$$ is not large, Bin($$n,p$$) behaves like $$n-$$ Poisson($$np$$) and again the normal approximation has a large error. (This is really the same statement as the one before it, because if $$X \sim$$ Bin($$n,p$$) then $$n-X \sim$$ Bin($$n,q$$)).
One way to anticipate this might happen in advance is to use a quantitative refinement of the CLT such as the Berry-Esseen theorem. The Berry-Esseen theorem for the binomial distribution gives an estimate for the difference of the CDFs as $$C \frac{1}{\sqrt{n}} \frac{pq^3+qp^3}{(pq)^{3/2}}$$ where $$0.4 is a constant. The important thing is that ratio involving $$p$$ and $$q$$, which behaves as $$p^{-1/2}$$ as $$p \to 0$$ and as $$q^{-1/2}$$ as $$q \to 0$$. Thus the Berry-Esseen theorem roughly speaking bounds the error by $$\frac{C'}{\sqrt{n \min \{ p,q \}}}$$ where $$C'$$ is a new constant.
Intuitively what the Berry-Esseen theorem is capturing is that the normal approximation to a distribution is symmetric about its mean, whereas the original distribution in general is not. Thus if a distribution (with the standard deviation scaled out) is highly skewed, then $$n$$ must become quite large in order to mitigate the effect of this skew.
Seems like if $$q=1–p$$, then $$nq=(n)(1–p)$$ is a scaled estimate of the variance of the binomial distribution.