Estimate of the speed of convergence to normal of the average of binomially distributioned variables

Question

Let us fix $n$, and consider a sequence of identical, independent random variables with binomial distributions with $n$ trials and probability of success $p=1/2$. Are there estimate on how fast the average of the sequence will converge to a normal distribution? (I mean, more accurate than application of the standard results.)

Answer 1

If $X_i$ are $\text{Bin}(n, \frac12)$ then each has mean $\frac{n}{2}$ and variance $\frac{n}{4}$

If you take $m$ i.i.d. samples of this binomial random variable then their sum $m\overline{X}$ is binomially distributed with parameters $mn$ and $\frac12$, with the consequence that $\overline{X}$ has mean $\frac{n}{2}$ and variance $\frac{n}{4m}$.

Consider the error in using a normal approximation to find $\Pr(\overline{X}\lt \frac{n}{2})$ or $\Pr(\overline{X}\le \frac{n}{2})$. The approximation will suggest $\frac12$ for each, but if $nm$ is even then each will be wrong by $\frac12{nm \choose nm/2}2^{-nm}$, which by Stirling's approximation to the factorial, this error will be about $\frac{1}{\sqrt{2\pi nm}}$

For example if $n=10$ and $m=100$, then the two probabilities are actually about $0.4873875$ and $0.5126125$ while $\frac{1}{\sqrt{2\pi nm}} \approx 0.0126157$

The error being proportionate to $\frac{1}{\sqrt{m}}$ is consistent with the Berry–Esseen theorem, showing that you do not get faster convergence