If the probability of $1$ is $p$ what is the probability that a length $2n-1$ random $0/1$ vector has weight in range $[n,n+\sqrt{n\pi}]$?
For $p=\frac12$ is it (with $\delta\rightarrow0+$) given by lower bound $\sum_{i=n}^{n+\sqrt{n\pi}} \frac{\binom{2n-1}{i}}{2^{2n-1}}\gg \sqrt{n\pi} \frac{\binom{2n-1}{n+\sqrt{n\pi}}}{2^{2n-1}}= \sqrt{n\pi}\frac{\binom{2(n+\sqrt{\pi n})-1-2\sqrt{\pi n}}{n+\sqrt{\pi n}}}{2^{2n-1}}\approx \sqrt{n\pi}\frac{\frac{{2^{2n-1}}}{\sqrt{n\pi+\pi\sqrt{\pi n}}}}{2^{2n-1}}\approx\frac{1}{\sqrt{1+\sqrt{\frac\pi n}}}?$
If not what is a reasonable approximation that can be written down easily?
For $p=\frac12$ solution below solves.
What about $p\neq\frac12$?
I am interested in $p\geq\frac12+\frac1{2^n}$ and $p<\frac12+\frac1{2^n}$.
The weight $W$ (number of ones) follows a Binomial distribution with $\mu_W=(2n-1)(1/2+\delta)$ and $\sigma_W^2=(2n-1)(1/2-\delta)(1/2+\delta)$
As $n \to \infty$ this can be approximated by a gaussian $X$, with the above parameters. And the probability in question would (approximately) correspond to $P(n-\frac{1}{2} < X < n+ \sqrt {n \pi})$. In terms of $Z$ (normalized gaussian) the limits correpond to
$$Z_1 = \frac{(n-1/2)-\mu_W}{\sqrt{\sigma_W^2}}\approx -\delta \sqrt{8n}$$
$$Z_2 = \frac{(n+\sqrt {n \pi})-\mu_W}{\sqrt{\sigma_W^2}}\approx -\delta \sqrt{8n} + \sqrt{2 \pi}$$
Then, assuming $\delta \to 0$ fast enought (so that,eg, $\delta^2 n \to 0$) the probability area corresponds to $\sqrt{2 \pi} \approx 2.5$ (not far from the usual "three deviations" rule, so the probablity is approximately 0.5.