# Standard deviation of sum of samples from Bernoulli/Binomial

This should be simple I hope. In lecture notes from a physics course, the professor writes:

$$\text{Prob(success)}=\dfrac{k}{N}$$. So query $$q$$ times, see some number $$\mathcal{l}$$ successes and estimate probability $$\dfrac{k}{N}=\dfrac{\mathcal{l}}{q}$$. i.e. estimate $$\tilde{k}$$ of $$k$$ is $$\tilde{k}=\dfrac{\mathcal{l}N}{q}$$

Everything up to this point makes sense to me. But...

Now by properties of binomial distribution (with probability $$\dfrac{k}{N}$$), the standard deviation of $$\mathcal{l}$$ is $$\sqrt{\dfrac{k(N-k)}{q}}$$.

From what I recollect, the standard deviation of a binomial distribution $$\text{bin}(n,p)$$ is $$\sqrt{npq}\equiv\sqrt{np(1-p)}$$. With respect to the lecture notes, if we let $$n\rightarrow q$$ and $$p\rightarrow \dfrac{k}{N}$$, then that gives us stddev $$=\sqrt{\dfrac{qk(N-k)}{N^2}}\neq\sqrt{\dfrac{k(N-k)}{q}}$$.

Where am I messing up in the calculation?

You are estimating the variance of the statistic, $$p$$ in this case (proportion), this is a Bernoulli trial, which has variance $$p(1-p)$$.
• Well this is for the standard deviation (or variance) of $\mathcal{l}$, not $p$ (where we are using $\dfrac{\mathcal{l}}{q}$ to estimate $p$). $\mathcal{l}$ is a regular sample from a binomial distribution and should have the regular variance $np(1-p)$. I do agree that $\dfrac{\mathcal{l}}{q}$ should have variance $p(1-p)=\dfrac{k(N-k)}{N^2}$, but that still does not match up with the above conclusion. – Dan Jan 18 '19 at 9:05
I realized that the professor had a typo, and put down the standard deviation for $$\tilde{k}$$ where he was instead writing about the standard deviation for $$l$$.