Why is the standard deviation described as $\sqrt{pqn}$ sometimes and sometimes as $\sqrt{\frac{pq}{n}}$?

Question

I assume it has something to do with whether we start with a distribution or with samples, but why is the standard deviation increasing with $n$ in one case and decreasing in the other?

Answer 1

Be careful! $\bar{x}=\frac{1}{n}\sum x_i$ has a $1/n$.

For $x_i\sim\text{Bernoulli}(p)$, $\mathbb{E}\bar{x}=p$, and $\operatorname{Var}(\bar{x})=n^{-2}\operatorname{Var}(\sum x_i)=n^{-2}\cdot npq=\frac{pq}n$ if $x_i$s are independent.