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

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?

• Can you provide more description, i.e., how did you end up with that formula and how did you find out about the other formula/what do you think it represents? – Stan Tendijck May 16 at 13:09
• Please add details. Confidence interval for what? – StubbornAtom May 16 at 13:12
• The difference between the standard deviation of the sum and the standard error of the mean, namely a factor of $n$ – Henry May 16 at 13:19

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.
• Thank you! Could you also show me how in normal distributions we get $Var = pqn?$ – Asfangen May 16 at 13:32
• @Asfangen The $x_i$ are iid so that $\mathsf{Var}(\sum_ix_i)=\sum_i\mathsf{Var}(x_i)=n\mathsf{Var}(x_1)=npq$ – drhab May 16 at 13:38
• @Asfangen you are confusing $\operatorname{Var}(\sum x_i)=npq$ with $\operatorname{Var}(x_i)=pq$. $\sum x_i$ is a Binomial(n,p), not Bernoulli(p). – user10354138 May 16 at 13:47