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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?

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    $\begingroup$ 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? $\endgroup$ – Stan Tendijck May 16 at 13:09
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    $\begingroup$ Please add details. Confidence interval for what? $\endgroup$ – StubbornAtom May 16 at 13:12
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    $\begingroup$ The difference between the standard deviation of the sum and the standard error of the mean, namely a factor of $n$ $\endgroup$ – Henry May 16 at 13:19
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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.

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  • $\begingroup$ Thank you! Could you also show me how in normal distributions we get $Var = pqn?$ $\endgroup$ – Asfangen May 16 at 13:32
  • $\begingroup$ @Asfangen The $x_i$ are iid so that $\mathsf{Var}(\sum_ix_i)=\sum_i\mathsf{Var}(x_i)=n\mathsf{Var}(x_1)=npq$ $\endgroup$ – drhab May 16 at 13:38
  • $\begingroup$ en.wikipedia.org/wiki/Bernoulli_distribution Also, this says the variance is supposed to be pq only $\endgroup$ – Asfangen May 16 at 13:42
  • $\begingroup$ @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). $\endgroup$ – user10354138 May 16 at 13:47

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