I'm so used to seeing standard deviations being the sqrt of the entire formula so it's weird to me to see a SD that is defined as sd/sqrt(n). Can someone explain why?
2026-03-25 03:23:00.1774408980
Why is sampling distribution of means's standard deviation sd/sqrt(n)?
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It is the standard error of the mean which is $\sigma/\sqrt{n}$
If a sample of $n$ i.i.d. random variables each have mean $\mu$ and variance $\sigma^2$
then their sum has mean $n\mu$ and variance $n\sigma^2$ since they are independent
so their mean has mean $\dfrac{n \mu}{n}=\mu$ and variance $\dfrac{n\sigma^2}{n^2}=\dfrac{\sigma^2}{n}$ since you are dividing by a constant
leading to their mean having standard deviation $\sqrt{\dfrac{\sigma^2}{n}} = \dfrac{\sigma}{\sqrt{n}}$ by taking the square root of the variance