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I'm looking through my notes and something my lecture said seems off, just want to clarify.

"Let $Z_1,\ldots, Z_n$ be iid N(0,1) random variables and let $\overline{Z}$ be their average. Then $\overline{Z}$ is normal with mean 0 and deviation $\frac{1}{n}$. So

$$\frac{\overline{Z}-0}{\frac{1}{\sqrt{n}}}=\sqrt{n}\overline{Z}$$ is N(0,1). Is this true? would it not be that the variance of $\overline{Z}$ would be $\frac{1}{n}$?

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  • $\begingroup$ Variance of $\bar{Z}$ is $1/n$. $\endgroup$
    – StubbornAtom
    Apr 1, 2018 at 18:52
  • $\begingroup$ Thank you so mistake was "deviation $\frac{1}{n}$" but everything else is correct? $\endgroup$
    – user24907
    Apr 1, 2018 at 18:57
  • $\begingroup$ Unless 'deviation' is another way of saying variance there is no mistake. $\endgroup$
    – StubbornAtom
    Apr 1, 2018 at 18:59
  • $\begingroup$ but he says deviation is $\frac{1}{n}$ but you say variance is $\frac{1}{n}.$ So he did make a mistake and wrote deviation when he meant to write variance I think. $\endgroup$
    – user24907
    Apr 1, 2018 at 19:04

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You probably have found an error in your notes. Most likely, your lecturer meant to say “and variance $\frac{1}{n}$” or “and standard deviation $\frac{1}{\sqrt{n}}$.”

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