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The version of the CLT in my book states that if $X_1,...,X_n$ is a random sample, with mean $\mu$ and standard deviation $\sigma^2$, then $W=\frac{\bar X-\mu}{\sigma/\sqrt n}$~$N(0,1)$ as $n\rightarrow\infty$. I also hear this phrased as "the sample mean is approximately normally distributed for large $n.$" I'm having trouble justifying this second characterization (which I guess is just a corollary of the CLT). I know that $\mu_{\bar X}=\mu$ and $\sigma^2_{\bar X}$, and I know that if $X$~$N(\mu,\sigma^2)$, then $Z=\frac{X-\mu}{\sigma}$~$N(0,1).$ It seems like the second characterization is using the converse of this result. Is the converse also true? That is, if $Z$~$N(0,1)$, does that imply $\sigma Z+\mu$~$N(\mu,\sigma^2)?$

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    $\begingroup$ For your final question, yes. If $Z$~$N(0,1)$ then $\sigma Z+\mu$~$N(\mu,\sigma^2)$. It is just a change in location and scale parameters $\endgroup$ – Henry Sep 18 '14 at 8:02
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When $n$ is large $$ W=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1) $$ approximately, or equivalently, $$ \bar{X}= \mu+\frac{\sigma}{\sqrt{n}}W\sim N(\mu,\sigma^2/n) $$ approximately. That is, the sample mean $\bar{X}$ is approximately normal distributed with mean $\mu$ and variance $\sigma^2/n$ when $n$ is large.

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