I found this definition of central limit theorem in the book: Intro to probability and statistics using R:
I thought the sample mean $\bar X$ itself followed the normal distribution and not the calculated quantity as shown.
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6$\begingroup$ (1) $\bar X$ never follows a normal distribution unless the population distribution is normal. (2) The limiting distribution of $\bar X$ is a constant (the value $\mu$). $\endgroup$– whuber ♦Commented Sep 29, 2015 at 23:40
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2 Answers
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Well the sample mean $\bar{X}$ approaches a normal distribution with mean $\mu$ and variance $\frac{\sigma^2}{n}$, i.e. $\bar{X}\sim N(\mu,\frac{\sigma^2}{n}) $ however if you standardize the sample mean (by writing it in the form in the question), you find that approaches a Normal distribution with mean $0$ and variance $1$, i.e. $Z \sim N(0,1)$.
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1$\begingroup$ Welcome to CV! I suggest you replace "follows a normal distribution" with "approaches a normal distribution", since the CLT does not claim that the distribution is exact. $\endgroup$ Commented Sep 29, 2015 at 21:24
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1$\begingroup$ Thanks for that. (+1) Hope you continue contributing to our site. $\endgroup$ Commented Sep 29, 2015 at 21:29
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2$\begingroup$ In the context of "approaches", "$n$" is meaningless. $\endgroup$– whuber ♦Commented Sep 29, 2015 at 23:41
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2$\begingroup$ This answer is problematic, e.g. it's not true that $\bar X\sim\mathcal{N}\left(\mu,\frac{\sigma^2}{n}\right)$. The distribution of $\bar X$ looks kinda like normal when $n$ is large. $\endgroup$– AksakalCommented Sep 30, 2015 at 1:16
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It's a very common mistake - which I make sometimes myself - to have the non-constant on the right hand side of the arrow: $$f_n\to \mathcal{N}\left(\mu,\frac{\sigma^2}{n}\right)$$, where $f_n$ is the probability distribution of $\bar X$.
Since your right hand side is changing, it's difficult to prove statements about this convergence relation, and I mean impossible by difficult.
Hence, we want to have the constant right hand side to apply our convergence proving tricks, e.g. $\mathcal{N}(0,1)$ would be a non-changing distribution. So, we formulate the convergence statement using it: $$Z=\frac{\bar X-\mu}{\sigma/\sqrt n}\sim\tilde f_n$$, where $\tilde f_n$ - sequence of distributions (functions), which converges to the standard normal: $$\tilde f_n\to\mathcal{N}(0,1)$$
This one we can handle: the right hand side is not changing, only the left hand side is.
It's still tempting to think about the distribution of $\bar X$ as one that looks more and more like normal, i.e. taking a shape of the normal distribution. However I wouldn't use the term converges, or even approaches, as they are loaded in mathematics. You'll run into people who will be confused with this language for they will expect to see the non changing right hand side.
