Question regarding the derivation of the distribution of $(n-1)S^2/(\sigma^2)$ I will quote from my statistics manual:
"Consider a random sample $x_1,x_2...x_n$ taken from a population with distribution $N(\mu,\sigma^2)$, whose average $\mu$ is unknown; [through the central distribution theorem] it is then known that $\sum(x_i-\overline X )^2/\sigma^2$ is a [random variable with distribution] chi-squared with $(n-1)$ degrees of freedom."
($\overline X$ is the sample average.)
Does this imply, by the definition of a chi-squared distribution, that $x_i$ has distribution $N(\overline X,\sigma^2)$ ?
Because the only distribution I can see $x_i$ having is the original population distribution, $N(\mu,\sigma^2)$.
My confusion mostly comes from the fact that $\mu$ is a number, while $\overline X$ is a random variable itself. How can a random variable be centered around a random variable? 
(The title is there because this is then used to derive the formula shown.)
 A: If $\mu$ were known, then 
$$\frac{1}{\sigma^2}\sum_{i=1}^n (X_i - \mu)^2
= \sum_{i=1}^n \left(\frac{X_i - \mu}{\sigma}\right)^2 = \sum_{i=1}^n Z_i^2
\sim Chisq(df=n),$$
where $Z_i$ are iid standard normal.
In your case, $\mu$ is not known, and it needs to be estimated by $\bar X.$ The random variable
$$\frac{1}{\sigma^2}\sum_{i=1}^n (X_i - \bar X)^2 \sim Chisq(df = n-1),$$
but this does not imply that $$\left(\frac{X_i - \bar X}{\sigma}\right)^2$$ 
is standard normal. There do exist $n-1$ iid standard normal
random variables $Z_i^\prime$ such that
$\sum_{i=1}^{n-1} (Z_i^\prime)^2 = 
\frac{1}{\sigma^2}\sum_{i=1}^n (X_i - \bar X)^2,$ but it takes
a bit of effort to show that this is true. (One proof uses
a transformation with matrix algebra; another uses moment
generating functions.) Informally, one says "a degree of
freedom is lost in estimating $\mu.$"
You have not given any information on the level of your course
or text, so I don't know how to explain the details as an
appropriate mathematical level. If you want to pursue this further,
you might start with just $n=2$ independent normal random variables $X_1$ and $X_2,$ each with mean $\mu$ and variance $\sigma^2.\,$
It is not difficult to show that $Y_1 = X_1 + X_2$ and $Y_2 = X_1 - X_2$
are independent normal random variables. Then the first of these is
proportional to $\bar X:\;$$Y_1/2 = \bar X.$ The second is proportional to
the sample variance $S^2$ and, multiplying by an appropriate constant,
$(aY_2)^2 \sim Chisq(df = 1).$ 
Note: Notice that for $n = 2,$ this implies (stochastic) independence of $\bar X$ and sample variance $S^2.$ For normal data only,
this holds for any $n.$ Independence cannot be functional
independence because $\bar X$ appears in the definition of $S^2$.
(This note is my way of pointing out that you have asked about something
that is nontrivial and nonintuitive. You should congratulate
yourself for wondering about issues raised in your Question.)
