Distribution of the coefficient of variation of independent normal variables? Let $X_1,\dots,X_n$ be Gaussian random variables with mean $\mu$ and variance $\sigma^2$. What is known about the distribution of
$$\frac{\sqrt{\frac1{n-1}\sum_{i=1}^n(X_i-\bar X)^2}}{\bar X} $$
with $\bar X = \frac 1n \sum_{i=1}^n X_i$?
Does this distribution have a name? I am especially interested in its cumulative distribution function.
If this is too complicated, I'd settle for 
$$\frac{\sqrt{\frac1{n}\sum_{i=1}^n(X_i-\mu)^2}}{\mu} $$
 A: This is a bit long to be a comment and hence posting this as an answer. Look up Fisher distribution for the first one and chi-squared distribution for the second one though both the random variables don't exactly fit in the distributions. (For the first one, you need to subtract of the mean from the denominator and square it and rescale it to be a Fisher distribution while in the second case, you need to square it and rescale it to be chi-squared distribution).
A: The second one, as Sivaram indicated, is straightforward. First, however, note that it is not defined for $\mu = 0$. 
Assume that $\mu > 0$ (the case $\mu < 0$ being analogous). Then, for any $x \geq 0$,$$
{\rm P}\bigg[\frac{{\sqrt {n^{ - 1} \sum\nolimits_{i = 1}^n {(X_i  - \mu )^2 } } }}{\mu } \le x \bigg] = {\rm P}\bigg[\sum\limits_{i = 1}^n {\bigg(\frac{{X_i  - \mu }}{\sigma }\bigg)^2 }  \le \frac{{n\mu ^2 x^2 }}{{\sigma ^2 }}\bigg].
$$
Now, $\frac{{X_i  - \mu }}{\sigma }$ are i.i.d. ${\rm N}(0,1)$, hence their sum of squares is chi-squared with $n$ degrees of freedom. So just substitute $\frac{{n\mu ^2 x^2 }}{{\sigma ^2 }}$ in the cdf of the latter (which is given by $
\frac{1}{{\Gamma (n/2)}}\gamma (n/2,x/2)$).
A: For the first one, if you consider instead of $S/\bar{X}$:
$$\frac{\bar{X}-\mu}{S/\sqrt{n}}$$
then, this is a Student t-variable with $n-1$ degrees of freedom.
If you don't want the $\mu$, you can use the non-central t-distribution.
