What is the distribution of the z-score of the minimum of 3 normally distributed variables? Let $x_1, x_2, x_3$ come from a normal distribution with mean 0 and standard deviation 1.
Let $m=\frac{x_1+x_2+x_3}{3}$ and $s=\sqrt{\frac{(x_1-m)^2+(x_2-m)^2+(x_3-m)^2}{3}}$.
What is the distribution of $\frac{\textrm{min}(x_1,x_2,x_3)-m}{s}$?
I am also interested in the distribution of the z-score of the $k$th lowest term of $n$ normally distributed variables.
When $n=0$ or $1$ it is trivial, but I can't seem to make too much progress for larger $n$.
 A: To the credit of the OP's question, this appears to be a novel and challenging problem.

The $n = 2$ case
Let $X_1$ and $X_2$ denote the 2 ordered variables, such that $X_1 < X_2$.
Find the pdf of $Z =  \frac{X_1-m}{s}$
After some algebra:
$$Z = \frac{X_1-m}{s} = \frac{X_1-X_2}{\sqrt{\left(X_1-X_2\right){}^2}} = -1$$
i.e. in the $n=2$ case, $Z =-1$ with probability 1.

The $n=3$ case
Things rapidly get more complicated.
Let $\{X_1,X_2,X_3\}$ denote the 3 ordered variables, such that $X_1 < X_2 <X_3$.
Then, the joint pdf of the first 3 order statistics from a standard Normal parent is:
$$g(x_1,x_2,x_3) =\frac{3 e^{\frac{1}{2} \left(-x_1^2-x_2^2-x_3^2\right)}}{\sqrt{2} \pi ^{3/2}}  \quad \text{for} \quad x_1<x_2<x_3$$
The random variable of interest is (again after some algebra):
$$Z \quad = \quad \frac{X_1-m}{s} \quad = \quad \frac{2 X_1-X_2-X_3}{\sqrt{2} \sqrt{X_1^2+X_2^2+X_3^2-X_1 X_2 - X_2 X_3 - X_1 X_3}}$$
The problem is now to find the pdf of $Z$, when $(X_1,X_2,X_3)$ have joint pdf $g(x_1,x_2,x_3)$. This does not seem trivial.
Nevertheless, to get a flavour, I decided to do a Monte Carlo simulation of the pdf of $Z$ which may be of interest in itself, and perhaps also provide a useful check for any symbolic answer.
Monte Carlo simulation of the pdf of $Z = \frac{\textrm{min}(x_1,x_2,x_3)-m}{s}$ (here with 1 million samples):

By considering the limit of $Z$ as $X_1$, $X_2$ and $X_3$ tend to $\infty$ or $-\infty$, one can further show that the domain of support is $(-\sqrt{2}, -\frac{1}{\sqrt{2}})$.
