I'm doing a data analysis where the data is right-skewed, similar to a $\chi^2$-distribution. The data is then normalised to their z-scores.
$z = \frac{X-\mu}{\sigma}$
I want to do an outlier analysis of this scaled data, by looking at their quantile probabilities. So the question is, If I have a $\chi^2$-distribution of any $n$ degree of freedom, normalised. What would be the resulting distribution?
$\frac{\chi^2 - \mu}{\sigma} \sim \text{Some Distribution}$
Let $X\sim \chi^2_n$ $$ \chi^2_n\equiv Gamma(n/2, 1/2)\to f_X(x)= \frac{(1/2)^{n/2}}{\Gamma(n/2)}e^{-x/2}x^{n/2-1} $$ define $$ Y = X/\sigma - \mu/\sigma = aX+b, $$ then $$ F_Y(y) = P(Y\le y)= P(aX + b \le y)= P(X\le (y-b)/a) = F_X((y-b)/a) $$ thus $$ f_Y(y) = f_X((y-b)/a)\cdot 1/a =\frac{(1/2)^{n/2}}{\Gamma(n/2)}e^{-(y-b)/(2a)}((y-b)/a)^{n/2-1}\frac{1}{a} $$ plug in back that $a=1/\sigma$ and $b = -\mu/\sigma$, and simplify the result to get tighter density function