Distribution of the inverse square of a non-standard normal random variable multiplied by a constant It's a somewhat complicated situation and sorry about my phrasing, but it's my first time here. Suppose I have random normal variable $X$ ~ $N( \mu, \sigma^2)$, which represents some true effect(s). Then I try to estimate the mean of this effects using the estimates from a few (sometimes underpowered) trials -> Meta-Analysis, and do the following transformation:

$Y$ = $c/X^2$, where $c$ is a constant $> 0$ and $Y$ must be $> 0$ and therefore x>0. 
As a side note, $\sigma$ is for example 90% of $\mu$ and divided by the 97.5%-quantile of the standard normal distribution. What distribution has $Y$? But most importantly how can I determine/estimate the central tendency of a sample from this resulting distribution? 
What I found out til now, just squaring that $X$ would result in a non-central chi-square distribution with a non centrality parameter and then just do the inverse (maybe?) and still not knowing the central tendency. Also with that 90%-$\sigma$, I have some x<0 and with very small x I get a super long tail (which doesn't reflect any reality anymore, it's just a theoretical number). Tested with $\mu=6$ and $\sigma=2.76$, c=11128.6 and 3 trials 99%,80% and 10% powered depending on the trial size in a simulation with n=10000. 
I worked with the Median, but it doesn't seem to catch the center well.
 A: The description of your experiment is somewhat unclear, so I will limit to answer the question about distributions.
With your notation, $X/\sigma \sim N(\mu/\sigma,1)$ so the square of that, $(X/\sigma)^2$ has a noncentral chisquare distribution with 1 df (degree of freedom) and noncentrality parameter $\lambda=(\mu/\sigma)^2$. 
Then we can write $Y=c/X^2 = \frac{c/\sigma^2}{(X/\sigma)^2}$ so it is distributed like a constant ($c/\sigma^2$) times an inverse chi-squared rv (random variable) with 1 df and the noncentrality parameter $\lambda$. 
You will find implementations of the noncentral chi-squared many places, so it will be useful to express the density (cdf, ...) for the inverse noncentral chi-square using that. Let $U$ be a nonnegative rv and $t>0$, the density of $U$ is $f$. Then we find the density of $1/U$ as $f_{1/U}(t)=f_U(1/t)/t^2$ so you can apply that. 
In R there is an implementation (on CRAN) in package invgamma.  But it is very easy to write yourself:
dinvchisq  <-  function(x,df,ncp=0,log=FALSE) {
    res <- dchisq(1/x,df,ncp,log=TRUE) - 2*log(x)
    if(log) res else exp(res)
}

