Let $X\sim U[-c,c]$ and Z be the mean of 2n+1 readings of X. Show the probability density function of Z is $f(z)=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n$
What I think is as follows. The probability of getting $\zeta$ in a sample is $\frac{(2n+1)\delta}{2c}$ for a small delta. The probability of n values are less than $\zeta$ is $(\frac{c+\zeta}{2c})^n$. The probability the n values are greater than $\zeta$ given n values are less is $(\frac{c-\zeta}{2c})^n$. Multiplying these together we get $f(\zeta)=\frac{2n+1}{(2c)^{2n+1}}(c^2-z^2)^n$. But where does does the ${2n\choose n}$ factor come from?
I can conjecture how it could be, which would be: I have all $2n+1$ readings in a bucket. In front of my bucket I have a row of $2n+1$ chairs, which is the order of my readings. I take out $\zeta$ and put it in the middle chair. There are $2n\choose n$ ways to arrange $n$ readings on the chairs left of $\zeta$(e.g. the 1st 3rd and 7th readings are less than $\zeta$) and this determines which ones are greater than zeta.
I can't figure out why this kind of ordering needs to be accounted for. In my derivation, the order of the readings did not make a difference, we knew that one reading must be the median and found the probability it equals zeta.
Could someone explain where my reasoning goes wrong? I am less interested in alternative solutions as I already have a written solution, which uses a different method.