I am trying to find the 0.5 (mean) percentile of a CDF function.
$$ F(X) = 1 - e^{-(x/3)^2} $$
In my book's example it says
$$ m = 3[-ln(1-0.5)]^{1/2} = 3\sqrt{ln2}=2.498$$
I am not sure how to get to the above equation even using the definition of a percentile of $$F(x_p) = p $$
The $50$-th percentile is the median, not the mean. Call it $m$. We want to solve the equation $$1-e^{-(m/3)^2}=\frac{1}{2},$$ or equivalently $$ e^{-(m/3)^2}=\frac{1}{2}.$$ Take the (natural) logarithm of both sides. We get $$-\left(\frac{m}{3}\right)^2=\ln(1/2)=-\ln 2.$$ So we are solving $\frac{m^2}{9}=\ln 2$. That gives (since $m$ is positive) $m=3\sqrt{\ln 2}$.
