I am looking for a way to determine the mean of a normal distribution (with given variance), where e.g. $z = 0,37 = 37\% $ of values should be above a certain value $a$ (e.g. 0,2)?
My first idea was setting $\int_{0.2}^\inf \frac{1}{\sqrt{\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2}) = a = 0,2$ and solve after $\mu$, however this seems rather complicated to me.
Also, looking at $P(X\geq a = 0.2) = 1-\Phi(x)$ did not really help me.
Is there an easier way to do this or am I missing something fundamental? An approximation for the mean would also be okay. Any help is much appreciated.