# Compute Mean of Normal Distribution where x% of Values are over y

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.

• Can you clarify, do you actually have the variance? – mdewey Oct 22 '18 at 12:49
• Yes, variance is given. Lets say it is $\sigma = 0.05$ – bk_ Oct 22 '18 at 12:50
• In that case you know how many standard deviations above the mean $a$ must be from the standard normal and then using the known sd you can back-calculate the mean. If I have completely mis-understood (which is quite likely) then perhaps edit the question to clarify why that does not work? – mdewey Oct 22 '18 at 12:53

Using a Z-table you can see that $$37\%$$ of a normal distribution is $$0.34+$$ standard deviations above the mean. http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf
$$0.34 sd * 0.05 = 0.017$$
so $$\mu + 0.017 = 0.2$$
$$\mu = 0.183$$
The $$0.34$$ figure can be refined further, using a matlab appoximation, I get more like $$0.331$$, but it didn't affect $$\mu$$ very much.
• That's right. Indeed, you can solve this problem in a single line with 0.2 - qnorm(0.37, sd=0.05, lower.tail=FALSE) . – whuber Oct 22 '18 at 15:26