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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.

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  • $\begingroup$ Can you clarify, do you actually have the variance? $\endgroup$
    – mdewey
    Commented Oct 22, 2018 at 12:49
  • $\begingroup$ Yes, variance is given. Lets say it is $\sigma = 0.05$ $\endgroup$
    – bk_
    Commented Oct 22, 2018 at 12:50
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    $\begingroup$ 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? $\endgroup$
    – mdewey
    Commented Oct 22, 2018 at 12:53

1 Answer 1

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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.

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  • $\begingroup$ Perfect, thats what I was looking for. In R, qnorm(1-0.37) is the way to identify 0.331 as the factor of standard deviations. $\endgroup$
    – bk_
    Commented Oct 22, 2018 at 15:22
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    $\begingroup$ 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) . $\endgroup$
    – whuber
    Commented Oct 22, 2018 at 15:26

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