0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Can you clarify, do you actually have the variance? $\endgroup$
    – mdewey
    Oct 22, 2018 at 12:49
  • $\begingroup$ Yes, variance is given. Lets say it is $\sigma = 0.05$ $\endgroup$
    – bk_
    Oct 22, 2018 at 12:50
  • 1
    $\begingroup$ 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
    Oct 22, 2018 at 12:53

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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_
    Oct 22, 2018 at 15:22
  • 1
    $\begingroup$ 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
    Oct 22, 2018 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.