I'm studying Statistical Methods and trying to revise for the upcoming exam. Looking at the past exam paper, there is a question:

 X ~ N (µ, 10²), and P ( X > 50) = 0.9, find µ

I've only before done the calculations finding z = (x - mean)/standard deviation and looking at the values from the Normal distribution function table. I have no idea how to approach this question. Any help would be much appreciated.

  • $\begingroup$ Is 100 the variance or the standard deviation? $\endgroup$ Commented Jan 16, 2017 at 1:22
  • $\begingroup$ That is all that is said in the question - the answer is meant to be 62.816 but I have no idea how that answer was achieved. $\endgroup$
    – camnesia
    Commented Jan 16, 2017 at 1:24
  • $\begingroup$ I think it should be the sd $\endgroup$
    – camnesia
    Commented Jan 16, 2017 at 1:31
  • $\begingroup$ I took the sd to be 10. If you work it out using the steps in my answer and do not get the right answer try sd=100. $\endgroup$ Commented Jan 16, 2017 at 1:41
  • $\begingroup$ Thank you very much, I'll try using the formula below. I'll let you know if I reach the correct answer :) $\endgroup$
    – camnesia
    Commented Jan 16, 2017 at 1:43

1 Answer 1


Let us suppose that the variance is 100. Then the standard deviation is 10.

P$(X>50)$=P($X-mu>50-mu)$=P($(X-mu)/10 > (50-mu)/10)$=

P$(Z> (50-mu)/10)$=0.9.

So now you can go to the standard normal table and find what value of Z say a gives


After you find a you solve for mu by setting $(50-mu)/10$ = a and then solving for $ mu$.

  • $\begingroup$ As I check it you should get mu approximately equal to 62.18 when the sd=10. If you do not get that let me know what you did? $\endgroup$ Commented Jan 16, 2017 at 2:02
  • $\begingroup$ My answer was 62.816 so it did match the answer $\endgroup$
    – camnesia
    Commented Jan 16, 2017 at 2:16
  • $\begingroup$ This is a great answer. But suppose that we change the problem so that it is "X ~ N (µ, 10²), and P(–50 < X < 50) = 0.9, find µ." Am I right to think that we can't use the strategy given above, or any closer variant of it, to solve this new problem? We might start by noting that P(–50 < X < 50) = P(X < 50) – P(X < –50). But that doesn't seem to make the problem more tractable. $\endgroup$
    – user697473
    Commented Mar 10, 2020 at 0:54

Your Answer

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

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