# Normal Distribution - finding mean

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.

• Is 100 the variance or the standard deviation? Commented Jan 16, 2017 at 1:22
• 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. Commented Jan 16, 2017 at 1:24
• I think it should be the sd Commented Jan 16, 2017 at 1:31
• 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. Commented Jan 16, 2017 at 1:41
• Thank you very much, I'll try using the formula below. I'll let you know if I reach the correct answer :) Commented Jan 16, 2017 at 1:43

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

P($Z>$a)=0.9.

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

• 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? Commented Jan 16, 2017 at 2:02
• My answer was 62.816 so it did match the answer Commented Jan 16, 2017 at 2:16
• 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. Commented Mar 10, 2020 at 0:54