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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.

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  • $\begingroup$ Is 100 the variance or the standard deviation? $\endgroup$
    – Michael R. Chernick
    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. $\endgroup$
    – Michael R. Chernick
    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

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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$.

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  • $\begingroup$ 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? $\endgroup$
    – Michael R. Chernick
    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

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