I have a data of 30 values that represent returns on investment and which have a mean of 7 and standard deviation of 14. We are talking here about normal distribution. Now I am being required to do a following:

If you are given a choice between the top 6% of returns and a return of 25%, which option would you choose? Between what two values of returns (symmetrically distributed around the mean) will 32.6% of all possible returns contained?

I was thinking to sum top 6% of the values of a given data, and divide it by number of it, and to find out z score of 25, and to compare those two results. Is my approach good?

I am a bit confused about how to find between what two values of returns data I will get 32.6%?


1 Answer 1


First, using a normal table or the function NORMSINV(0,94) in MS Excel or OpenOffice/LibreOffice Calc, you should verify that $P(Z < 1.555) = 0.94$ (which imply $P(Z > 1.555) = 0.06$). By comparing 1.555 with the standardize value that correspond to 25, that is to the number $$ \frac{25 - 7}{14} = 1.286, $$ you see that the best choice is the first option.

For the second question you should find the $a\in \mathbb{R}$, such that $P(-a < Z < a) = 0.326$ thus $1 - 2P(Z > a) = 0.326$ or $P(Z > a) = 0.337$ or $P(Z < a) = 1 - 0.337 = 0.663$ from which you get $a = 0.421$, (by applying one more time the NORMINV function) thus the desired returns should have standardized values lying between -0.421 and 0.421. Then, it is an easy exercise to use one more time the transformation in order to calculate the point in the original distribution $N(\bar{x}, s^2)$ that has a standardized value equal to 0.421, in particular $$ \frac{x_a - 7}{14} = 0.421, $$ from where you get $x_a = 14 \cdot 0.421 + 7 = 12.9$, and in similar fashion $$ \frac{x_{-a} - 7}{14} = -0.421, $$ from where you get $x_{-a} = 14 \cdot (-0.421) + 7 = 1.1$, thus all returns lying between 1.1 and 12.9 will be 32.6% of all possible returns.

  • 1
    $\begingroup$ Please review the wiki for the self-study tag. I would like to suggest the OP might be better served by a thoughtful discussion of the assumptions you made to arrive at these results (rather than being given the results themselves) along with a clear interpretation of them. In particular, it would be interesting to see an analysis of how a mere 30 data values could be considered "normal" and to what extent your calculations based on a Normal distribution would be expected to differ from what is really in the dataset. $\endgroup$
    – whuber
    Apr 4, 2014 at 16:10
  • $\begingroup$ @whuber I see your point. Actually, it was a reaction of the moment to answer this undergraduate exercise, although I admit that the way I did it was not appropriate since my answer is not guiding the asker to the right path towards knowledge. $\endgroup$ Apr 4, 2014 at 18:14

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.