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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%?

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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.

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    $\begingroup$ 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 ♦
    Commented 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. $\endgroup$
    – Epameinondas
    Commented Apr 4, 2014 at 18:14

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