Probability on Z-score 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%?
 A: 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.
