[original post removed]
Because it is asking for the probability that a single observation (return in a given year) will lie in that range; for a single observation, you use the standard deviation as your measure of dispersion.
If it had asked the probability that the mean of a number of observations (average return over a number of years) would lie in that range, you would use the standard error (= σ / √_n_) as the measure of dispersion, so you would need to know n.
That makes sense, thanks. And I am assuming n would be the number of years in that case?
Cool!
My pleasure.
Yes: if they were annual returns, then n would be the number of years: the number of observations.
How did you get 0.6026?
*Looked at this post using my phone
0.6026 = Pr (Z <=0.26666) = Pr (Z <= (12 - 8)/15)
I think OP meant z-scores are 0.2666 and 0.4666 with corresponding cumulative probabilities of 0.6772 and 0.6026.
Yes, we are using the population mean as our expected return so 12-8/15 = 0.2666 and 15-8/15 = 0.4666; so the corresponding probability values on the z table are 0.6772 and 0.6026. The difference between them is the probability that returns lie between 12% and 15%.