Question

ch 15 #14. (15.35) Andrew plans to retire in 60 years. He is thinking of investing his retirement funds in stocks, so he seeks out information on past returns. He learns that over the 101 years from 1900 to 2000, the real (that is, adjusted for inflation) returns on U.S. common stocks had mean 9.4% and standard deviation 21.6%. The distribution of annual returns on common stocks is roughly symmetric, so the mean return over even a moderate number of years is close to Normal.

a. What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 60 years will exceed 9%? (±0.0001)

b. What is the probability (±0.0001) that the mean return will be less than 6%?

Answer 1

mean=9.4%   std=21.6% normal distribution

a)we are interested in a return greater than 9% so we find the z value 9 first

z=x-mean/std

z=9-9.4/21.6=-0.4/21.6=-0.01851 use z table and we see for this z score its 0.5074 which is the proababilty of it being 9 or less. But we want when its greater than 9 so we take this and subtract from 1.

1-0.5074=0.4926 or 49.26%

b)same procedure as above find the z score first

z=6-9.4/21.6=-0.1574 or about -.11          using the ztable the value is 0.4375 or 43.75%

we don't have to subtract from 1 since this value is for the probability of 6 or less.