Over the entire 20th century the real that is adjusted for...

Over the entire 20th century, the real (that is, adjusted for inflation) annual returns (in percent) on U.S. common stocks had mean 8.7 and standard deviation 20.25. 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 10 years will be between 5 and 10 (%)? (b) Over the next 10 years, the top 5% of returns would be above what average return?

Transcript

00:01 So for this problem, to begin in part a, we're looking for the probability that the mean annual return over the next 10 years, so this is a sample mean value, will be between 5 and 10%.

00:13 So probability that x bar is between 5 and 10.

00:16 The way that we'll do this is we'll translate those boundaries into the corresponding z scores.

00:23 So that would be, i'll write this out explicitly for the lower bound, that would be 5 minus the population mean value, divided by the population standard deviation over the square root of the sample size, where the sample size here is 10, or actually i'll keep the generic form theme there.

00:42 So it's 5 minus mu divided by sigma over root n, will be less than x bar, which is in turn less than 10 minus mu over sigma divided by root n.

00:53 And i'll note that the way that we actually calculate this out is we would find the probability that z is less than the upper bound, and then subtract off the probability that z is less than the lower bound.

01:05 So for the upper bound z score, that's 10 minus 8 .7, divided by 20 .25 over the square root of 10.

01:13 So that gives us an upper bound z score of 0 .203.

01:18 Then we subtract off the probability of z less than the lower bound value.

01:24 So that's 5 minus 8 .7 over 20 .25 divided by root 10, which gives us a result of roughly negative 0 .5.

01:32 Now you can find those values using a table of values, or i'm just going to use my software here.

01:38 So i'm doing cdf, so this gives me the probability to the left of the value.

01:43 So normal distribution with no arguments is the standard normal.

01:46 Valuating that out, we get 0 .203.

01:50 So we get the probability to the left of the upper bound is 0 .5404, roughly.

01:58 Then we subtract off the proportion to the left of negative 0 .58, which would be equal to about 0 .281 .1.

02:05 Roughly...

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