The distribution of the mean of a sample of size n from a normal distribution

with mean mu and SD sigma, has mean m=mu and SD sm=sigma/sqrt(n).

So in this case the mean of the sample of 10 has mean 172 lb and SD 29/sqrt(10) ~= 9.17 lb.

Now the z score of a mean of 150 lb is:

z = (150 - m)/sm = (150 - 172)/9.17 ~= -2.40

Now we look this z-score up in a table of the cumulative standard normal distribution to

get the probability of observing a z-score this low or lower. Doing this we find the probability

is 0.0082 (0.82%), which is also the probability that the mean weight of 10 of the men is <= 150 lb.

The probability that the mean weight of a sample of 90 of the men is >150lb is derived

by observing for a sample of 90 the SD of the mean is sm = sigma/sqrt(90) = 29/sqrt(90) =3.06.

So the z-score is now:

z = (150 - m)/sm = (150 - 172)/3.06 ~= -7.19

and there is no need to look this up the required probability is so close to 1 that my tables

do not go that far.

RonL