Net profits for small retail stores are normally distributed with a
mean of $60,000 and a standard deviation of $10,000.
Find the probability that a sample of 15 randomly selected stores will
have a mean profit greater than $70,000.

Hi!!

I am not sure if this is the way to solve this, so I post this in the answer box:
To answer this problem we need to use the Cental Limit Teorem:
"The mean of a sampling distribution of the mean is always equal to
the population mean. The standard error of the mean is always equal to
the population standard deviation divided by the square root of the
sample size:
Mu_d = Mu
StD_d = Std/sqrt(n)  
When the population is normally distributed, and the value of the
population standard deviation is known, the sampling distribution of
the mean will be normally distributed. If the population is not
normally distributed, and the value of the population standard
deviation is known, the sampling distribution is still normally
distributed."
http://espse.ed.psu.edu/statistics/Chapters/Chapter9/Chap9.html#Sampling
Distribution of the Mean Characteristics


In this case:
n=15

Mu_d = Mu = $60,000

Std_d = Std/sqrt(n)
      = $10,000/sqrt(15) =
      = $2,581.99

Now we have a normal distributed ramdom variable X (Net annual profits)
with mean Mu_d = $60,000 and standard deviation StD_d = $2,581.99 .

We want to know the probability to get a sample with a mean greater
than X=$70,000. So we need to find the probability for a sample to
have a mean greater than X.

The first step is to normalize the variable X:
Z = (X-Mu_d)/StD_d = (70,000-60,000)/2,582 = 3.9

Then:
P(X > 70,000) = P(Z > 3.9) =
              = 0.5 - P(0 < Z < 3.9) =
              = 0.5 - 0.499952 =
              = 0.000048


Let me know if this answers your question correctly.

I am not sure about the final answer. When I look up the table for
3.9, I get .4990. When I calculate 0.5 - 0.4990, I get 0.001. Does
that then that the probability that a randomly selected store having a
mean profit greater than $70 000 is 0.1%?

Yes, the meaning of the answer is that the probability that a sample
of 15 randomly selected stores will have a mean profit greater than
$70,000 is 0.1% (the diference between your result and mine is due
rounding errors, it is not significative).

Great - thanks so much for your help