# Normal Distribution sample mean and population mean?

Assume that house prices in an area are normally distributed with a standard deviation of \$20,000. A random sample of 16 houses is taken. What is the probability that the sample mean differs from the population mean by more than \$ 5,000.

I can write two equations as follows

For case I - $N(\mu_1, 20000)$

For case II - $N\left(\mu_2, \sqrt{\frac{20000^2}{16}}\right)$

But in both cases $\mu_1 = \mu_2$ right? Can anyone give me a tip?

The arithmetic average of $K$ equally normal-distributed quantities has the same mean $\mu$; but the standard deviation decreases by the factor of $\sqrt{K}$. In this case, $K=16$ so the standard deviation of the mean will be $20,000/\sqrt{16} = 5,000$ dollars. It just happens to be the value equal to the deviation in the last, main question you asked.
• Thank you for the answer. Let me check if I got it correctly :) So if the sample size is 20, then the standard deviation decreases by a factor of 4.472 and that means $\frac{5000}{4472} = 1.118$ standard deviations? – Blogger Jan 15 '16 at 7:56
• Sorry, the last step is $4472/5000=0.894$ standard deviations. – Luboš Motl Jan 15 '16 at 16:25