Answer
0.0143
Work Step by Step
Given $\mu=82000,\sigma=5000,n=50,N=800$
we have $n/N=50/800=0.0625>0.05$, we should use
a correction factor for the standard error, hence
$\sigma_{\bar X}=\frac{5000}{\sqrt {50}}\times\sqrt {\frac{800-50}{800-1}}=685.08$
so $z=\frac{83500-82000}{685.08}=2.1895$. Use table E or a calculator,
the probability that the mean of the values of these homes is greater
than 83,500 is given by $P(X>83500)=1-P(z)=1-0.9857=0.0143$
