Question

If 100 people were samples, what is the probability their average income will be above 45 thousand dollars?

Mean= 41 standard deviation= 29

Answer 1

The variance of the distribution of the sampling mean is equal to the variance of the parent population divided by the sample size, $n.$

$$Z={X-\mu \over \sigma/\sqrt n}$$

$$\mu=41, \sigma=29, n=10$$

$$P(X>45)=1-P(X<45)=1-P(Z<{45-41 \over 29/\sqrt 100})=$$

$$=1-P(Z<1.3793)=1-0.9161=0.0839$$