Question

The mean salary of people living in a certain city is $37500 with a standard deviation of $1561. a sample of 33 people is selected at random from those living in the city. find the probability that the mean income of the sample is less than $38000.

Answer 1

The probability is 0.9671.

The z-score for this is given by

z = (x - μ)/(σ/√n)
z = (38000-37500)/(1561/(√33)) = 1.84.

Using a z-table (http://www.z-table.com) we see that the area under the curve to the left of, less than, this is 0.9671.