posted by Michael .
Assume that the average annual salary for a worker in the United States is $30,500 and that the annual salaries for Americans are normally distributed with a standard deviation equal to $6,250. Find the following:
(A) What percentage of Americans earn below $19,000?
(B) What percentage of Americans earn above $40,000?
This is what I am getting so maybe someone can tell me if I am right and if not where am I going wrong.
Z = (score-mean)/SD
A. Z(19,000)= (19,000-30,500)/6250= -1.84
P(x<19,000)=P(z< -1.84) = 23.40 23%
B. z(40,000) = (40,000-30,500)/6250= 1.52
P(x>40,000) = P (z> 1.52) = + 23.40 23%