Posted by **Michael** on Sunday, October 28, 2012 at 4:22pm.

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?

- Math -
**Michael**, Sunday, October 28, 2012 at 4:46pm
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%

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