1. Assume the average annual salary for a worker in the United States is $27500 and that the annual salaries for Americans are normally distributed with a standard deviation equal to $6250. Find a. the percentage of americans working below $18000
b. the percentage of americans working below $40000


2. Find the value of z such that 40% of the distribution lies between it and the mean.


:confused:

Do you have minitab or SAS?
Excel will work
How far have you gone through this problem?

Or just plain ol' math? :P

or just plain ol' math? :P

That's what we asians call it... :p ;)

for part a:

Use the formula and then look up the probability which corresponds to the z-score you just found. I will show you this one and you do the other. Okey-doke?

z=\frac{x-{\mu}}{\sigma}

\frac{18000-27500}{6250}=-1.52

Look up -1.52 in the z-table and we find 0.064...

A little over 6% have have a salary below $18000.


If you have a fancy-schmancy calculator or Excel you can find the z-score without the table.

If you want to show off, here's the formula the values are derived from:

\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{\frac{-t^{2}}{2}}dt