# Finding Expected Value and Standard deviation...Help please :(

• Dec 4th 2008, 05:04 PM
Darkechidna
Finding Expected Value and Standard deviation...Help please :(
The question has a table that contain shows the approximate number of males of Hispanic origin employed in the U.S. in 2005 broken down by age group

Age: 15-24.9 25-54.9 55-64.9
Employment (thousands):16,000 13,000 1,600

It then asks that using the midpoints of the given measurement classes, compute the expected value and the standard deviation of the age X of a male Hispanic worker in the U.S...

The back of the book tells me that the Mean is 30.2 years and a standard deviation of 11.78 years but I have absolutely NO idea how these were obtained. The chapter that covers this question is a chapter about Measures of Dispersion if that helps at all. Thank you for all your help.
• Dec 4th 2008, 06:36 PM
nzmathman
Total no. of people working $= 16000 + 13000 + 1600 = 30600$

expected value, $E(X) = \frac{24.9 + 15}{2} \times \frac{16000}{30600} + \frac{54.9 + 25}{2} \times \frac{13000}{30600} + \frac{64.9 + 55}{2} \times \frac{1600}{30600} = 30.5 years$

standard deviation, $\sigma = \sqrt{E(X^2) - [E(X)]^2}$

To find $E(X^2)$,

$E(X^2) = \left( \frac{24.9 + 15}{2}\right)^2$ $\times \frac{16000}{30600} + \left( \frac{54.9 + 25}{2} \right)^2$ $\times \frac{13000}{30600} + \left( \frac{64.9 + 55}{2} \right)^2 \times \frac{1600}{30600}$

$\sigma = 11.895 years$

My answers are slightly different but I'm sure I've done the method correctly...hmmm....