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Math Help - Finding Expected Value and Standard deviation...Help please :(

  1. #1
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    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.
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  2. #2
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    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....
    Last edited by nzmathman; December 4th 2008 at 11:11 PM.
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