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Mean of a probability distribution

  1. Mar 31, 2008 #1
    [SOLVED]Mean of a probability distribution

    1. The problem statement, all variables and given/known data

    2. Relevant equations



    [tex]M_2=\int^{b}_{a} x^2p(x)dx[/tex]

    3. The attempt at a solution

    I found that [tex]c=\frac{1}{b}[/tex] which is a right answer.

    What I did next was:


    [tex]=\int^{0}_{-b}x(\frac{cx}{b}+c)dx\ + \int^{b}_{0}x(\frac{-cx}{b}+c)dx[/tex]

    [tex]= \int^{0}_{-b}\frac{cx^2}{b}\ +\ c\ dx\ + \int^{b}_{0}\frac{-cx^2}{b}\ +\ c\ dx[/tex]

    [tex]=\left[ \frac{cx^3}{3b}+cx \right]_{-b}^{0}+\left[ \frac{-cx^3}{3b}+cx \right]_{0}^{b}[/tex]




    But the answer says that [tex]\bar{x}=0[/tex]

    If I can manage to get x-bar, I can manage to get the variance and SD.
    Last edited: Mar 31, 2008
  2. jcsd
  3. Mar 31, 2008 #2


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    Science Advisor
    Homework Helper

    i) you are integrating cx, that should give you an antiderivative of cx^2/2. ii) put in the top limit and subtract the bottom limit. You are making sign errors. The cx^3/3b terms also cancel.
  4. Mar 31, 2008 #3
    What an amateur mistake. I feel foolish. Cheers, Dick.
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