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Homework Help: 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
    von5s8.jpg


    2. Relevant equations
    [tex]\int^{b}_{a}p(x)dx=1[/tex]

    [tex]V=M_2-\bar{x}^2[/tex]

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

    [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]
    \bar{x}=\int^{b}_{-b}xp(x)dx[/tex]

    [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]

    [tex]=\frac{-2cb^3}{3b}+{2cb}[/tex]

    [tex]=\frac{-2b^2}{3b}+\frac{2b}{b}[/tex]

    [tex]=\frac{-2b}{3}+2
    [/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

    Dick

    User Avatar
    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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