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Find the median of the probability distribution

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data

    f(x) = C|x-2| for 0 <= x <= 3
    f(x) = 0 otherwise

    2. Relevant equations



    3. The attempt at a solution
    Solved for C, found it to be (2/5).

    So.. I'm confused how to set up my integral here. I tried integral(2/3(x-2)dx) from m to 3 = 1/2. That didn't yield the correct result.
     
  2. jcsd
  3. Mar 8, 2013 #2

    SammyS

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    If you found C to be 2/5, then the integral should be
    [itex]\displaystyle \int_{M}^{3}\frac{2}{5}|x-2|\,dx \ .[/itex]​

    Set that equal to 1/2 and solve for M, although it might be easier to integrate from 0 to M.

    Graph the integrand to see why.
     
    Last edited: Mar 8, 2013
  4. Mar 8, 2013 #3
    Yes, actually that's what I meant to say. I don't understand how I can integrate the function having the absolute value there. I understand how to do it if it were say.. integrating from 0 to 3. But since we don't know what m is I don't understand.

    And neither does wolfram alpha lol
     
  5. Mar 8, 2013 #4

    SammyS

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    Did you graph the function?

    If x ≤ 2, then |x - 2| = 2 - x .

    If x ≥ 2, then |x - 2| = x - 2 .
     
    Last edited: Mar 8, 2013
  6. Mar 8, 2013 #5
    Yes, and I understand that. What I don't understand is where m lies. Is it in the range of <2 or >2
     
  7. Mar 8, 2013 #6

    Ray Vickson

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    That is what you need to figure out. If in doubt, try it both ways to see what happens! However, careful examination of your graph should be enough.
     
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