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Probability f(x) = (2/9)(3x -x^2)

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the probability P(x>1)
    The function is f(x) = (2/9)(3x -x^2)

    2. Relevant equations



    3. The attempt at a solution

    So,
    P(x>1) = ∫ (2/9)(3x-x^2) dx = (2/9)((3x^2/2) - (x^3/3))[1,∞]

    When you do the improper integral lim a->∞ [ (2/9)((3(a)^2/2) -((a)^3)/3)] this is just the first part of the evaluation I stopped because it goes to infinity. So If the integral does not converge then I say the event will never happen?
     
  2. jcsd
  3. Oct 22, 2013 #2

    Ray Vickson

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    Homework Helper

    Surely this cannot be the whole problem statement, because it is nonsense as written. Do you mean that f(x) is the probability density function? If so, say it. But that is a rather minor quibble. More serious the fact that for some value of x, f(x) becomes < 0 and that can never, ever happen for a legitimate probability density. Also: the density is supposed to integrate to 1, but yours gives a divergent integral.

    So, you need to tell us EXACTLY what the problem says.
     
  4. Oct 22, 2013 #3
    Problem says.
    f(x) = k(3x-x^2) if 0<= x <= 3 and f(x) = 0 if x < 0 or x>3
    a. for what value of k is f a probability density funciton?
    b. for that value of k find P(X>1)
    c. Find the mean.
    I'm asking about b. K was determined to be 2/9
     
  5. Oct 22, 2013 #4
    Yeah I found the mistake it is [0,3] not infinity. So I can do integrate from [0,1] then do 1- whatever the result is.
    to get P
     
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