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Continuous probability distribution

  1. Jul 29, 2010 #1
    1. The problem statement, all variables and given/known data

    A continuous random variable ,X represents the period, of a telephone call in the office. The cdf of x is given by

    F(x)= x^2/8 for 0<=x<=2

    =1-4/x^3 for x>2

    Find pdf , mean and variance.

    2. Relevant equations

    3. The attempt at a solution


    f(x) = 1/4 x for 0<=x<=2

    =12/x^4 for x>2

    = 0 , otherwise

    [tex]E(x)=\int^{2}_{0}x(\frac{1}{4})x dx +\int^{\infty}_{2}x(\frac{12}{x^4})dx[/tex]


    [tex]E(x^2)=\int^{2}_{0}x(\frac{1}{4})x^2 dx +\int^{\infty}_{2}x^2(\frac{12}{x^4})dx[/tex]



    Variance is negative here , how can that be? I have checked my working thoroughly but am not able to spot my mistakes.
  2. jcsd
  3. Jul 29, 2010 #2


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    Homework Helper

    Check your calculation of [tex] E(X) [/tex] once again - I get a different value.
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