1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    pdf:

    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]

    =20/3

    [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]

    =7

    Var(x)=7-(20/3)^2

    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

    statdad

    User Avatar
    Homework Helper

    Check your calculation of [tex] E(X) [/tex] once again - I get a different value.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Continuous probability distribution
Loading...