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Math Help - Probability pdf

  1. #1
    Junior Member
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    Probability pdf

    Hey guys, here's a problem I'm stuck on:

    The lifetime of X (years) has pdf:

    f(x) = 2x/(theta) *e^ -(x/theta) , when... 0<x<infinity

    If P(X>5) = 0.01, determine theta.

    --------------

    So I thought, P(x>5) = 1-P(x<5)

    so, 1- integral (from 0-->5) f(x) = 0.01

    Is this the right logic? If so, the problem becomes evaluating that integral.
    Can someone help me? It's been two years since Calculus, and I'd appreciate it.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by amor_vincit_omnia View Post
    The lifetime of X (years) has pdf:

    f(x) = 2x/(theta) *e^ -(x/theta) , when... 0<x<infinity

    If P(X>5) = 0.01, determine theta.

    --------------

    So I thought, P(x>5) = 1-P(x<5)

    so, 1- integral (from 0-->5) f(x) = 0.01

    Is this the right logic? If so, the problem becomes evaluating that integral.
    1-\int_0^5 \frac{2x}{\theta^2}\mathrm{e}^{-\frac{x^2}{\theta^2}}\,\mathrm{d}x=0.01

    What you've done is correct. To evaluate the integral, notice that \left(-\frac{x^2}{\theta^2}\right)'=-\frac{2x}{\theta^2} : you may try substituting u=\frac{x^2}{\theta^2}.
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  3. #3
    Junior Member
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    okay, for the final answer i'm getting


    1 - {1/ [e^(25/θ)]}

    anyone else? Thanks for the help, wish I could find all of my old Cal notes!
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by amor_vincit_omnia View Post
    okay, for the final answer i'm getting


    1 - {1/ [e^(25/θ)]}
    That's it
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