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Math Help - Mean and Variance of Continous distribution

  1. #1
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    Post Mean and Variance of Continous distribution

    A continuous probability density distribution for the random
    variable
    X has the form:

    f(
    x) = ke^x if 0 < x <2 (the < should be equal to or less than)
    =
    ke^(4-x) if 2 < x < 4
    = 0 otherwise

    Use integration to determine:

    the value of k
    (i) the mean of the distribution,
    (ii) the variance of the distribution.

    Your help would be very much appreciated

    Last edited by pip1690; May 10th 2008 at 08:48 AM. Reason: sorry forgot the k part
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Then mean is given by \mu=\int_{\mathbb{R}}xf(x)\,\mathrm{d}x

    Splitting the integral gives : <br />
\mu=\int_{-\infty}^0xf(x)\,\mathrm{d}x+\int_0^2xf(x)\,\mathrm  {d}x<br />
+\int_2^4xf(x)\,\mathrm{d}x<br />
+\int_4^{\infty}xf(x)\,\mathrm{d}x and as you know f on each interval, you can go on .

    Concerning the variance, it is defined by V=\int_{\mathbb{R}}x^2f(x)\,\mathrm{d}x-\mu^2.

    Again, split the integral into four parts and it will give you the result.

    note : in both cases, integration by parts might be helpful.

    Hope that helps.
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  3. #3
    Senior Member tukeywilliams's Avatar
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     \mu = \int_{0}^{2} xke^{x}\ dx + \int_{2}^{4} xke^{4-x} \ dx .

    Then  \text{Var}(X) = (E[X])^{2} - \mu^{2}
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  4. #4
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    Quote Originally Posted by tukeywilliams View Post
     \mu = \int_{0}^{2} xke^{x}\ dx + \int_{2}^{4} xke^{4-x} \ dx .

    Then  \text{Var}(X) = {\color{red}E[X^{2}]} - \mu^{2}
    I've corrected (in red) a small typo.
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