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Math Help - Chebyshev and uniform distribution

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    Chebyshev and uniform distribution

    I'm having a lot of trouble with this question:

    Find P(|Y-\sigma| \leq 2\sigma) for the uniform random variable distribution. Compare with the corresponding probabilistic statement given by Chebyshev's theorem and the empirical rule.

    according to the solution in the book the answer is 1; I have no idea what this number represents.

    The variance of a uniform distribution is Vr(X) = \sigma ^2 = \frac{(\theta_2-\theta_2)^2}{12} so in this case it should be 4\sigma = \frac{(\theta_2-\theta_2)^2}{12}. This is where I get stuck, plus I don't think that I'm even going in the right direction. Any help would be appreciated.
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    Quote Originally Posted by lllll View Post
    I'm having a lot of trouble with this question:

    Find P(|Y-\sigma| \leq 2\sigma) for the uniform random variable distribution. Compare with the corresponding probabilistic statement given by Chebyshev's theorem and the empirical rule.

    according to the solution in the book the answer is 1; I have no idea what this number represents.

    The variance of a uniform distribution is Vr(X) = \sigma ^2 = \frac{(\theta_2-\theta_2)^2}{12} so in this case it should be 4\sigma = \frac{(\theta_2-\theta_2)^2}{12}. This is where I get stuck, plus I don't think that I'm even going in the right direction. Any help would be appreciated.
    \Pr(|Y - \sigma| \leq 2 \sigma) = \Pr(-\sigma \leq Y \leq 3 \sigma).

    If Y is uniform over [a, b]:

    \bullet f(y) = \frac{1}{b-a} for a \leq y \leq b and zero otherwise.

    \bullet \sigma = \frac{b-a}{\sqrt{12}}.


    Note that - \sigma = \frac{a-b}{\sqrt{12}} < a for b > a. This means that:

    \Pr(-\sigma \leq Y \leq 3 \sigma) = \Pr(-\sigma \leq Y \leq a) + \Pr(a \leq Y \leq 3 \sigma) = \Pr(a \leq Y \leq 3 \sigma).

    Furthermore:

    3 \sigma = \frac{3(b-a)}{\sqrt{12}} < b if  a > \frac{-(\sqrt{12} - 3)b}{3}.

    3 \sigma = \frac{3(b-a)}{\sqrt{12}} > b if  a < \frac{-(\sqrt{12} - 3)b}{3}. In this case the required probability is obviously equal to 1.
    Last edited by mr fantastic; September 8th 2008 at 09:10 PM.
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