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Thread: [SOLVED] Mean and Variance Question...

  1. April 14th 2008, 07:51 AM #1
    looi76
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    [SOLVED] Mean and Variance Question...

    Question:
    The random variable T has the probability distribution given in the following table.


    Attempt:


    E(T) = 4 , Var(T) = 19.6

    Where did I go wrong?
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  2. April 14th 2008, 08:02 AM #2
    Peritus
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    your definition of variance is wrong it's supposed to be:


    VAR(t) = \sum\limits_i {\left( {t_i - < t > } \right)^2 p(t_i )}

    or equivalently:

    <br /> VAR(x) = \sum\limits_i {t_i ^2 p(t_i )} - \left[ {E(T)} \right]^2 = 19.6 - 16 = 3.6<br />
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