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

  1. #1
    Member looi76's Avatar
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    [SOLVED] Mean and Variance Question...

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


    Attempt:


    $\displaystyle E(T) = 4 , Var(T) = 19.6$

    Where did I go wrong?
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  2. #2
    Senior Member Peritus's Avatar
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    your definition of variance is wrong it's supposed to be:


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

    or equivalently:

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