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Math Help - Expected Value and Variance Again

  1. #1
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    Expected Value and Variance Again

    Hey guys, thanks for the help last time.

    I had another one that I'm not really sure how to start out.

    Suppose that a bag contains six slips of paper: one with the number 1 written on it, two with the number 2, and three with the number 3. What is the expected value and variance of the number draw if one slip is selected at random from the bag?

    I already have:

    S = {1, 2,2,3,3,3}
    Probability of selecting a specific number slip.
    PR[1] = 1/6
    PR[2] = 1/3
    PR[3] =

    I was thinking a good way to start this was:

    u = E[x] = 1(1/6) + 2(1/3) + 3(1/3) +4(1/2)+5(1/2)+ 6(1/2) to get the expected value but i wasn't sure if this is correct. Any ideas?

    Thanks
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  2. #2
    MHF Contributor

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    You have several misconceptions going on there.
    \begin{gathered}<br />
E(X) = \sum {xp(x)} = 1\left( {\frac{1}<br />
{6}} \right) + 2\left( {\frac{1}<br />
{3}} \right) + 3\left( {\frac{1}<br />
{2}} \right) \hfill \\<br />
V(X) = E(X^2 ) - E^2 (X) \hfill \\ <br />
\end{gathered}
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  3. #3
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    Thanks a bunch plato. I kinda realized a few minutes later where i went wrong after i posted it. But i didn't quite get it till you posted the equations.

    Much appreciated ^_^
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