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    • Thread Starter

    I've worked out the answers to the following questions but I don't know how I should set them out

    1a) (It's a table)

    When x=1, P(X=x) is 0.3-a,
    When x=2, P)X=x) is 2a,
    When x=3, P(X=x) is 0.7-a

    "Find the possible values of a" - 0<a<0.3 (thanks to paronomase )

    "Find the value of a when var(X)=0.64" - I think it's 0.1, but I found it by trial and error

    Can anyone help me with setting the answers out correctly, please?

    First of all,
    do you know that 0< (or equal) P(X=x) < or = 1
    Therefore, here you have to solve:

    0< 0,3-a < 1 (again it is inferior or equal)
    0 < 2a < 1
    0 < 0,7-a < 1

    this gives you

    0 < a < 0,5
    -0,3 < a < 0,7

    Now, you are looking for a value of a that satisfies all of these inequalities (the intersection of the intervals that satisfy the three inequalities)
    With the 2nd one, you are imposed a>0
    With the 1st one, you are imposed a<0,3

    Therefore, those two conditions have to be satisfied, therefore 0<a<0,3

    a is between zero and 0.3, as p(x=1,2,3) can not be less than zero. Sorted, right that

    so 0.3-a + 2a +0.7-a=1, (1(0.3-a)+4(2a) +9(0.7-a))-(1(0.3-a) + 2(2a) +3(0.7-a)) squared=0.64 (for variance)

    algebra it out and you get whatever the answer is
    • Thread Starter

    Thank you both for your help

    I still can't do it though

    I thought I had an answer, but when I used it to work out the next part, P(X1 + X2=4) = 0.28, I didn't get 0.28
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Updated: February 7, 2010
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