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Math Help - Var + e

  1. #1
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    Smile Var + e

    I'm stuck, can anyone help me with this "easy" statistics?

    X Y p(x,y)
    0 0 0,20
    0 1 0,25
    1 0 0,30
    1 1 0,25

    Var (X+Y) = ?
    E (Y(X=0) = ?

    Thanks a lot!!
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  2. #2
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    Re: Var + e

    Quote Originally Posted by emme1 View Post
    I'm stuck, can anyone help me with this "easy" statistics?
    X Y p(x,y)
    0 0 0,20
    0 1 0,25
    1 0 0,30
    1 1 0,25
    Var (X+Y) = ?
    E (Y[FONT=sans-serif][COLOR=#000000](X=0)
    Well, what kind of help? What had you done so far on each of them?
    What don't you understand on each?

    If you are expecting someone here to do them for you, that is not going to happen.
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  3. #3
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    Re: Var + e

    x and y, separately, can be 0 or 1 so x+ y can be 0, 1, or 2.
    What must x and y be so that x+ y= 0? What is the probability of that?
    What must x and y be so that x+ y= 1? What is the probability of that?
    What must x and y be so that x+ y= 2? What is the probability of that?
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  4. #4
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    Re: Var + e

    I did this...

    z pz(z)
    0 0,20
    1 0,55
    2 0,25

    Var(Z) = 0^2*0,20 + 1^2*0,55 + 2^2*0,25
    But it doesn't feel right
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  5. #5
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    Re: Var + e

    Quote Originally Posted by emme1 View Post
    I did this...

    z pz(z)
    0 0,20
    1 0,55
    2 0,25

    Var(Z) = 0^2*0,20 + 1^2*0,55 + 2^2*0,25
    But it doesn't feel right
    That is not right.

    Did you find E(Z)~?.

    How is \text{Var}(Z) defined in terms of E(Z)~?
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  6. #6
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    Re: Var + e

    Your probability distribution for z= x+ y, 0,20 that z= 0, 0,55 that z= 1, and 0,25 that z= 2, is correct. Now, the expected value is 0(0,20)+ 1(0,55)+ 2(0,25)= 1,10+ 0,50= 1,60. The variance, then, is 0,20(0- 1,60)^2+ 0,55(1- 1,60)^2+ 0,25(2- 1,60)^2. You used the values, not the variation from the expected value.
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