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    Can someone please tell me how to answer this?

    X-Po(2) Y-Po(3)
    W=XY

    Find P(W=4)
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    (Original post by Pinkdragon)
    Can someone please tell me how to answer this?

    X-Po(2) Y-Po(3)
    W=XY

    Find P(W=4)
    If we assume X and Y are two independent random variables we can see that the expected value of XY, i.e. E(XY) would be E(X)E(Y). In this case E(XY) or E(W) in other words would equal 6.

    Therefore we can write W as being Poisson distributed with a distribution of W~Po(6).

    From there we can use a table or calculator to evaluate: (e^-6)(6^4)/4! which gives:
    0.1339 (4dp).
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    (Original post by Pinkdragon)
    Can someone please tell me how to answer this?

    X-Po(2) Y-Po(3)
    W=XY

    Find P(W=4)
    I don't get what is "Po" ? Is this teletubbies?
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    (Original post by hiq)
    If we assume X and Y are two independent random variables we can see that the expected value of XY, i.e. E(XY) would be E(X)E(Y). In this case E(XY) or E(W) in other words would equal 6.

    Therefore we can write W as being Poisson distributed with a distribution of W~Po(6).

    From there we can use a table or calculator to evaluate: (e^-6)(6^4)/4! which gives:
    0.1339 (4dp).
    For part a) to this question, I worked out that E(XY)=6 and Var(XY)=36. Would this change the answer because for poison mean and variance are the same?
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    (Original post by Pinkdragon)
    For part a) to this question, I worked out that E(XY)=6 and Var(XY)=36. Would this change the answer because for poison mean and variance are the same?
    I think you've made a mistake calculating the combination variance:

    Var(W) = Var(XY) = Var(X)Var(Y) = (2)(3) = 6

    Therefore: Var(W) = E(W) and Poisson would be a justified model for this problem.
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    (Original post by hiq)
    I think you've made a mistake calculating the combination variance:

    Var(W) = Var(XY) = Var(X)Var(Y) = (2)(3) = 6

    Therefore: Var(W) = E(W) and Poisson would be a justified model for this problem.
    What exam board is this from?

    I have edexcel s2 in 2 days and kinda worried never seen questions of this type.
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    (Original post by Hamzah249)
    What exam board is this from?

    I have edexcel s2 in 2 days and kinda worried never seen questions of this type.
    I think the concept is first met in S1 (edexcel) with the idea of event independence with probability:

    "If two events are independent P(AnB) = P(A) x P(B)"

    You don't really get into combinations of distributions till S3 though, but even in S3 you're just combining normal distributions which is a little different.
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    (Original post by hiq)
    I think the concept is first met in S1 (edexcel) with the idea of event independence with probability:

    "If two events are independent P(AnB) = P(A) x P(B)"

    You don't really get into combinations of distributions till S3 though, but even in S3 you're just combining normal distributions which is a little different.
    Oh, that's a relief.
 
 
 
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