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Math Help - normal distributions

  1. #1
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    normal distributions

    I have two similar (but not identical) normal distrubutions. Let's say distribution "A" has a mean of a and an SD of A, and distribution "B" has a mean of b and an SD of B. QUESTION: What is the probability that a score from distribution A will be greater than a score from distribution B? Is there a formula in terms of a,b,A and B?
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  2. #2
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    Re: normal distributions

    Consider the distribution A - B. What is its mean? What is its Standard Deviation? What is p(A-B>0)?
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    Re: normal distributions

    Thanks for that. Got it sorted. Now I'm up against the next stage question. Consider 3 teams (A,B & C) in a league. The probability that A finishes the season higher than B is P1 and the probability that A finishes higher than C is P2. Q: What is the probability (P3) that A finishes higher in the league than BOTH B & C? (I know that the answer is not P1 x P2 by considering the case of 3 equal teams where you would have P1 = 0.5, P2 = 0.5 and P3 = 0.3333)
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  4. #4
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    Re: normal distributions

    Should we be on a different thread?

    How about something in the neighborhood of this: p(A>B|B>C) + p(A>C|C>B). Of course, one would want to generalize this before attempting too many teams.
    Last edited by TKHunny; July 19th 2011 at 03:23 AM.
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  5. #5
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    Re: normal distributions

    Quote Originally Posted by TKHunny View Post
    Should we be on a different thread?

    How about something in the neighborhood of this: p(A>B|B>C) + p(A>C|C>B). Of course, one would ant to generalize this before attempting too many teams.
    I'm new on here, so I'm not sure now how to move this converstaion to the right area.

    My maths degree is 25 years ago and I'm struggling with your notation. How do I actually calculate p(A>B|B>C) - (it cannot just be 0.5 x 0.5 becasue this would give a final answer of 0.25 + 0.25 = 0.5 for 3 equal teams which is clearly incorrect.

    Sorry to be a pain!
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  6. #6
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    Re: normal distributions

    Read p(A>C|B>C) "the probability that A finishes higher than C, given that B finished higher than C."

    Off the top of my head, it does not appear to be enough information. I'm looking for p(B>C) or p(C>B) and we've neither.

    Alternatively, you could hald a tournament. There are only three games.

    A plays B
    A plays C
    B plays C

    Each team has only two games and this, three outcomes (0,2), (1,1), (2,0). Can you assign probabilities to these outcomes?
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  7. #7
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    Re: normal distributions

    OK, thank you. I understand the "given that" concept. P(B>C) is 1-P(C>B) and lets call this P3 for the general situation. As stated earlier P1 is P(A>B) and P2 is P(A>C). So the question is:

    What is P(A> both B & C) in terms of P1, P2 and P3?

    A simulated tornament is tricky. In the situation I'm trying to grasp, teams A,B,C are any three chosen teams from a league with many more teams in it, and P1, P2, P3 are calulated using your previous help on comparing normal distributions.

    Please ignore earlier notation where I used "P3" to donote the answer.
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