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Math Help - Binomial Distribution

  1. #1
    Yan
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    Binomial Distribution

    In planning the operation of a new school, one school board member claims that four out of five newly hired teachers will stay with the school for more than a year, while another school board member claims that it would be correct to say three out of five. In the past, two board members have been about equally reliable in their predictions, so in the absence of any other information we would assign their judgments equal weight. If one or the other has to be right, what probabilities would we assign to their claims if it were found that 11 of 12 newly hired teachers stayed with the school for more than a year?
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    Quote Originally Posted by Yan View Post
    In planning the operation of a new school, one school board member claims that four out of five newly hired teachers will stay with the school for more than a year, while another school board member claims that it would be correct to say three out of five. In the past, two board members have been about equally reliable in their predictions, so in the absence of any other information we would assign their judgments equal weight. If one or the other has to be right, what probabilities would we assign to their claims if it were found that 11 of 12 newly hired teachers stayed with the school for more than a year?
    X ~ Binomial with n = 12.

    Calculate Pr(X > 10 | p = 4/5).

    Calculate Pr(X > 10 | p = 3/5).

    Draw a conclusion.
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  3. #3
    Yan
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    Quote Originally Posted by The Second Solution View Post
    X ~ Binomial with n = 12.

    Calculate Pr(X > 10 | p = 4/5).

    Calculate Pr(X > 10 | p = 3/5).

    Draw a conclusion.
    How to calculate this? Can you show me step by step? Thanks!!!
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  4. #4
    Flow Master
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    Quote Originally Posted by Yan View Post
    How to calculate this? Can you show me step by step? Thanks!!!
    X ~ Binomial(n = 12, p = 4/5). Calculate \Pr(X = 11).

    X ~ Binomial(n = 12, p = 3/5). Calculate \Pr(X = 11).

    Surely you're familar with how to do these sorts of calculation ....?
    Last edited by mr fantastic; February 11th 2009 at 10:18 PM.
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  5. #5
    Yan
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    Quote Originally Posted by mr fantastic View Post
    X ~ Binomial(n = 12, p = 4/5). Calculate \Pr(X > 10) = \Pr(X = 11) + \Pr(X = 12).

    X ~ Binomial(n = 12, p = 3/5). Calculate \Pr(X > 10) = \Pr(X = 11) + \Pr(X = 12).

    Surely you're familar with how to do these sorts of calculation ....?
    but the answer wasn't right! in my solution sheet, the answer is 0.9222. can you explain more detail?
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    Quote Originally Posted by Yan View Post
    but the answer wasn't right! in my solution sheet, the answer is 0.9222. can you explain more detail?
    You need to incorporate the prior probability ....

    Quote Originally Posted by Yan View Post
    How to calculate this? Can you show me step by step? Thanks!!!
    You need to calculate Pr(p = 4/5 | data).

    Prior probability: Pr(p = 4/5) = 1/2 and Pr(p = 3/5) = 1/2.

    Pr(data | p = 4/5) = Pr(X = 11 | p = 4/5) = 0.20616.

    Pr(data | p = 3/5) = Pr(X = 11 | p = 3/5) = 0.01741.

    Pr(p = 4/5 | data) = Pr(data | p = 4/5).Pr(p = 4/5)/Pr(data).

    Pr(data) = Pr(data | p = 4/5).Pr(p = 4/5) + Pr(data | p = 3/5).Pr(p = 3/5) = 1/2 (0.20616 + 0.01741).

    Therefore Pr(p = 4/5 | data) = 0.20616/(0.20616 + 0.01741) = 0.92213.

    Note that Pr(p = 3/5) = 0.01741/(0.20616 + 0.01741) = 0.07787.
    Last edited by mr fantastic; February 11th 2009 at 10:27 PM.
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