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Math Help - Poisson distribution question.

  1. #1
    Super Member Quacky's Avatar
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    Poisson distribution question.

    The question is about a mechanic.

    To summarise, he gets 7 calls per hour. The question wants to know what the probability is of him getting <5 callouts in one hour.
    I understand the formula for poisson, which is:
     \frac{e^{-\lambda}\times\lambda^r}{r!} where  \lambda is the mean. But I substitute \lambda =7, r=5 in and get the wrong answer. I then substiture \lambda = \frac{7}{60} and r=5 but this is still incorrect.

    The answer is 0.173 to 3 d.p.

    I have the poisson cumulative frequency tables to save you some working.

    Please help?

    Edit: I now understand what \lambda represents, and know how to get the correct answer from my cumulative frequency table. I just don't understand what the r! represents?

    I don't expect a huge, half a page formula, because I wouldn't want to waste any one's time, but how would I get the answer without the Poisson cumulative frequency diagram?
    Last edited by Quacky; February 8th 2010 at 02:55 PM.
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  2. #2
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    P( R < 5 ) = P(0) + P(1) + P(2) + P(3) + P(4)
    = 0.1729

    R is just the number of events occurring, its usually denoted X. In your CDF table of poisson probabilities, look up Lamda = 7, X (or R) = 4
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  3. #3
    Super Member Quacky's Avatar
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    Oh, that obvious. Thanks, your simple explanation helped greatly.
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