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Math Help - Poisson Distribution

  1. #1
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    Poisson Distribution

    A cement pylon is considered safe if the hairline cracks are spread out and occur on the average of 1.2 cracks per 10-ft length of cement. In a 40-ft section of the cement pylon, what's the probability of 4 or more hairline cracks?

    P(X=x)=\frac{\lambda^xe^{-\lambda}}{x!}

    So \lambda is the average rate (1.2 per 40ft?) and x is the # of events.

    ?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by JJ007 View Post
    A cement pylon is considered safe if the hairline cracks are spread out and occur on the average of 1.2 cracks per 10-ft length of cement. In a 40-ft section of the cement pylon, what's the probability of 4 or more hairline cracks?

    P(X=x)=\frac{\lambda^xe^{-\lambda}}{x!}

    So \lambda is the average rate (1.2 per 40ft?) and x is the # of events.

    ?
    I would think that \lambda=4.8 cracks per 40 ft.

    Thus, \mathbb{P}(X\geq 4) = 1-\mathbb{P}(X\leq 3)=1-\displaystyle\sum\limits_{x=0}^3\dfrac{(4.8)^xe^{-4.8}}{x!}\approx 0.7058

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    I would think that \lambda=4.8 cracks per 40 ft.

    Thus, \mathbb{P}(X\geq 4) = 1-\mathbb{P}(X\leq 3)=1-\displaystyle\sum\limits_{x=0}^3\dfrac{(4.8)^xe^{-4.8}}{x!}\approx 0.7058

    Does this make sense?
    How did you get 4.8? Also did you use 3! for x! ?

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  4. #4
    MHF Contributor harish21's Avatar
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    You have written 1.2 cracks per 10 feet. SO for 40 feet, you would have 1.2*4=4.8 cracks.

    And, for your 2nd question, Chris L T521's notation given above means:

     \displaystyle\sum\limits_{x=0}^3\dfrac{(4.8)^xe^{-4.8}}{x!} = \frac{(4.8)^0 e^{-4.8}}{0!}+\frac{(4.8)^1 e^{-4.8}}{1!}+\frac{(4.8)^2 e^{-4.8}}{2!}+\frac{(4.8)^3 e^{-4.8}}{3!}
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