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Math Help - Poisson Question

  1. #1
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    Poisson Question

    The number X of new cases of food poisoning reported in a random month in a cafeteria has a Poisson distribution with mean 0.5. The cafeteria is required by law to request a state inspection of its kitchen whenever 2 or more cases of food poisoning occur in a month.
    a) What is the probability that the cafeteria will have to request an inspection this month?
    b) The number of new cases of food poisoning in any given month is independent of the number of cases of food poisoning in all other months of the year. Let Z be the number of months (out of a twelve month year) that the cafeteria must be inspected by the state. Find the probability that the cafeteria will be inspected more than once in a year.
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  2. #2
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Janu42 View Post
    The number X of new cases of food poisoning reported in a random month in a cafeteria has a Poisson distribution with mean 0.5. The cafeteria is required by law to request a state inspection of its kitchen whenever 2 or more cases of food poisoning occur in a month.
    a) What is the probability that the cafeteria will have to request an inspection this month?
    b) The number of new cases of food poisoning in any given month is independent of the number of cases of food poisoning in all other months of the year. Let Z be the number of months (out of a twelve month year) that the cafeteria must be inspected by the state. Find the probability that the cafeteria will be inspected more than once in a year.
    a) Well, we know \lambda=0.5, the average number food poisoning cases over a month period. We want to know the probability of two or more cases, since we are asked to find the probability of a cafeteria inspection. So, you want to find

    P(X\geq\\2)=\sum_{x=2}^{\infty}\frac{\lambda^{x}\\  e^{-\lambda}}{x!} or equivalently <br />
1-\sum_{x=0}^{1}\frac{\lambda^{x}\\e^{-\lambda}}{x!}, with \lambda=0.5.
    Last edited by Danneedshelp; December 11th 2009 at 10:51 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    I assume you want the first sum to start at 2.

    (b) is Binomial, n=12 and p is the probability from part (a).

    You want P(Z> 1)=1-[P(Z=0)+P(Z=1)]
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    MHF Contributor matheagle's Avatar
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    I just noticed the exponent of e is off, you need -\lambda
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by matheagle View Post
    I just noticed the exponent of e is off, you need -\lambda
    Wow, good save. Sorry about that.
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    MHF Contributor matheagle's Avatar
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    NP, that's what my mathEAGLE eyes r4.
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    Quote Originally Posted by matheagle View Post
    I assume you want the first sum to start at 2.

    (b) is Binomial, n=12 and p is the probability from part (a).

    You want P(Z> 1)=1-[P(Z=0)+P(Z=1)]

    So p from part (a) is the probability it gets inspected in a random month. n is obviously 12 for months of the year, so I use the simple binomial formula right? To get P ( Z = 0) and P (Z = 1)?
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  8. #8
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    Quote Originally Posted by Janu42 View Post
    So p from part (a) is the probability it gets inspected in a random month. n is obviously 12 for months of the year, so I use the simple binomial formula right? To get P ( Z = 0) and P (Z = 1)?
    What you have done is stated exactly what is in the quote. matheagle is crystal clear on what yuo have to do.
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