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Math Help - Poisson distribution

  1. #1
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    Poisson distribution

    Hi, I got this question here and I found it is really tricky to solve.

    20000 people are watching football match at the stadium. The number of cans of beer that a spectator wants to drink is distributed according to a Poisson distribution with parameter \lambda =0.4. How many cans of beer do organizer need to stock so that the probability that there is not enough beer is less than 0.05?

    I have no idea how to start, can anyone please give me some advice? Thanks a lot.
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  2. #2
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    The Poisson Distribution is perfectly scalable.

    1 guy -- \lambda_{1} = 0.40

    2 folks -- \lambda_{2} = 2*0.40 = 0.80

    35 homies -- \lambda_{35} = 35*0.40 = 14.00

    9432 peeps -- \lambda_{9432} = 9432*0.40 = 3772.80

    Where does that leave us?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    The Poisson Distribution is perfectly scalable.

    1 guy -- \lambda_{1} = 0.40

    2 folks -- \lambda_{2} = 2*0.40 = 0.80

    35 homies -- \lambda_{35} = 35*0.40 = 14.00

    9432 peeps -- \lambda_{9432} = 9432*0.40 = 3772.80

    Where does that leave us?

    Thanks a lot, I got what you mean. But I still don't know how to solve the question. So, is this the kind of situation that Poisson distribution with "large" parameter, so should I use Normal approximation? The one that use Normal distribution to solve the problem? Please kindly give me more advice. Thanks a lot.
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  4. #4
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    You could use a normal approximation with mean = standard deviation = 20000*0.40. In fact, due to the difficulty in calculating directly such Poisson probabilities, I would recommend the Normal Approximation. Just leave 0.05 in the right tail.

    Why is it difficult to calculate directly the Poisson probabilities?
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  5. #5
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    I'm still not quite sure what to do, so I should try to use Normal Approximation. But, I would be able to calculate the probability that enough beers, so the should the probability of not enough beers equal to 1 minus the probability of enough beers?
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  6. #6
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    ??? p(too many customers) = 0.05. This is a z-score lookup.

    Using the empirical rule:

    p(customers exceeds the mean) = 50%
    p(customers exceed mean + 1 standard deviation) = 84%
    p(customers exceed mean + 2 standard deviation) = 97.5%

    What will it take to get 99.5%?
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  7. #7
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    I tried to do something like this:

    P(Not\ enough\ beers)=1-P(Enough\ beers)<0.05
    1-P(z\leq \frac{x-8000}{40\sqrt{5}})<0.05

    Then I solved for x and I got the answer is x=8147, by using table of standard normal for z values.

    So that means there are 8147 cans of beer need to be stocked.

    Did I do right things? I just kind of feel the number 8147 is not reasonable enough.
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  8. #8
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    What is 40\cdot\sqrt{5}?

    It's Poisson! mean = standard deviation = 20000*0.40
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  9. #9
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    Quote Originally Posted by TKHunny View Post
    What is 40\cdot\sqrt{5}?

    It's Poisson! mean = standard deviation = 20000*0.40
    Sorry I'm confused, because I saw my notes say for Poisson mean=variance=parameter, which is 20000*0.4, that's why I squared it to get standard deviation.
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  10. #10
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    It appears I need a nap. You have it right. Many apologies.
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