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Poisson Distribution Problem

  1. Aug 7, 2013 #1
    1. The problem statement, all variables and given/known data

    On the average, a grocer sells 4 of a certain article per week. How many of these should he have in stock so that the chance of his running of stock within a week will be less than 0.01? Assume Poisson distribution.

    2. Relevant equations

    3. The attempt at a solution

    So I set λ = 4, plugged it into e^(-λ)λ^(x) / x! set it <0.01 and looked for an x that brought the equation to <0.01.

    I was unsure how solve this for x, because of the x! in the bottom, so I just started with x = 0 and plugged and chugged till I came across x = 10, P(X=10)=0.0053 < 0.01.

    However this is incorrect. Any suggestions?
  2. jcsd
  3. Aug 7, 2013 #2


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    What event does that represent the probability of in the context of this question?
    Also, there is a possible ambiguity in the question. If the storekeeper stocks N items and gets requests for exactly N then technically she has 'run out', but it is not a concern to her unless there is a demand for N+1 or more.
  4. Aug 7, 2013 #3
    Thanks for the quick reply...

    From my limited knowledge of Poisson distribution, lambda represents the average rate of success, in the case of the problem at hand, an average of 4 successful sales per week.

    So my thought was that given the average rate of sales = 4 per week, find the amount of stuff he should stock, 'x', so that its probability is less that 0.01.

    P(X=x)<0.01 given average 4 sales per week
    Last edited: Aug 7, 2013
  5. Aug 7, 2013 #4


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    Yes, except that I would take it as the rate of opportunities to sell, i.e. requests for the item. The actual number of sales will depend on the number stocked.
    Sure, but that's the probability of exactly x requests for the item. There could be more.
    Last edited: Aug 8, 2013
  6. Aug 8, 2013 #5

    Ray Vickson

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    If X~Po(4) is the demand and the grocer starts the week with N in stock, then he requires P(X > N) < 0.01, or P(X ≤ N) > 0.99.
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