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Poisson DIstribution

  1. Dec 26, 2011 #1
    1. The problem statement, all variables and given/known data
    A hardwhere store sells on average 8 drills per week.
    The store recieves ONE delivery of drills at the same time each week.
    Find the no. of drills that need to be in stock after a delivery for there to be at most a 5% chance of the store NOT having sufficent drills to meet demand before the next delivery.


    2. Relevant equations
    3. The attempt at a solution

    Ok so we know that Y≈Po(8)
    where Y = no. expected to sell
    lets say n = demand no.

    thus we want
    p(n > Y) < 0.05
    so
    P(Y<n) < 0.05

    BUT THE ANSWER SAYS
    P(Y>n) < 0.05

    How is this possible
    thanks
     
  2. jcsd
  3. Dec 26, 2011 #2

    micromass

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    Aren't your Y and n thesame thing. The number expected to sell surely equals the demand??

    I think your n should be something like "drills in store".
     
  4. Dec 26, 2011 #3
    how can no. expected = demand

    the Poisson value = E(X) but demand could be greater than or smaller than this, surely
     
  5. Dec 26, 2011 #4
    ahh
    Y is a distribution which means it could take on OTHER values but not def.
    so Y = demand
    n = no. in stock
    with Y-P(8)

    So
    P(Y>n) <= 0.05

    I see
    ok thanks
     
  6. Dec 26, 2011 #5

    Ray Vickson

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    You need to distinguish between sales and demand. If demand = X~Po(8) and we start the week with stock N, then sales S = min(X,N); that is, sales = demand if demand <= N, and sales = N otherwise. You need to choose N so that P(X > N) <= 0.05; that is, there is a chance of no more than 5% that demand exceeds on-hand stock. (By the way, that is <= 0.05, not < 0.05, because that is what 'no more than' means.)

    RGV
     
  7. Dec 27, 2011 #6
    cheers :)
     
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