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Thread: Poisson Distribution - Optimisation

  1. #1
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    Poisson Distribution - Optimisation

    A store offers a new seasonal product featured. Let N be the random variable which means the number of clients who come to the store during the season, where N ~ Poi (19). It is estimated that the probability that a customer buys the new product is 0.38 regardless from a customer to another.

    a) It is assumed here that the store has an unlimited supply of product . variables are random X and Y such that

    X: The number of customers who purchase the product ;

    Y: The number of customers who do not buy the product.

    Are the variables X and Y independent ? Justify .

    Answer: I think they are dependent, because there is a fixed number of costumers, and when the number of buyers changes, the number of non buyers changes too.

    b) The store has a profit of 63millionforeachunitsold.Eachunsoldunitshouldbestoredfornextyearatacostof63millionforeachunitsold.Eachunsoldunitshouldbestoredfornextyearatacostof38 M .

    Determine what the number of units stored should be to maximize its average profit.

    Answer: Lets start with the pdf of Poisson distribution.

    px(k)=exp(−α)α^k/k! , for k=0,1,...


    px(k)=exp(−19)19^k/k! , for k=0,1,...

    Now i don't know how to relate my 0.38 to the equation i just came up with...

    Since 38% of the people buy, i need to first estimate how many costumers ill have with the Poisson distribution, then assume that 38% only will buy. So to maximize my profits, lets say there is 100 costumers estimated, ill only put 38, so ill sell everyone of them. But i don't know how to apply this logic to the problem

    Your help is really appreciated!

  2. #2
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    Re: Poisson Distribution - Optimisation

    The poisson distribution: p(k;\alpha)=\frac{\alpha^ke^{-\alpha}}{k!} for k=0,1,2,...

    Based on what you've written it looks like you have \alpha=19. But we know \alpha=n\theta. Seems to me \theta=.38 in this case?
    Last edited by Buckeye; 07-24-2016 at 05:20 PM.

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