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Math Help - Poisson distrubution

  1. #1
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    Poisson distrubution

    Nr of people entering a store per min. is 3.8.

    If 30 non-overlapping minutes are picked, then what is the probability that in at least 20 of these minutes, at least 4 people arrived to the store.

    I've calculated for P(X>=4) , which was approximately 0.5265.

    But how do you go on from here. Is it binomial now? Should I now approximate or use the central limit theorem or something?

    Btw, answer to question is 0.09, but how to solve?
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  2. #2
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    Quote Originally Posted by Allan89a View Post
    Nr of people entering a store per min. is 3.8.

    If 30 non-overlapping minutes are picked, then what is the probability that in at least 20 of these minutes, at least 4 people arrived to the store.

    I've calculated for P(X>=4) , which was approximately 0.5265.

    But how do you go on from here. Is it binomial now? Should I now approximate or use the central limit theorem or something?

    Btw, answer to question is 0.09, but how to solve?
    Let Y be the random variable 'number of non-overlapping minutes in which at least 4 people arrived'.

    Y ~ Binomial(n = 30, p = 0.5265).

    Calculate \Pr(Y \geq 20).

    I get 0.0868.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Let Y be the random variable 'number of non-overlapping minutes in which at least 4 people arrived'.

    Y ~ Binomial(n = 30, p = 0.5265).

    Calculate \Pr(Y \geq 20).

    I get 0.0868.
    Hi,

    Thanks for your post! I seem to have mixed up some concepts, but now I finally got to 1 -F(1.35)= 0.0885.

    And yeah, the answer in the book's probably approximated to two decimals.
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