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Poisson distribution

  1. Dec 22, 2009 #1
    Nr of customers arriving at a shop follow Poisson.
    In 15, an average of 4 customers arrive.

    a)
    A customer has just arrived. Then a minute passed and no one arrived. What is the probability of it takoing at least 5 more min. until another customer arrives?

    b)
    Consider 40 non-overlapping periods of 15 min.

    What is the probability that
    at least 7 and at most 15 of those intervals have at most 2 customers arriving?

    In book, answer to
    a) is 0.2636
    b) is F(2.22) - F(-1.12) = 0.855

    In a) although I don't see why, I understande that it's something about memorylessness or something. But how do you get to the answer in question b)?
     
  2. jcsd
  3. Dec 22, 2009 #2
    via binomial distribution
     
  4. Dec 22, 2009 #3
    Ok, I see now.Thank you.
     
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