Suppose a store expects to have 2 customers in any 15 minute interval, and that customers arrive according to a Poisson distribution. a) What is the probability that the store will have no customers from 10:00 to 10:15? b) What is the probability that the store will have more than 2 customers from 10:00 to 10:15? c) What is the probability that the store will have no customers from 9:30 to 10:00? d) What is the probability that the store will have 9 customers in any hour?
Added by Debbie S.
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So, we have: P(X = 0) = e^(-2) * 2^0 / 0! = e^(-2) ≈ 0.1353 Show more…
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