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**lllll** A cashier passes an average of 7 customers an hour. Using Poisson distribution, what is probability that she will pass 2 customers in 2 different 1 hour time frames (i.e between 1:00-2:00 & 3:00-4:00)?

Solution:

since they it's 2 diffent one hour periods, I would think that you would have: $\displaystyle 2 \left( \frac{7^1}{1!} e^{-7} + \frac{7^2}{2!} e^{-7} + e^{-7} \right) = 0.05927237$ , since she can have 0 in the first interval but 2 in the second one, or one in the first and another in the second. Is this correct?