1. Poisson Distribution Question

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?

2. Originally Posted by 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?
If this mean what is the probability that she passess 2 customers total in the
two time slices, then it does not matter that they are disjoint, this is the
same probability that she passess two customers in a two hour interval.

RonL

3. I'm not sure I fully understand. Are you saying that it doesnt make a difference if the intervals are split or joined, implying that the mean would be 14=2*7?

4. Originally Posted by lllll
I'm not sure I fully understand. Are you saying that it doesnt make a difference if the intervals are split or joined, implying that the mean would be 14=2*7?

Yes the mean would be that, and the number has a Poisson distribution
with that mean.

RonL