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the time between the arrival of two consecutive customers at a postoffice is 3 minutes, on average. Assuming that customers arrive in accordance with a Poisson process, find the probability that tomorrow during the lunch hour (between noon and 12:30 pm) fewer than seven customers arrive

Oct 14th, 2015

Thank you for the opportunity to help you with your question!

the average number of customers arrive in 30 minutes is 30/3 = 10

so the probability is 

P (N<=6) = (sum of n from 0 to 6) 

exp(-10)10^n/n! = exp(-10) (1+10+100/2+1000/6+10000/24+10^5/120+10^6/720)

=   0.130

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 14th, 2015

I'm a little confused... because the arrival of 2 customers is 3 minutes, on average. So, lambda should be 2/3 instead of 1/3??? can you explain me how do you assume that?

Oct 14th, 2015

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Oct 14th, 2015
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Oct 14th, 2015
Jun 26th, 2017
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