Sunday, February 14, 2016

Sally Monilla has started serving lunch at her new business called Salmonilla’s Diner. During lunch hours, they serve an average of 10 customers...

This is a question of queuing theory and the Poisson process would be applicable in this question. Sally typically serves 10 customers per hour.


The number of customers served in a given 15 minute window


= 10 customer/hour x 15 minutes x 1/60 hour/minute = 2.5 customers


This can also be termed as mean serving rate (number of customers served per unit time) or . Thus,


= 2.5 customers


The probability of serving less...

This is a question of queuing theory and the Poisson process would be applicable in this question. Sally typically serves 10 customers per hour.


The number of customers served in a given 15 minute window


= 10 customer/hour x 15 minutes x 1/60 hour/minute = 2.5 customers


This can also be termed as mean serving rate (number of customers served per unit time) or . Thus,


= 2.5 customers


The probability of serving less than 3 customer per block of 15 minutes can be determined as


P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)


where, x is the number of customers served in the 15 minute time period.


Thus, P (x < 3) = 


= 0.544.


 Thus, the probability that less than 3 customers are served in a given 15 minute time period is 0.544.


Hope this helps. 

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