Customers arrive at a restaurant according to a Poisson distribution at a rate of 20 customers per hour. The restaurant opens for business at 11:00 am. (Assume that customers will stay long enough). Find the probability of having 20 customers in the restaurant at 11:12 am given that there were 17 customers at 11:07 am.

0.087

Step-by-step explanation:

Given that there were 17 customers at 11:07, probability of having 20 customers in the restaurant at 11:12 am could be computed as:

= Probability of having 3 customers in that 5 minute period. For every minute period, the number of customers coming can be modeled as:

X₅ ~ Poisson (20 (5/60))

X₅ ~ Poisson (1.6667)

Formula for computing probabilities for Poisson is as follows:

P (X=ₓ) = ((e^(-λ)) λˣ)/ₓ!

P(X₅= 3) = ((e^(-λ)) λˣ)/ₓ! =  (e^-1.6667)((1.6667²)/3!)

P(X₅= 3) = (2.718^(-1.6667))((2.78)/6)

P(X₅= 3) = (2.718^(-1.6667))0.46

P(X₅= 3) = 0.1889×0.46

P(X₅= 3) = 0.086894

P(X₅= 3) = 0.087

Therefore, the probability of having 20 customers in the restaurant at 11:12 am given that there were 17 customers at 11:07 am is 0.087.

add an image !

step-by-step explanation:

answer is : muchas/tipicas

have a nice day!

~recklesschicken