I am the only bank teller on duty at my local bank. I need to run out for 10 minutes, but I don’t want to miss many customers. Suppose the arrival of customers can be modeled by a Poisson distribution with mean 2 customers per hour.
a) What is the probability that no one will arrive in the next 10 minutes?
b) What’s the probability that 2 or more people arrive in the next 10 minutes?
c) you just served 2 customers who came in one after the other, is this a better time to run out?
a) What is the probability that no one will arrive in the next 10 minutes?
b) What’s the probability that 2 or more people arrive in the next 10 minutes?
c) you just served 2 customers who came in one after the other, is this a better time to run out?
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Gergr =
Poisson with λ = (2 / hr)(10/60) = 1/3
a) P(X=0) = e^-(1/3) (1/3)^0 / 0! = e^(-1/3) = 0.7165
b) P(X > or equal to 2) = 1 - P(X=0) - P(X=1) = 1 - 0.7165 - e^-(1/3) (1/3)^1 / 1!
= 1 - 0.7165 - 0.2388 = 0.0447
c) The Poisson has "independent" time intervals, so it doesn't matter if you just served 2, 100 or zero customers, the probability of what happens in the next 10 minute interval is totally independent of the past.
Hope that helps
Poisson with λ = (2 / hr)(10/60) = 1/3
a) P(X=0) = e^-(1/3) (1/3)^0 / 0! = e^(-1/3) = 0.7165
b) P(X > or equal to 2) = 1 - P(X=0) - P(X=1) = 1 - 0.7165 - e^-(1/3) (1/3)^1 / 1!
= 1 - 0.7165 - 0.2388 = 0.0447
c) The Poisson has "independent" time intervals, so it doesn't matter if you just served 2, 100 or zero customers, the probability of what happens in the next 10 minute interval is totally independent of the past.
Hope that helps