SOLUTION: suppose that the arrival of patients at a herbalist clinic in a certain rural town is a poison process, with rate parameter =2.8 per hour. Calculate the probability that a herbal

Question 1088383: suppose that the arrival of patients at a herbalist clinic in a certain rural town is a poison process, with rate parameter =2.8 per hour.
Calculate the probability that a herbalist operating in this town for two hours;
(i)Will receive no patients (4marks)
(ii) Will wait for at least 30 minutes before receiving the first patient (5marks)
(i) What is the mean time lag between any two patients visiting the herbalist

Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
suppose that the arrival of patients at a herbalist clinic in a certain rural town is a poison process, with rate parameter =2.8 per hour.
Calculate the probability that a herbalist operating in this town for two hours;
(i)Will receive no patients
(ii) Will wait for at least 30 minutes before receiving the first patient
(i) What is the mean time lag between any two patients visiting the herbalist

Solution:
Given the arrival is a Poisson process (note spelling) with rate parameter 2.8 arrivals per hour. Note that Poisson processes have no memory, meaning that the arrival of one does not affect subsequent arrival times.

The pmf (probability mass distribution) of a Poisson process is given by the Poisson Distribution, with λ=mean number of arrivals per unit time.
P(X=k)=(λ^k)(e^(-λ))/k! [k ∈Z ≥ 0]

The given situation is:
λ=2.8 [period is 1 hour]
(i) k=0 [no patients]
P(X=0)=(λ^k)(e^(-λ))/k!
=(2.8^0)(e^(-2.8))/0!
=0.0608

(ii) λ=1.4, [period is 30 minutes]
P(X=0)=P(X=0)=(λ^k)(e^(-λ))/k!
=(1.4^0)(e^(-1.4))/0!
=0.247

(iii) λ=2.8 [period = 1 hour]
Mean time lag = 60 minutes /2.8 = 21.4 min.

Note that for a Poisson distribution, λ IS the mean rate.

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