In: Statistics and Probability
Pr9.
Suppose the emergency room at Mass General opens at 6am and has a mean arrival rate throughout the day of 6.9 patients per hour (that is λ = 6.9).
(A) What is the probability that 12 patients arrive between 6am and 7am?
(B) What is the probability that no patient arrives before 7am?
(C) What is the probability that the first patient arrives between 6am and 7am?
(D) What is the probability that the first patient arrives between 6:15 and 6:45?
(E) Suppose it is 6:15 and no patient has arrived yet; now what is the probability that the first patient arrives between 6:15 and 6:45?
Hint: Use the Poisson for (A) and (B) and the Exponential for (C), (D), and (E). Note carefully how (D) and (E) are different: in (D) it is possible that the first patient arrives between 6am and 6:15am, whereas for (E) you know this has not happened. The result for the two will be different!
Pr.9
λ = 6.9 patients per hour
(a) Here p(x) = e-λ λx/x!
p(x = 12) = e-6.9 6.912/12!
p(x = 12) = 0.0245
(b) P(x = 0) = e-6.9 6.90/0! = 0.0010
(c) P(First patient arrive between 6:00 AM to 7:00 AM) = 1 - P(No patient in 6:00 AM to 7:00 AM)
= 1 - 0.0010 = 0.9990
(d) P(First patient arrive between 6:15 AM to 6:45 AM) = P(No patient between 6:00 AM to 6:15 AM) * P(first patient is in between 6:15 AM to 6:45 AM)
Expected Number of patients arrive in between 6:00 AM to 6:15 AM = 6.9/4 = 1.725
P(No patient between 6:00 AM to 6:15 AM) = e-1.725 = 0.1782
P(first patient is in between 6:15 AM to 6:45 AM) = 1- P(No patient arrives in between 6:15 AM to 6:45 AM)
=1 - e-1.725*2 =1 - 0.0317 = 0.9682
P(First patient arrive between 6:15 AM to 6:45 AM) = 0.1782 * 0.9682 = 0.1725
(E) As poisson and exponential distribution has memoryless property, so whatever happened in the past doesn't affect the future.
P(First patient arrives in between 6:15 AM to 6:45 AM when there is no patient in First 15 minutes) = 1 - P(No patient arrives in between 6:15 AM to 6:45 AM when there is no patient in First 15 minutes)= 1 - e-3.45 = 0.9682