Question

Exercise 5.49 (Algorithmic)}

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 11 passengers per minute.

1. Compute the probability of no arrivals in a one-minute period (to 6 decimals).
   
   
2. Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals).
   
   
3. Compute the probability of no arrivals in a 15-second period (to 4 decimals).
   
   
4. Compute the probability of at least one arrival in a 15-second period (to 4 decimals).

Answer 1

= 11

It is a Poisson distribution.

P(X = x) = e/x!​​​

A) P(X = 0) = e^(-11) * (11)^0/0! = 0.0000

B) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= e^(-11) * (11)^0/0! + e^(-11) * (11)^1/1! + e^(-11) * (11)^2/2! + e^(-11) * (11)^3/3! = 0.0049

C) Here, = 11/60 * 15 = 2.75

P(X = 0) = e^(-2.75) * (2.75)^0/0! = 0.0639

D) P(X > 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 -  (e^(-2.75) * (2.75)^0/0!))

= 1 - 0.0639 = 0.9361