Airline passengers arrive randomly and independently at the passenger-screening

Question

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

Explanation / Answer

Solution:-

The mean arrival rate is 11 passengers per minute = = 11

a) The probability of no arrivals in a one-minute period is 0.00000.

x = 0

By applying poisons distribution:-

P(x; ) = (e-) (x) / x!

P(x = 0) = 0.00000

b) The probability that three or fewer passengers arrive in a one-minute period is 0.00492.

x = 3

By applying poisons distribution:-

P(x; ) = (e-) (x) / x!

P(x < 3) = 0.00492.

c) The probability of no arrivals in a 15-second period is 0.06393.

The mean arrival rate for 15 seconds = = 11/4 = 2.75

By applying poisons distribution:-

P(x; ) = (e-) (x) / x!

P(x = 0) = 0.06393

d) The probability of at least one arrival in a 15-second period is 0.9361.

The mean arrival rate for 15 seconds = = 11/4 = 2.75

By applying poisons distribution:-

P(x; ) = (e-) (x) / x!

P(x > 1) = 0.9361

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