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

Anonym

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Answered question

2021-06-16

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.

Answer & Explanation

Margot Mill

Margot Mill

Skilled2021-06-17Added 106 answers

a. Compute the probability of no arrivals in a one-minute period (to 6 decimals) = .000017 
b. Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals) = .0049 
c. Compute the probability of no arrivals in a 15-second period (to 4 decimals) = .0639 
d. Compute the probability of at least one arrival in a 15-second period (to 4 decimals) = .9361

Jeffrey Jordon

Jeffrey Jordon

Expert2021-10-12Added 2605 answers

A) probability of no arrivals in a one-minute period f(0)=[100×e10]0!

f(0) = 0.00004539993

To six decimal places: f(0) = 0.000045

 

B) the probability that three or fewer passengers arrive in a one-minute period:

P(x3)=f(0)+f(1)+f(2)+f(3)

f(1)=[101×e10]1!

f(1) = 0.000454

 

f(2)=[102×e10]2!

f(2) = 0.00227

 

f(3)=[103×e10]3!

f(3) = 0.007567

 

P(x3)=0.000045+0.00227+0.007567=0.009882

To 4 decimal places  P(x3)=0.0099

 

C) no arrivals in a 15-second period

μ=10×1560

μ=2.5

The probability of no arrivals in a 15-second period ia;

f(0)=[2.50×e2.5]0!

f(0) = 0.082085

To 4 decimal places is:

f(0) = 0.0821

 

D) P(x1)=1f(0)

P(x1)=10.0821

P(x1)=0.9179

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