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

Anonym 2021-06-16 Answered
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.
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Margot Mill
Answered 2021-06-17 Author has 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
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Jeffrey Jordon
Answered 2021-10-12 Author has 2047 answers

A) probability of no arrivals in a one-minute period will be;

f(0)=[100×e10]0!

f(0) = 0.00004539993

To six decimal places gives;

f(0) = 0.000045

 

B) the probability that three or fewer passengers arrive in a one-minute period is given by;

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

We already have f(0) = 0.000045

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

 

Thus;

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

To four decimal places gives;

P(x3)=0.0099

 

C) no arrivals in a 15-second period means that;

μ=10×1560

μ=2.5

Thus;

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

f(0)=[2.50×e2.5]0!

f(0) = 0.082085

To four decimal places is;

f(0) = 0.0821

 

D) the probability of at least one arrival in a 15-second period will be gotten from the complement rule. Thus;

P(x1)=1f(0)

P(x1)=10.0821

P(x1)=0.9179

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