Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 9 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).
A) probability of no arrivals in a one-minute period will be; f(0)=[100×e−10]0! f(0)=[100×e−10]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(x≤3)=f(0)+f(1)+f(2)+f(3) P(x≤3)=f(0)+f(1)+f(2)+f(3) We already have f(0) = 0.000045 f(1)=[101×e−10]1! f(1)=[101×e−10]1! f(1) = 0.000454
Thus; P(x≤3)=0.000045+0.00227+0.007567=0.009882 P(x≤3)=0.000045+0.00227+0.007567=0.009882 To four decimal places gives; P(x≤3)=0.0099 P(x≤3)=0.0099
C) no arrivals in a 15-second period means that; μ=10×1560μ=10×1560μ=2.5μ=2.5 Thus; the probability of no arrivals in a 15-second period ia; f(0)=[2.50×e−2.5]0! f(0)=[2.50×e−2.5]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(x≥1)=1−f(0)P(x≥1)=1−f(0)P(x≥1)=1−0.0821P(x≥1)=1−0.0821P(x≥1)=0.9179P(x≥1)=0.9179
