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.

Question

Margot Mill · Accepted Answer

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

Jeffrey Jordon · Accepted Answer

A) probability of no arrivals in a one-minute period&amp;nbsp;f(0)=[100×e−10]0!f(0) = 0.00004539993To six decimal places: f(0) = 0.000045&amp;nbsp;B) the probability that three or fewer passengers arrive in a one-minute period:P(x≤3)=f(0)+f(1)+f(2)+f(3)f(1)=[101×e−10]1!f(1) = 0.000454&amp;nbsp;f(2)=[102×e−10]2!f(2) = 0.00227&amp;nbsp;f(3)=[103×e−10]3!f(3) = 0.007567&amp;nbsp;P(x≤3)=0.000045+0.00227+0.007567=0.009882To 4 decimal places &amp;nbsp;P(x≤3)=0.0099&amp;nbsp;C) no arrivals in a 15-second periodμ=10×1560μ=2.5The probability of no arrivals in a 15-second period ia;f(0)=[2.50×e−2.5]0!f(0) = 0.082085To 4 decimal places is:f(0) = 0.0821&amp;nbsp;D)&amp;nbsp;P(x≥1)=1−f(0)P(x≥1)=1−0.0821P(x≥1)=0.9179

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

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