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

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
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

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

$f\left(0\right)=\frac{\left[{10}^{0}×{e}^{-10}\right]}{0!}$

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;

$P\left(x\le 3\right)=f\left(0\right)+f\left(1\right)+f\left(2\right)+f\left(3\right)$

We already have f(0) = 0.000045

$f\left(1\right)=\frac{\left[{10}^{1}×{e}^{-10}\right]}{1!}$

f(1) = 0.000454

$f\left(2\right)=\frac{\left[{10}^{2}×{e}^{-10}\right]}{2!}$

f(2) = 0.00227

$f\left(3\right)=\frac{\left[{10}^{3}×{e}^{-10}\right]}{3!}$

f(3) = 0.007567

Thus;

$P\left(x\le 3\right)=0.000045+0.00227+0.007567=0.009882$

To four decimal places gives;

$P\left(x\le 3\right)=0.0099$

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

$\mu =10×\frac{15}{60}$

$\mu =2.5$

Thus;

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

$f\left(0\right)=\frac{\left[{2.5}^{0}×{e}^{-2.5}\right]}{0!}$

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;

$P\left(x\ge 1\right)=1-f\left(0\right)$

$P\left(x\ge 1\right)=1-0.0821$

$P\left(x\ge 1\right)=0.9179$