Mathematics, 15.04.2020 02:30 gracie0818

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 6 passengers per minute. Compute the probability of no arrivals in a one-minute period (to 6 decimals).

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Answer from: officialrogerfp3gf2s

The probability of no arrivals in a one-minute period is 0.002479.

Step-by-step explanation:

We are given that Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 6 passengers per minute.

The above situation can be represented through Poisson distribution because a random variable which includes the arrival rate is considered as Poisson random variable.

The Probability distribution function for Poisson random variable is ;

$P(X=x) = \frac{e^{-\lambda }\times \lambda^{x} }{x!} ;x=0,1,2,3,.....$

where, $\lambda$ = arrival rate

Let X = Arrival of Airline passengers

The mean of the Poisson distribution is given by = E(X) = $\lambda$ , which is given to us as 6 passengers per minute.

So, X ~ Poisson ( $\lambda=6$ )

Now, the probability of no arrivals in a one-minute period is given by = P(X = 0)

P(X = 0) = $\frac{e^{-6}\times 6^{0} }{0!}$

= $e^{-6}$

= 0.002479

Hence, the probability of no arrivals in a one-minute period is 0.002479.

Answer from: nick607

0.002478 is the probability of no arrivals in a one-minute period.

Step-by-step explanation:

We are given the following information in the question:

Mean arrival rate = 6 passengers per minute.

$\lambda = 6$

The airline passengers can be treated as a Poisson distribution.

Poisson distribution.

The Poisson distribution is the discrete probability distribution of the number of events occurring in a given time period, given the average number of times the event occurs over that time period.

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

P( no arrivals in a one-minute period)

$P( x =0) \\\\= \displaystyle\frac{6^0 e^{-6}}{0!}= 0.002478$

0.002478 is the probability of no arrivals in a one-minute period.

Answer from: Quest

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step-by-step explanation:

Answer from: Quest

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step-by-step explanation:

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