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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).

User PKumar
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2 Answers

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Answer:

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

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(\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(6^0 e^(-6))/(0!)= 0.002478

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

User Nick Pampoukidis
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7.2k points
0 votes

Answer:

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

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) = (e^(-\lambda )* \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) =
(e^(-6)* 6^(0) )/(0!)

=
e^(-6)

= 0.002479

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

User Aefxx
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