Respuesta :
Answer:
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 ;
[tex]P(X=x) = \frac{e^{-\lambda }\times \lambda^{x} }{x!} ;x=0,1,2,3,.....[/tex]
where, [tex]\lambda[/tex] = arrival rate
Let X = Arrival of Airline passengers
The mean of the Poisson distribution is given by = E(X) = [tex]\lambda[/tex], which is given to us as 6 passengers per minute.
So, X ~ Poisson ([tex]\lambda=6[/tex])
Now, the probability of no arrivals in a one-minute period is given by = P(X = 0)
P(X = 0) = [tex]\frac{e^{-6}\times 6^{0} }{0!}[/tex]
= [tex]e^{-6}[/tex]
= 0.002479
Hence, the probability of no arrivals in a one-minute period is 0.002479.
Answer:
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.
[tex]\lambda = 6[/tex]
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:
[tex]P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}[/tex]
P( no arrivals in a one-minute period)
[tex]P( x =0) \\\\= \displaystyle\frac{6^0 e^{-6}}{0!}= 0.002478[/tex]
0.002478 is the probability of no arrivals in a one-minute period.
