Airline passengers arrive randomly and independently at a passenger screening fa

Question

Airline passengers arrive randomly and independently at a passenger screening facility in a major international airport. The mean arrival rate is 5 passengers per minute. What is the probability of 0 arrivals in a fifteen second period? (Round to whole number )

a. 19%     b. 48%     c. 33%     d. 29%     e. none of these

For the problem above what is the variance of x for a fifteen second period?

hint: there is something special about the distribution used above in regards to the relationship between the mean and standard deviation

a. 4.0        b. 1.25           c. 1.45         d. 0.5         e. none of these

Explanation / Answer

Solution:

Here X : counts the total number of passengers arrive during a given time period

µ = 5

X ~ Poisson(5)

The mean arrival rate is 5 passengers per minute hence in 15 seconds 5/4 = 1.25 passangers will be arrive

P ( X = 0 ) = (e-) (x) / x!

=((e^-1.25)*(1.25^0))/0!

=0.2865

   0.29

That is 29% is the probability of 0 arrivals in a fifteen second period.

Hence Answer is option d.

We know that, If is the average number of successes occurring in a given time interval or region in the Poisson distribution, then the mean and the variance of thePoisson distribution are both equal to . here for 15 sec average number os success is 1.25.

Hence standard deviaation is 1.25

Hence option b. is correct.

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