Question

A toll plaza has 5 booths, with each booth capable of servicing 50 cars per hour. Cars arrive at the plaza at the rate of 225 cars per hour.

Make the standard assumptions of a Poisson distribution for arrivals, and an Exponential distribution for service times, and calculate the following:

a) What is the probability of zero cars in the toll plaza?

b) What is the average length (in cars) of the (total) queue?

Answer 1

The probability of zero cars in the toll plaza is e^-3.75. The average length of the total queue is 8.93 cars.

Step-by-step explanation:

To calculate the probability of zero cars in the toll plaza, we need to use the Poisson distribution. The average arrival rate is 225 cars per hour or 3.75 cars per minute. We can calculate the probability of zero cars using the formula: P(X = 0) = (e^-lambda) * (lambda^k) / k!, where lambda is the average arrival rate and k is the number of arrivals. In this case, lambda = 3.75 and k = 0, so the probability is P(X = 0) = (e^-3.75) * (3.75^0) / 0! = (e^-3.75) * 1 = e^-3.75. To calculate the average length of the total queue, we need to consider the arrival rate and the service rate. The service rate is 5 booths * 50 cars per hour = 250 cars per hour or 4.17 cars per minute. The average length of the queue can be calculated using the formula: L = lambda / (mu - lambda), where lambda is the arrival rate and mu is the service rate. In this case, lambda = 3.75 and mu = 4.17, so the average length of the queue is L = 3.75 / (4.17 - 3.75) = 3.75 / 0.42 = 8.93 cars.