Question

A morning traffic volume on a major road was 640 vehicles during 30 minutes, and a Poisson distribution can describe the arrival of traffic. The YIELD signs control unsignaled intersections, where a car from a minor road require a 7 seconds minimum acceptable gap to cross the major road. During this period, estimate the number of acceptable gaps on the major roads for the cars on the minor road to cross.

Answer 1

To estimate the number of acceptable gaps on the major road for cars on the minor road to cross we can use the Poisson distribution.

The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space given the average rate of occurrence.

In this case we are given that the average traffic volume during a 30-minute period is 640 vehicles on the major road. We can use this average rate to calculate the average number of vehicles passing through the major road in 7 seconds.

To do this we need to convert the 30-minute average to a 7-second average. Since there are 60 minutes in an hour and 60 seconds in a minute there are 60*60 = 3600 seconds in an hour. Therefore the average rate of vehicles in a 7-second interval would be:

rate = (640 vehicles / 30 minutes) * (1 minute / 60 seconds) * 7 seconds = 3.111 vehicles

Now we can use the Poisson distribution to estimate the number of acceptable gaps on the major road for cars on the minor road to cross.

Let's denote the number of acceptable gaps by the random variable X. Given the rate of 3.111 vehicles per 7 seconds we can use the Poisson distribution formula:

P(X = k) = (e^-λ * λ^k) / k!

where λ is the average number of occurrences in the interval and k is the number of occurrences we want to estimate.

For k = 0 to 10 (we can choose any appropriate range we can calculate the probabilities:

P(X = 0) = (e^-3.111 * 3.111^0) / 0! = 0.043

P(X = 1) = (e^-3.111 * 3.111^1) / 1! = 0.134

P(X = 2) = (e^-3.111 * 3.111^2) / 2! = 0.209

P(X = 3) = (e^-3.111 * 3.111^3) / 3! = 0.206

P(X = 4) = (e^-3.111 * 3.111^4) / 4! = 0.170

P(X = 5) = (e^-3.111 * 3.111^5) / 5! = 0.113

P(X = 6) = (e^-3.111 * 3.111^6) / 6! = 0.063

P(X = 7) = (e^-3.111 * 3.111^7) / 7! = 0.030

P(X = 8) = (e^-3.111 * 3.111^8) / 8! = 0.013

P(X = 9) = (e^-3.111 * 3.111^9) / 9! = 0.005

P(X = 10) = (e^-3.111 * 3.111^10) / 10! = 0.002

To estimate the number of acceptable gaps we sum all the probabilities:

Estimated number of acceptable gaps = Σ P(X = k) for k = 0 to 10

= 0.043 + 0.134 + 0.209 + 0.206 + 0.170 + 0.113 + 0.063 + 0.030 + 0.013 + 0.005 + 0.002

≈ 1.078

Therefore we estimate that there are approximately 1.078 acceptable gaps on the major road for the cars on the minor road to cross during this 7-second period.