Question

Cars arrive at a toll booth according to a Poisson process with mean 70 cars per hour. Suppose the attendant makes a phone call. How long, in seconds, can the attendant's phone call last if the probability is at least 0.5 that no cars arrive during the call

Answer 1

35.6 seconds

Explanation:

The Poisson distribution has a probability model that follows the given relationship:

`P(X=x) = (\lambda^xe^(-\lambda))/(x!)`

For x = zero calls at a probability of 50%, the arrival rate during the call is:

`P(X=0)=0.50 = (\lambda^0e^(-\lambda))/(0!)\\e^(-\lambda)=0.50\\\lambda=0.6931\ cars/ n\ seconds`

Since cars arrive at a rate of 70 per hour, the length of call 'n' in seconds is:

`r = (70)/(h)*(1h)/(3,600s) = 0.019444\ cars/s\\n= (\lambda)/(r)=(0.6931)/(0.019444) =35.6\ seconds`

The call should be at most 35.6 seconds long to ensure a 50% probability that no cars arrive during the call.