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

Answer:

35.6 seconds

Step-by-step explanation:

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

$P(X=x) = \frac{\lambda^xe^{-\lambda}}{x!}$

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

$P(X=0)=0.50 = \frac{\lambda^0e^{-\lambda}}{0!}\\e^{-\lambda}=0.50\\\lambda=0.6931\ cars/ n\ seconds$

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

$r = \frac{70}{h}*\frac{1h}{3,600s} = 0.019444\ cars/s\\n= \frac{\lambda}{r}=\frac{0.6931}{0.019444} =35.6\ seconds$

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