Question

Say that the number of requests for towing from OU Parking Services follows a Poisson distribution with a rate of four per hour. a. What is the probability that exactly 10 students, staff, and faculty get their cars towed in a two-hour period? b. If the towing operators take a 30-minute lunch break, what is the probability that they do not miss any calls from OU Parking?

Answer 1

We know that the mean number of requests, lambda= 4 per hour.

a.

We must find the probability that exactly 10 students get their cars towed in a 2 hour period.

The mean for 1 hour is 4. Thus, the mean for 2 hours is 2*4= 8.

Lambda=8

P(X=x)= (lambda^x * e^-lambda ) / x!

x= 10

Thus, P(X=10)= (8¹⁰ * e^-8 ) / 10!

= 0.09926

b.

We must find the probability that exactly 0 students get their cars towed in a 30 minute period.

The mean for 1 hour is 4. Thus, the mean for half hour is 4/2= 2.

Lambda=2

P(X=x)= (lambda^x * e^-lambda ) / x!

x= 0

Thus, P(X=0)= (2⁰ * e^-2 ) / 0!

= 0.13534