Question

The number of requests for assistance received by a towing service is a Poisson process with rate ? = 4 per hour.

a) If the operators take a 30 min break for lunch, what is the probability that they do not miss any calls for assistance?

b) Assuming they work 10 total hours on a particular day. What is the probability that assist less than 30 people (consider and approximation to help you solve this part)?

c) Calculate the average waiting time between requests (consider a continuous distribution).

Answer 1

a) Here, ? = 2 in 30-mins. The operators will not miss any calls if none came in. Thus, the required probability is

Here, ? = 2 in 30-mins. The operators will not miss any calls if none came in. Thus, the required probability is P(X = 0) = e^-2 2^0/0! (P (X = x) = e^-lambda lambda^x/x!) = 0.135335 Here, ? = 40 in 10-hours. Therefore, P(X < 30) = P(X lessthanorequalto 29) = sigma^29_x 0 P(X = x) = sigma^29_x 0 e^-40 40^x/x! = 0.043228682 Mu = 1/4 = 0.25

b) Here, ? = 40 in 10-hours. Therefore,

c) Mu = 1/4 = 0.25