xample 4 (borrowed from Durrett's book). There's a tailor shop in Annapolis (called Naval Tailor). Suppose the arrival rate starts at 0 at 10:00, increases to 4 at 12:00, to 6 by 2:00, drops to 2 by 4:00 and decreases to 0 by the time the store closes at 6:00, and that the arrival rates are linear in between these time points. 1. If the tailor decides to close the store at 5:30 instead of 6:00, what is the expected number of customers that are lost? Answer: Recall that the rate is linear and that the number of customers in between time periods is Poisson with rate t s (r)dr. In this case, N (6 : 00) N (5 : 30) is Poisson with parameter 6:00 5:30 (r)dr = .25/2 = 1/8. Draw the linear rates and prove this to yourself. Since X Poisson() = E(X) = , it follows that the expected number of customers that are lost is 1 8 . 2. What is the probability that at least one customer arrives to find the store closed (if the tailor decides to close the store at 5:30)? Answer: For the first part of this