Automobiles arrive at the drive-through window at the downtown Baton Rouge, Louisiana. post office at the rate of 4every 10 minutes. The average service time is 2 minutes. The Poisson distribution is appropriate for the arrival rate andservice times are exponentially distributed. a) What is the average time a car is in the system? b) What is the average number of cars in the system? c) What is the average number of cars waiting to receive service? d) What is the average time a car is in the queue? e) What is the probability that there are no cars at the window? f) What percentage of the time is the postal clerk busy?9) What is the probability that there are exactly 2 cars in the system? h) By how much would your answer to part (a) be reduced if a second drive-through window, with its own server, wereadded?

Answer:a) 3/4 minutes, (b) 3/-2 (c) 3/8 (d) 3/-2minutes (e) 0.0996 (f) 37.5% (g) 0.0187 (h) 2 minutes

Explanation:

Arrival rate = 60/10 = 6 minutes, service rate = 2minutes

(a) ti = Arrival rate/service rate (service rate - Arrival rate)

=6/2 (2 - 6)

=6/2 (-4)

= 6/8

= 3/4minutes

(b) ts = Arrival rate / service rate - Arrival rate

=6/2-6

=6/-4

= 3/-2

(c) ti = Arrival rate / 2 (service rate) (service rate - Arrival rate)

=6/2 (2) (2-6)

= 6/4 (-4)

=6/16

=3/8

(d) ts = Arrival rate / service rate - Arrival rate

=6/2-6

=6/-4

=3/-2 minutes

(e) P (0) using Poisson formula P (r) = e∧-mm∧r/r!

Where P (r) is the probability of event r, m = the mean of event, e = a constant with value of 2.71828, e∧-m = 1/em

P (0) = (2.71828) ∧-3/2 (3/2) ∧0/0!

=1 / (2.71828) ∧3/2

=1/10.0427

=0.09957

= 0.0996 Approximately

(f) 3/8 * 100

= 0.375 * 100

= 37.5%

(g) using Poisson formul P (2) = (2.71828) ∧-3/2 (3/2) ∧2/2!

=1/10.0427 * 0.75/4

= 0.75/40.1708

=0.01867

= 0.0187 Approximately

(h) standard deviation = Arrival rate / C where C = 2 (second drive - through window with its own server added)

= standard deviation = 6/2

Standard deviation = 4

Taking the square root of 4

= 2

= 2minutes