Automobiles arrive at the drive-through window at a post office at the rate of 4

Question

Automobiles arrive at the drive-through window at a post office at the rate of 4 every 10 minutes. The average service time is 2 minutes. The Poisson distribution is appropriate for the arrival rate and service 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 time cars spend waiting to receive service? (d) What is the average number of cars in line behind the customer receiving service? (e) What is the probability that there are no cars at the window? (f) What percentage of the time is the postal clerk busy? (g) What is the probability that there are exactly two cars in the system?

I have the answer but I need explanation for this question. basically, I'm not sure how to find the mew, and the other symbol (I cant recall the name but looks somehwat like y)

Explanation / Answer

a) Average time a car is in the system = Ws = 1/( µ - )

= 1/ (0.5 – 0.4

= 10 minutes

b) Average number of cars in the system =  Ls = /(µ - )

= 0.4/ (0.5 – 0.4)

= 4 cars

c) Average time cars spend waiting to receive service = Lq = 2 / {µ (µ - )}

= 0.42/ {0.5(0.5 – 0.4)}

= 3.2 minutes

d) Average number of cars in line behind the customer receiving service = Wq = / {µ (µ - )}

= 0.4/ {0.5(0.5 – 0.4)}

= 8 cars

e) Probability that there are no cars at the window = P0 = 1 - /µ

= 1 –0.8

= 0.2

f) Percentage of the time is the postal clerk busy = = /µ

= 0.4/0.5

= 0.8 or 80%

g) Probability that there are exactly two cars in the system = Pn=2 = (/µ)kx (1- /µ)

= (0.4/0.5)2x 0.2

= 0.64 x 0.2

= 0.128 or 12.8%

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