Automobiles arrive at the drive-through window at a post office at the rate of 7 every 10 minutes. The average service time is 7 minutes. The Poisson distribution is appropriate for the arrival rate and service times are exponentially distributed. (Note: the given equations are NOT necessarily in the correct order.) What is the average time a car is in the system?Formulas for 13.17 What is the average number of cars in the system? What is the average time cars spend waiting to receive service? What is the average number of cars in line behind the customer receiving service? What is the probability that there are no cars at the window? What percentage of the time is the postal clerk busy? What is the probability that there are exactly two cars in the system?

1) the average time a car is in the system is 0.1587 min

2) the average number of cars in the system 0.1111

3) the average time cars spend waiting to receive service is 0.01587 min

4) the average number of cars in line behind the customer receiving service is 0.01111

5) the probability that there are no cars at the window is 0.9

6) the percentage of time that the postal clerk is busy is 10%

7) the probability that there are exactly two cars in the system is 0.1217

Step-by-step explanation:

Given that;

Arrival rate λ = the rate of 7 every 10 minutes = 7/10 = 0.7

Average service time μ = 7 minutes

The Poisson distribution is appropriate for the arrival rate and service times are exponentially distributed.

1) What is the average time a car is in the system?

Average in the system W = 1 / (μ - λ)

= 1 / (7 - 0.7) = 1 / 6.3 = 0.1587 min

therefore the average time a car is in the system is 0.1587 min

2) What is the average number of cars in the system?

Average number of cars L = λ / (μ - λ)

= 0.7 / (7 - 0.7) = 0.7/6.3 = 0.1111

Therefore the average number of cars in the system 0.1111

3) What is the average time cars spend waiting to receive service?

Average time spend waiting for services  Wq = λ / μ(μ - λ)

= 0.7 / 7(7 - 0.7) = 0.7/44.1 = 0.01587 min

Therefore the average time cars spend waiting to receive service is 0.01587 min

4) What is the average number of cars in line behind the customer receiving service?

average number of cars in line behind the customer receiving service

Lq = λ² / μ(μ - λ) = (0.7)² / 7(7 - 0.7) = 0.49 / 44.1 = 0.01111

Therefore the average number of cars in line behind the customer receiving service is 0.01111

5) What is the probability that there are no cars at the window?

probability that there are no cars at the window = 1 -  (λ / μ)

= 1 -  (0.7 / 7) = 1 - 0.1 = 0.9

Therefore the probability that there are no cars at the window is 0.9

6) What percentage of the time is the postal clerk busy?

percentage of the time is the postal clerk busy = (λ / μ)

= (0.7 / 7) = 0.1 ≈ 10%

Therefore the percentage of time that the postal clerk is busy is 10%

7) What is the probability that there are exactly two cars in the system?

the probability that there are exactly two cars in the system = (λ²e^-λ) / 2!

= (0.7²e^-0.7) / 2! = 0.2433 / 2 = 0.1217

Therefore the probability that there are exactly two cars in the system is 0.1217

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9