automoniles arrive at the drive-through windom at a post office at the rate of 4 every 10 minutes. the average service time is 2 minutes. the poisson distributimes 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 2 cars in the system?

To solve these queations, we can use the principles of queuing theory. There are several formulas and concepts we can use, such as Little's Law and the Poisson distribution. Let's go through each question step by step:

(A) Average time a car is in the system:
To find the average time a car is in the system, we can use Little's Law. Little's Law states that the average number of customers in a system (L) is equal to the average arrival rate (λ) multiplied by the average time spent in the system (W).

Formula: L = λ * W

Given that the arrival rate (λ) is 4 every 10 minutes, we can calculate the arrival rate per minute: λ = 4 / 10 = 0.4 cars per minute.
The service time (W) is given as 2 minutes.

Plugging these values into Little's Law:
L = 0.4 * 2 = 0.8 minutes

Therefore, the average time a car is in the system is 0.8 minutes.

(B) Average number of cars in the system:
To find the average number of cars in the system, we can use Little's Law:

Formula: L = λ * W

Given that the arrival rate (λ) is 0.4 cars per minute and the average service time (W) is 2 minutes, we can plug these values into the formula:

L = 0.4 * 2 = 0.8 cars

Therefore, the average number of cars in the system is 0.8 cars.

(C) Average time cars spend waiting to receive service:
To find the average waiting time, we need to subtract the average service time from the average time a car is in the system.

Average waiting time = Average time in the system - Average service time
Average waiting time = 0.8 - 2 = -1.2 minutes

Since the result is negative, it means there is no waiting time. The average time cars spend waiting to receive service is 0 minutes.

(D) Average number of cars in line behind the customer receiving service:
To find the average number of cars in line, we can subtract 1 from the average number of cars in the system.

Average number of cars in line = Average number of cars in the system - 1
Average number of cars in line = 0.8 - 1 = -0.2 cars

Since the result is negative, it means there are no cars in line behind the customer receiving service.

(E) Probability that there are no cars at the window:
To find the probability that there are no cars at the window, we can use the Poisson distribution.

Formula: P(X = 0) = e^(-λ) * (λ^0 / 0!)

Given that the arrival rate (λ) is 0.4 cars per minute, we can plug this value into the formula:

P(X = 0) = e^(-0.4) * (0.4^0 / 0!) = e^(-0.4)

Using a calculator or a math library, we can find that e^(-0.4) is approximately 0.6703.

Therefore, the probability that there are no cars at the window is approximately 0.6703 or 67.03%.

(F) Percentage of the time the postal clerk is busy:
To find the percentage of time the postal clerk is busy, we can use the formula:

Busy time percentage = (λ * W / λ) * 100%

Using the values we already calculated, the formula becomes:

Busy time percentage = (0.4 * 2 / 0.4) * 100% = (0.8) * 100%
Busy time percentage = 80%

Therefore, the postal clerk is busy 80% of the time.

(G) Probability that there are exactly 2 cars in the system:
To find the probability that there are exactly 2 cars in the system, we can use the Poisson distribution.

Formula: P(X = k) = e^(-λ) * (λ^k / k!)

Given that the arrival rate (λ) is 0.4 cars per minute and we want to find the probability of 2 cars (k = 2), we can plug these values into the formula:

P(X = 2) = e^(-0.4) * (0.4^2 / 2!)

Using a calculator or a math library, we can find that e^(-0.4) * (0.4^2 / 2!) is approximately 0.0733.

Therefore, the probability that there are exactly 2 cars in the system is approximately 0.0733 or 7.33%.