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Customers arrive at a one-window drive according to a Poisson distribution with mean of 10 minutes and service time per customer is exponential with mean of 6 minutes. The space in front of the window can accommodate only three vehicles including the serviced one. Other vehicles have to wait outside this space. What will be the Probability that an arriving customer can drive directly to the space in front of the window, the Probability that an arriving customer will have to wait outside the directed space and expected waiting time (in minutes) of a customer before getting the service respectively?
  • a)
    0.784, 0.216, 18
  • b)
    0.216, 0.784, 9
  • c)
    0.216, 0.784, 18
  • d)
    0.784, 0.216, 9
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Customers arrive at a one-window drive according to a Poisson distrib...
Given:
- Arrival rate of customers: Poisson distribution with mean of 10 minutes
- Service time per customer: exponential with mean of 6 minutes
- Capacity of space in front of the window: 3 vehicles

Probability that an arriving customer can drive directly to the space in front of the window:
- Probability that there are 0, 1, or 2 vehicles in the space when the new customer arrives
- Using the formula for Poisson distribution, the probability of 0, 1, or 2 vehicles in the space are 0.0498, 0.1493, and 0.2240 respectively
- Therefore, the probability that the new customer can drive directly to the space is 1 - 0.0498 - 0.1493 - 0.2240 = 0.5769

Probability that an arriving customer will have to wait outside the directed space:
- Probability that there are already 3 vehicles in the space when the new customer arrives
- Using the formula for Poisson distribution, the probability of 3 vehicles in the space is 0.1126
- Therefore, the probability that the new customer will have to wait outside the directed space is 0.1126

Expected waiting time (in minutes) of a customer before getting the service:
- This can be found using Little's Law, which states that the average number of customers in the system (waiting and being serviced) is equal to the arrival rate multiplied by the average time in the system
- The average time in the system is the sum of the average waiting time and the average service time
- The arrival rate is 1/10 customers per minute, and the average service time is 6 minutes
- To find the average waiting time, we need to find the average number of customers in the system
- We can do this by considering the different cases: 0, 1, 2, or 3 vehicles in the space when the new customer arrives
- Using the probabilities found earlier, the average number of customers in the system are 0.0498*0 + 0.1493*1 + 0.2240*2 + 0.1126*3 = 0.9244
- Therefore, the average waiting time is (0.9244-1)*6 = -0.4554 minutes, which means that the customer is expected to receive service 0.4554 minutes earlier than the average time in the system.
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Community Answer
Customers arrive at a one-window drive according to a Poisson distrib...
From the Problem, we have λ = 6 customers per hour; and μ = 10 customers per hour.
(a) Probability that an arriving customer can drive directly to the space in front of the window:
(b) Probability that an arriving customer will have to wait outside the directed space:
(c) Expected waiting time of a customer before getting the service is:
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Customers arrive at a one-window drive according to a Poisson distribution with mean of 10 minutes and service time per customer is exponential with mean of 6 minutes. The space in front of the window can accommodate only three vehicles including the serviced one. Other vehicles have to wait outside this space. What will be the Probability that an arriving customer can drive directly to the space in front of the window, the Probability that an arriving customer will have to wait outside the directed space and expected waiting time (in minutes) of a customer before getting the service respectively?a)0.784, 0.216, 18b)0.216, 0.784, 9c)0.216, 0.784, 18d)0.784, 0.216, 9Correct answer is option 'D'. Can you explain this answer?
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Customers arrive at a one-window drive according to a Poisson distribution with mean of 10 minutes and service time per customer is exponential with mean of 6 minutes. The space in front of the window can accommodate only three vehicles including the serviced one. Other vehicles have to wait outside this space. What will be the Probability that an arriving customer can drive directly to the space in front of the window, the Probability that an arriving customer will have to wait outside the directed space and expected waiting time (in minutes) of a customer before getting the service respectively?a)0.784, 0.216, 18b)0.216, 0.784, 9c)0.216, 0.784, 18d)0.784, 0.216, 9Correct answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Customers arrive at a one-window drive according to a Poisson distribution with mean of 10 minutes and service time per customer is exponential with mean of 6 minutes. The space in front of the window can accommodate only three vehicles including the serviced one. Other vehicles have to wait outside this space. What will be the Probability that an arriving customer can drive directly to the space in front of the window, the Probability that an arriving customer will have to wait outside the directed space and expected waiting time (in minutes) of a customer before getting the service respectively?a)0.784, 0.216, 18b)0.216, 0.784, 9c)0.216, 0.784, 18d)0.784, 0.216, 9Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Customers arrive at a one-window drive according to a Poisson distribution with mean of 10 minutes and service time per customer is exponential with mean of 6 minutes. The space in front of the window can accommodate only three vehicles including the serviced one. Other vehicles have to wait outside this space. What will be the Probability that an arriving customer can drive directly to the space in front of the window, the Probability that an arriving customer will have to wait outside the directed space and expected waiting time (in minutes) of a customer before getting the service respectively?a)0.784, 0.216, 18b)0.216, 0.784, 9c)0.216, 0.784, 18d)0.784, 0.216, 9Correct answer is option 'D'. Can you explain this answer?.
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