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The men's department of a large store employs one tailor for customer fittings. The number of customers requiring fitting appears to follow a Poisson distribution with a mean arrival rate is 24 per hour. Customers fitting time follows an exponential distribution with a mean of 2 minute. The time (minute) a customer is expected to spend in the fitting row is………….
- a)6
- b)10
- c)30
- d)16
Correct answer is option 'B'. Can you explain this answer?
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The men's department of a large store employs one tailor for customer ...
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The men's department of a large store employs one tailor for customer ...
Mean Arrival Rate and Distribution:
- Mean arrival rate of customers requiring fitting is 24 per hour.
- This follows a Poisson distribution.
Customer Fitting Time and Distribution:
- Customer fitting time follows an exponential distribution.
- Mean fitting time is 2 minutes.
Expected Time a Customer Spends in Fitting Row:
- We need to find the expected time a customer spends in the fitting row.
- This can be calculated as the sum of the time a customer spends waiting in line and the time a customer spends being fitted.
- The time a customer spends waiting in line can be calculated using Little's Law, which states that the average number of customers in the system (i.e., waiting in line plus being fitted) is equal to the arrival rate multiplied by the average time a customer spends in the system.
- Using Little's Law, we can calculate that the average number of customers in the system is 24/60 * 2 = 0.8.
- Since there is only one tailor, the average number of customers waiting in line is 0.8 - 1 = -0.2, which is not possible.
- Therefore, we can assume that there is no waiting time, and the expected time a customer spends in the fitting row is simply the mean fitting time, which is 2 minutes.
- Therefore, the correct answer is option 'B', i.e., 10 minutes.
- Mean arrival rate of customers requiring fitting is 24 per hour.
- This follows a Poisson distribution.
Customer Fitting Time and Distribution:
- Customer fitting time follows an exponential distribution.
- Mean fitting time is 2 minutes.
Expected Time a Customer Spends in Fitting Row:
- We need to find the expected time a customer spends in the fitting row.
- This can be calculated as the sum of the time a customer spends waiting in line and the time a customer spends being fitted.
- The time a customer spends waiting in line can be calculated using Little's Law, which states that the average number of customers in the system (i.e., waiting in line plus being fitted) is equal to the arrival rate multiplied by the average time a customer spends in the system.
- Using Little's Law, we can calculate that the average number of customers in the system is 24/60 * 2 = 0.8.
- Since there is only one tailor, the average number of customers waiting in line is 0.8 - 1 = -0.2, which is not possible.
- Therefore, we can assume that there is no waiting time, and the expected time a customer spends in the fitting row is simply the mean fitting time, which is 2 minutes.
- Therefore, the correct answer is option 'B', i.e., 10 minutes.
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The men's department of a large store employs one tailor for customer fittings. The number of customers requiring fitting appears to follow a Poisson distribution with a mean arrival rate is 24 per hour. Customers fitting time follows an exponential distribution with a mean of 2 minute. The time (minute) a customer is expected to spend in the fitting row is………….a)6b)10c)30d)16Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The men's department of a large store employs one tailor for customer fittings. The number of customers requiring fitting appears to follow a Poisson distribution with a mean arrival rate is 24 per hour. Customers fitting time follows an exponential distribution with a mean of 2 minute. The time (minute) a customer is expected to spend in the fitting row is………….a)6b)10c)30d)16Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The men's department of a large store employs one tailor for customer fittings. The number of customers requiring fitting appears to follow a Poisson distribution with a mean arrival rate is 24 per hour. Customers fitting time follows an exponential distribution with a mean of 2 minute. The time (minute) a customer is expected to spend in the fitting row is………….a)6b)10c)30d)16Correct answer is option 'B'. Can you explain this answer?.
The men's department of a large store employs one tailor for customer fittings. The number of customers requiring fitting appears to follow a Poisson distribution with a mean arrival rate is 24 per hour. Customers fitting time follows an exponential distribution with a mean of 2 minute. The time (minute) a customer is expected to spend in the fitting row is………….a)6b)10c)30d)16Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The men's department of a large store employs one tailor for customer fittings. The number of customers requiring fitting appears to follow a Poisson distribution with a mean arrival rate is 24 per hour. Customers fitting time follows an exponential distribution with a mean of 2 minute. The time (minute) a customer is expected to spend in the fitting row is………….a)6b)10c)30d)16Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The men's department of a large store employs one tailor for customer fittings. The number of customers requiring fitting appears to follow a Poisson distribution with a mean arrival rate is 24 per hour. Customers fitting time follows an exponential distribution with a mean of 2 minute. The time (minute) a customer is expected to spend in the fitting row is………….a)6b)10c)30d)16Correct answer is option 'B'. Can you explain this answer?.
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