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A suburban specialty restaurant has developed a single drive-thru window. customers order, pay, and pick up their food at the same window. arrivals follow a poisson distribution while service times follow an exponential distribution. if the average number of arrivals is 6 per hour and the service rate is 2 every 15 minutes, how much time will elapse (in hours) from the time a customer enters the line until he/she leaves the restaurant?

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Final answer:

The average time a customer will spend from entering the line to leaving the restaurant is 0.5 hours or 30 minutes, considering the arrival rate (λ) is 6 per hour and the service rate (μ) is 8 per hour.

Step-by-step explanation:

The question involves queueing theory, where the arrivals follow a Poisson distribution and service times follow an exponential distribution. Given the average number of arrivals is 6 per hour, we can find the average time between arrivals. Also, with a service rate of 2 every 15 minutes, or 8 per hour, we can determine the average service time as well.

a. The average time between two successive arrivals (arrival rate λ) can be calculated as:
1 hour / 6 arrivals = 0.1667 hours (or 10 minutes).

b. The average service rate (μ) is:
2 services / 15 minutes = 8 services / 1 hour.
Hence, the average service time is:
1 hour / 8 services = 0.125 hours (or 7.5 minutes).

The total average time a customer spends in the system (system time) is the sum of the average service time and the time waiting in line. This is often modeled as 1 / (μ - λ) for M/M/1 queue:

System Time = 1 / (μ - λ) = 1 / (8 - 6) = 1 / 2
= 0.5 hours (or 30 minutes).

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