Question

A single-server waiting line system has an arrival pattern characterized by a Poisson distribution with 3 customers per hour. The average service time is 12 minutes. The service times are distributed according to the negative exponential distribution. Determine the following: 1. The expected number of customers in the waiting line. 2. The average number of customers in the system. 3. The average wait time in the queue.

          A single-server waiting line system has an arrival pattern
characterized by a Poisson distribution with 3 customers per hour.
The average service time is 12 minutes. The service times are
distributed according to the negative exponential distribution.
Determine the following:
1. The expected number of customers in the waiting line.
2. The average number of customers in the system.
3. The average wait time in the queue.
        
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Added by Evelyn R.

Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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A single-server waiting line system has an arrival pattern characterized by a Poisson distribution with 3 customers per hour. The average service time is 12 minutes. The service times are distributed according to the negative exponential distribution. Determine the following: 1. The expected number of customers in the waiting line. 2. The average number of customers in the system. 3. The average wait time in the queue.
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Transcript

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00:01 So for this problem, to begin, we know that we have that the parameter, our lambda, is equal to three customers per hour.
00:11 Then we know that mu, our mean service rate, is 12 minutes per customer, which then means that that would be five customers per hour.
00:32 So the first thing that we need to do is figure out the average number of customers who will be waiting in the line, which would be lq.
00:39 We find that out using, oh, pardon me, actually, i need to back up for a second here.
00:44 No, actually, the first thing that we need to do is figure out our utilization, row, which is equal to lambda over mu.
00:52 So we have that row would be equal to 3 over 5 or 0 .6.
00:57 So we have 60 % utilization.
01:00 Then we need lq, the average number of customers wading in line, which is going to be equal to row squared, divided by 1 minus row.
01:09 So 0 .6 squared is going to be 0 .36, divided by 1 minus 0 .6, gives us a result of 0 .9 as the average number of customers waiting in line, which i'll note here that this is the answer to part 2.
01:28 Part 2, the average number of customers waiting in line or in the q is 0 .9.
01:35 So we do actually end up answering part 2 sort of on our way through the different parts of the problem.
01:42 Then, the next thing that we want to do is find the average time spent waiting.
01:48 So that's going to be equal to lq, average time waiting in line, divided by lambda.
01:55 Or pardon me, average number of customers waiting in line divided by lambda.
01:59 So that's going to be 0 .9 divided, let me fix that there, 0 .9 divided by 3, which gives a result of 0 .3 hours.
02:11 Then having that, we want to find the average time spent in the system in total, which is going to be the average amount of time waiting in line, plus the serving time, which would be equal to 0 .3 plus 1 over 5...
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