A certain grocery store has noted the following figures with regard to the numbe
ID: 2748085 • Letter: A
Question
A certain grocery store has noted the following figures with regard to the number of people who arrive at its three checkout stands ready to check out, and the time it takes to check out the individuals.
Arrivals/Min.
Frequency
Service Time
in Min
Frequency
0
0.3
1
0.1
1
0.5
2
0.3
2
0.2
3
0.4
4
0.2
Create an appropriate table of interval of random numbers for both variables.
Answer:
Arrivals
Interval of
Random #s
Service Time
Interval of
Random #s
0
01-30
1
01-10
1
31-80
2
11-40
2
81-00
3
41-80
4
81-00
The time between arrivals at a drive-through window of a fast-food restaurant follows the distribution given below. The service time distribution is also given in the table below. Use the random numbers provided to simulate the activity of the first five arrivals. Assume that the window opens at 11:00 a.m. and the first arrival after this is based on the first interarrival time generated.
Time Between
Arrivals
Probability
Service Time
Probability
1
0.2
1
0.3
2
0.3
2
0.5
3
0.3
3
0.2
4
0.2
Random numbers for arrivals: 14, 74, 27, 03
Random numbers for service times: 88, 32, 36, 24
What times does the fourth customer leave the system?
11:06.
None of the alternatives are correct.
11:09
11:04.
11:08.
Arrivals/Min.
Frequency
Service Time
in Min
Frequency
0
0.3
1
0.1
1
0.5
2
0.3
2
0.2
3
0.4
4
0.2
Explanation / Answer
The 1st question is already answered.
The 2nd question: The time between arrivals at a drive-through window of a fast-food restaurant follows the distribution given below. The service time distribution is also given in the table below. Use the random numbers provided to simulate the activity of the first five arrivals. Assume that the window opens at 11:00 a.m. and the first arrival after this is based on the first interarrival time generated.
Solution:
First arrival (random 14) = 11:01. Service time 3 (random 88). Thus leave time = arrival time+service time = 11:01+3 = 11:04.
2nd arrival (random 74) = 11:04 i.e 3 minutes after the 1st arrival. service time = 2 (random 32). Leave time = arrival time+service time = 11:04+2 = 11:06
3rd arrival (random 27) = 11:06 i.e 2 minutes after the 2nd arrival. service time = 2 (random 36). leave time = arrival time+service time = 11:06+2 = 11:08
4th arrival (random 03) = 11:07 i.e 1 minute after the 3rd arrival. wait time will be 1 as the 3rd arrival is being serviced till 11:08. service time = 1 (random 24). leave time = arrival+wait+service = 11:07+1+1 = 11:09.
Thus the answer is 11:09
Time between arrivals Probability Random interval Service time Random interval 1 0.2 01-20 1 01-30 2 0.3 21-50 2 31-80 3 0.3 51-80 3 81-100 4 0.2 81-100Related Questions
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