Every time a machine breaks down at the Dynaco Manufacturing Company (Problem 3), either 1, 2, or
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Question:
Every time a machine breaks down at the Dynaco Manufacturing Company (Problem 3), either 1, 2, or 3 hours are required to fix it, according to the following probability distribution:
Simulate the repair time for 20 weeks and then compute the average weekly repair time.
Simulation | |||||
Week | RN | Breakdown # | RN | Repair time/breakdown | Repair Time/week |
1 | 0/1 | #VALUE! | |||
2 | 50/2 | #VALUE! | |||
3 | 44/2 | #VALUE! | |||
4 | 86/3 | #VALUE! | |||
5 | 58/3 | #VALUE! | |||
6 | 38/2 | #VALUE! | |||
7 | 1-May | 0 | |||
8 | 15/1 | #VALUE! | |||
9 | 94/3 | #VALUE! | |||
10 | 8 1 | #VALUE! | |||
11 | 16/1 | #VALUE! | |||
12 | 50/2 | #VALUE! | |||
13 | 88/3 | #VALUE! | |||
14 | 29/1 | #VALUE! | |||
15 | 79/3 | #VALUE! | |||
16 | 92/3 | #VALUE! | |||
17 | 26/1 | #VALUE! | |||
18 | 36/2 | #VALUE! | |||
19 | 1-Jun | 0 | |||
20 | 41/2 | #VALUE! |
1.9 hours | Average repair time |
Table from P3 | ||
Breakdown | ||
P(x) | Cumulative | Breakdown |
0.1 | 0.1 | 0 |
0.2 | 0.3 | 1 |
0.15 | 0.45 | 2 |
0.3 | 0.75 | 3 |
0.15 | 0.9 | 4 |
0.1 | 1 | 5 |
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