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
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 |
Repair Time (hr.) Probability 1 0.20 0.50 3 0.30 1.00 2.
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