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TWEET # Untitled a guest Oct 9th, 2019 111 Never
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1. Let X be the number of failed modules in t months. Average number of modules failure per month is 1.3
2.
3. Assuming X follows Poisson distribution, X ~ Poisson(\lambda = 1.3t)
4.
5. For 3-month period, t = 3 and X ~ Poisson(\lambda = 1.3 * 3 = 3.9)
6.
7. Probability that the maintenance crew have to go to the solar installation before the end of the regular 3-month period =
8.
9. P(X > 7) = 1 - P(X \le 7)
10.
11. = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)]
12.
13. = 1 - [exp(-3.9) * 3.90 / 0! + exp(-3.9) * 3.91 / 1! + exp(-3.9) * 3.92 ​​​​​​​/ 2! + exp(-3.9) * 3.93 ​​​​​​​/ 3! + exp(-3.9) * 3.94 ​​​​​​​/ 4! + exp(-3.9) * 3.95 ​​​​​​​/ 5! + exp(-3.9) * 3.96 ​​​​​​​/ 6! + exp(-3.9) * 3.97 ​​​​​​​/ 7! ]
14.
15. = 1 - (0.02024191 + 0.07894345 + 0.15393974 + 0.20012166 + 0.19511862 + 0.15219252 + 0.09892514 + 0.05511543)
16.
17. = 0.04540153
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