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Answer: 47.51%
## Explanation This is a Poisson distribution problem where: - Weekly average errors = 8 - Workdays per week = 5 - Daily error rate λ = 8/5 = 1.6 We need P(K > 1) = 1 - P(K ≤ 1) = 1 - [P(K=0) + P(K=1)] **Poisson probability formula:** P(K=k) = (λ^k * e^(-λ)) / k! **Calculations:** - P(K=0) = (1.6^0 * e^(-1.6)) / 0! = 1 * e^(-1.6) = 0.2019 - P(K=1) = (1.6^1 * e^(-1.6)) / 1! = 1.6 * e^(-1.6) = 0.3230 **Final probability:** P(K > 1) = 1 - (0.2019 + 0.3230) = 1 - 0.5249 = 0.4751 = 47.51% Therefore, the nearest probability is **47.51%**, which corresponds to option C.
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A certain low-severity administrative (operational) process tends to produce an average of eight errors per week (where each week is five workdays). If this loss frequency process can be characterized by a Poisson distribution, which is NEAREST to the probability that more than one error will be produced tomorrow; i.e., P(K > 1 | λ = 8/5)?
A
20.19%
B
32.30%
C
47.51%
D
66.49%
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