Statistics - Please help. Show work. Thank you!
ANSWER::
6Q)
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6Q. Assume the life of a roller bearing follows a Weibull distribution with parameters
β = 2 and
δ = 7,500 hours.
Solution:
a) The probability that a bearing lasts at least 8000 hours:
P(X ≥ 8000) = 1 - e-(X/δ)^β
P(X ≥ 8000) = 1 - e-(8000/7500)^2
P(X ≥ 8000) = 1 - e-(1.138)
P(X ≥ 8000) = 1 - 0.3205
P(X ≥ 8000) = 0.6795
Therefore, the probability that a bearing lasts at least 8000 hours would be 0.6795.
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