Question

Assume that the life of a roller bearing follows a Weibull distribution with parameters beta=2 and delta=10,000 hours.

a) Determine the probability that a bearing lasts at least 8000 hours. Round answer to 4 decimal places.
b) Determine the mean time until failure of a bearing. Round answer to 4 significant figures.
c) If 10 bearings are in use and failures occur independently, what is the probability that all 10 bearings last at least 8000 hours? Round answer to 4 decimal places

Answer 1

The probability that a roller bearing lasts at least 8000 hours is 0.1353, the mean time until failure is approximately 8862 hours, and the probability that all 10 bearings last at least 8000 hours is approximately 0.0000 when failures are independent.

Step-by-step explanation:

The life of a roller bearing is assumed to follow a Weibull distribution with parameters β=2 and δ=10,000 hours.

1. To determine the probability that a bearing lasts at least 8000 hours, we use the cumulative distribution function (CDF) of the Weibull distribution, P(X ≥ x) = e^{-(x/δ)^{β}} for x ≥ 0. Substituting x with 8000, we get P(X ≥ 8000) = e^{-(8000/10000)^2} which is approximately 0.1353, rounded to four decimal places.
2. The mean time until failure, which is the expected value of a Weibull distribution, is given by δΓ(1 + 1/β) where Γ is the gamma function. For β=2, Γ(1.5) equals approximately 0.8862. Therefore, the mean time until failure is 10000 × 0.8862, which is approximately 8862 hours, rounded to four significant figures.
3. For independent failures of 10 bearings, the joint probability is the product of the probabilities for individual bearings lasting at least 8000 hours. Hence, the probability that all 10 bearings last at least 8000 hours is (0.1353)^{10} which is approximately 0.0000, rounded to four decimal places.