Determine the probability that a bearing lasts at least 6000 hours. Now, EXAMPLE 4.21 I Bearing Wear The time to failure (in hours) of a bearing in a mechanical shaft is satisfactorily modeled as a Weibull random variable with B = 1/2 and 8 = 5000 hours. Determine the mean time until failure. From the expression for the mean, 6000 P(X > 6000) = 1 – F(6000) = exp| 5000 = e-1.44 = 0.237 E(X) = 5000T[1+ (1/2)] = 500or[1.5] = 5000×0.5/T ni anoitonul vpienob vilidedong botoolaz lo adasn od nodw isi) so2 nso sw.nonomt ytiens Practical Interpretation: Consequently, only 23.7% of all ai nobudinalb lludisW orl = 4431.1 hours olludiialb dglolaH odt.oalA.gotudi bearings last at least 6000 hours.

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section9.3: Binomial Probability
Problem 2E: If a binomial experiment has probability p success, then the probability of failure is...
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I have a question about this practice problem from my textbook. I included an image. I'm having some trouble understanding the gamma function. I don't understand how they calculated the mean in the first part of the problem. For the second part of the problem, when I plug in the numbers, I'm getting 0.3343, but they got 0.237. I'm using beta = 0.5, but it seems like they are using beta = 2.

Determine the probability that a bearing lasts at least
6000 hours. Now,
EXAMPLE 4.21 I Bearing Wear
The time to failure (in hours) of a bearing in a mechanical
shaft is satisfactorily modeled as a Weibull random variable
with B = 1/2 and 8 = 5000 hours. Determine the mean time
until failure.
From the expression for the mean,
6000
P(X > 6000) = 1 – F(6000) = exp|
5000
= e-1.44
= 0.237
E(X) = 5000T[1+ (1/2)] = 500or[1.5] = 5000×0.5/T ni anoitonul vpienob vilidedong botoolaz lo adasn od
nodw isi) so2 nso sw.nonomt ytiens Practical Interpretation: Consequently, only 23.7% of all
ai nobudinalb lludisW orl
= 4431.1 hours
olludiialb dglolaH odt.oalA.gotudi bearings last at least 6000 hours.
Transcribed Image Text:Determine the probability that a bearing lasts at least 6000 hours. Now, EXAMPLE 4.21 I Bearing Wear The time to failure (in hours) of a bearing in a mechanical shaft is satisfactorily modeled as a Weibull random variable with B = 1/2 and 8 = 5000 hours. Determine the mean time until failure. From the expression for the mean, 6000 P(X > 6000) = 1 – F(6000) = exp| 5000 = e-1.44 = 0.237 E(X) = 5000T[1+ (1/2)] = 500or[1.5] = 5000×0.5/T ni anoitonul vpienob vilidedong botoolaz lo adasn od nodw isi) so2 nso sw.nonomt ytiens Practical Interpretation: Consequently, only 23.7% of all ai nobudinalb lludisW orl = 4431.1 hours olludiialb dglolaH odt.oalA.gotudi bearings last at least 6000 hours.
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