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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.

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.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter2: Exponential, Logarithmic, And Trigonometric Functions
Section2.CR: Chapter 2 Review
Problem 111CR: Respiratory Rate Researchers have found that the 95 th percentile the value at which 95% of the data...
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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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