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 least6000 hours. Now,EXAMPLE 4.21 I Bearing WearThe time to failure (in hours) of a bearing in a mechanicalshaft is satisfactorily modeled as a Weibull random variablewith B = 1/2 and 8 = 5000 hours. Determine the mean timeuntil failure.From the expression for the mean,6000P(X > 6000) = 1 – F(6000) = exp|5000= e-1.44= 0.237E(X) = 5000T[1+ (1/2)] = 500or[1.5] = 5000×0.5/T ni anoitonul vpienob vilidedong botoolaz lo adasn odnodw isi) so2 nso sw.nonomt ytiens Practical Interpretation: Consequently, only 23.7% of allai nobudinalb lludisW orl= 4431.1 hoursolludiialb dglolaH odt.oalA.gotudi bearings last at least 6000 hours.

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Description: 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 th

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Answered: Determine the probability that a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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.

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