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.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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