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Question 380 / 2 ptsproduction machine that consists of three components connected in series The first component follows Weibull probability distribution with shape...

Question

Question 380 / 2 ptsproduction machine that consists of three components connected in series The first component follows Weibull probability distribution with shape parameter of 1.3 and scale parameter (characteristic life) of 24,000 hr: The second follows Weibull probability distribution with shape parameter of 1.9 and scale parameter (characteristic life) of 18,000 hr The third component follows exponential distribution with MTTF = 48,000 hr: Find the reliability of the system at t = 60000.847

Question 38 0 / 2 pts production machine that consists of three components connected in series The first component follows Weibull probability distribution with shape parameter of 1.3 and scale parameter (characteristic life) of 24,000 hr: The second follows Weibull probability distribution with shape parameter of 1.9 and scale parameter (characteristic life) of 18,000 hr The third component follows exponential distribution with MTTF = 48,000 hr: Find the reliability of the system at t = 6000 0.8479 0.66104 0.88338 0.9999



Answers

A very important distribution for analyzing the reliability of manufactured goods is the Weibull distribution, whose probability density function is defined by
$$
f(x)=a b x^{b-1} e^{-a x^{b}} \quad \text { for } x \text { in }[0, \infty)
$$
where $a$ and $b$ are constants. Notice that when $b=1$ , this reduces to the exponential distribution. The Weibull distribution is more general than the exponential, because it applies
even when the failure rate is not constant. Use Simpson's rule with $n=100$ , or the integration feature on a graphing calculator, to approximate the following for the Weibull distribution with $a=4$ and $b=1.5 .$ Use a limit of 3 in place of $\infty$ .

We're showing the hazard function given here and for part, they were asked to find the reliability, function and geographic. Now, the reliability function is given by this formula. So we want to break this into two sections. So there's a section where time is between zero and 200. So this is the reliability function for time is between 0 200 and then for time greater than 200. We have the following. Now, in order to reach a time of greater than 200 we must first make up to 200 hours. The probability of of exceeding 200 is reliability of 200. And then we multiply that on the reliability for every duration after that time. So that's from 200 to T. To find the reliability of 200 hours, we simply plug 200 hours into T here that comes out to eat of the minus 0.4. And so this is the reliability for time is greater than 200. And this is for between zero and 200 and actually this should be 0.2 and we're also asked for a plot of the reliability function. So it looks like this. So the reliability plot is shown here. You can see that at 200 hours the reliability function begins to drop off less deeply, which is indicative of the lower hazard function. For time greater than 200 hours next for Part B were asked to find the density function. The density function is simply given by the negative first derivative of the reliability function, and that comes out as follows. So that answers Part B and then for part C, we're looking for the mean time to failure. This is given by this formula. It's the integral from zero to infinity of the reliability function. And so we'll split it into the two components first from 0 to 200 hours. Your reliability function for that time period is E to the minus 0.2 t and then for the second period, just 200 hours plus and this comes out to 835 0.16 So the mean time to failure for this component is 835.16 hours, and I just point out that this plot goes on to infinity

In this exercise were given a Waibel random variable with shape parameter Alfa equals two and scale parameter beta equals three and were asked to solve just a few things. So for part, they were asked to find the expected value and the variance. Now we have the expected value formula from the textbook. It's given as this so substituting in Alfa and Beta, we have the following and then using the rule for the gamma function that we can re express this as three times one half time Scam of one half and gamma of one half is equal to square it up I So we have 3/2 times the square root of pi. So that's the mean now for the variance of X. So the formula for the variance is given as follows. And so then using this formula and substituting in Alfa and Beta get nine on the following now using the following fact, we have a one factorial for the first term. And for the second term we have already solved this just equal thio this So we have one half times the square root of pi all squared and this all comes out to 1.935 So that finishes part a next for part B. We want to solve the probability that X is less than or equal to six. Yeah. Now, generally speaking, the CDF of Waibel random variable is given as the following. So if we use this formula, we have one minus e to the exponents. Negative 6/3 squared. And this comes out to 0.982 And finally, for part C, we want the probability that X is between 1.5 and sex. Now we've already solved for this right here, 0.982 and the second term comes out to 0.2 to 1, which gives us a probability of 0.761 of being between 1.5 and six.

All right, We have this probability distribution function, so it's a slight generalization of the exponential distribution each minus X to me. All right. And so this is from zero to infinity X in that range. Now we're going to set A to B for B B one four five, and then we're going to put the effective Ah, women. We can say this is about three. So instead of integrating all the way to infinity, we could just integrate two three, because after that point, it'Ll the distribution will essentially be zero. Okay, So first of all, I want to find the average the expected dying, and this is just well, it's approximately equal to it's here to three. And then we just plug in all of our numbers. Six. So you put our weight of X and there on six next to the B minus one, which is Europe one five, and then e to the minus for next one point five. And just using the integration feature on micro graphing calculator. Get this is your a point three five eight three now for the standard distribution. So sigma squared, which is the variance, is just well is approximately equal to your girls. Your three. And then we have X minus. Well, in this case, mu squared times six times x to the zero point five each of the negative for extra the one point five the ex. And this is, um So this is going to give us the variance in approximately for the variance and everyday square root we'LL get the The standard deviation is about zero point two four three two.

Yeah, this problem we are given the following exponential distribution. Now we're told that we have three of these. Let's just focus on the one for now. Now we're told that the equipment fails if at least two of these components failed and we want the probability that the equipment will operate for at least 200 hours without failure. Now let's begin by finding the probability that Y is greater than or equal to 200. And so this density function, this is equal to the integral from 200 to infinity of one over 100. E to the negative Y over 100 dy. And so this is equal to negative E to the negative Y over 100 evaluated from y is 200 to infinity plan. Your infinite is just not any of us zero if you date that limit. And so then we're left with E to the 19 of 200 over 100. Mhm. So if we evaluate that just to get it as a decimal, it was approximately 0.1353 And what that means is that in general we have a 0.1353 chance that an individual component will last at least 200 hours. Now we want the probability that at least two of the components make it 200 hours. We have three components and so we need at least 202 of the components to make it three hours. So now let's treat this, I can binomial distribution or this is R P back. This means in general P fx is 3 to 6 because we have three of them 0.1353 to the x times one minus 10.1353 to the three minus x. Now we want the probability that x is greater than or equal to two because we need to at least two of them to make it 200 hours. And so this is three. Well it would be the sum from Mexico is from 2 to 3. Again, we're just going to plug this in, put two in and part three in and then add them together. Yeah, a 0.13 Unless it's a combination of three, choose X times 0.1353 to the arts times one minus 10.1353 23 minus X. And so then we're just going to plug in two point and three here and then add them together. I don't know what we do sort of us are probability of point oh 500 So that's the problem.


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