consider-the-six-component-system-displayed-below-let-r_1t-ldots-r_6t-denote-the-component-reliabili

Question

Consider the six-component system displayed below. Let $R_{1}(t), \ldots, R_{6}(t)$ denote the component reliability functions. Assume the six components operate independently.
(a) Find the system reliability function.
(b) Assuming all six components have exponentially distributed lifetimes with mean 100 $\mathrm{h}$ ,
find the mean time to failure for the system.

Chris Trentman · Accepted Answer

were given a six component system with component reliability functions are one through our six, and we&#x27;re told that the six components operate independently. In part, they were asked to find the system reliability function. So to see the six component system, look at the graph in exercise 1 27 of this section. Looking at the graph, we see that components one and two are in parallel. So the reliability function of the 12 subsystem is by the formula from this section one minus one minus our one of t times one minus are two of tea and likewise for the reliability functions of the 34 and 56 subsystems. Now we have that three subsystems air connected in Siris. Therefore, the overall system reliability is our tea, which is gonna be the product of their liabilities of the subsystems. So one minus one minus our one of t times one minus are two of tea times one minus one minus are three of tea times one minus are four of tea times one minus one minus are five of tea. Times one minus are six of tea. Next in part B. We&#x27;re told that all six components had exponentially distributed lifetimes with mean value of 100 hours and Rask defying the mean time to failure for the system. Now, because Celtics components have exponentially distributed lifetimes, it follows that they all have reliability function. R i f t is equal to each of the negative t over Lambda, which in this case is 100 so follows that the overall reliability of the system are of tea is plugging in to our result from part a one minus one minus e to the negative t over 100 and this is going to be squared and this whole expression here is going to be cute. Now you want to write this out as a linear combination of exponential terms, So we have first applying the expert on the inside. This is the same as one minus ones. Those cancel out and then plus to either the negative t over 100 and minus each of the negative to t over 100 she must be written Well, what does keep it in that form off cute And then using binomial the&#x27;re, um this is going to be two cubes which is eight times eat to the negative t over 100 cube E to the negative three t over 100 and then minus. And this that was zero. Choose 3 to 0. Another three choose one is three so three times and then three times two squared Just force. This is 12 times eat to the negative t over 100 squared which is e to the negative to to over 100 times each of the negative to t over 100. Just eat the negative fourty over 100 and then the next term This has x coefficient three choose to which is again three. So we have three times to to the first power so six times negative one to the second power just one So plus six and then e to the negative t every 100 to the first power times e to the negative to t over 100 to the second power Such e to the negative five t over 100. And finally we have three Choose three is one. Then times negative e to the negative to t every 100 to the third power which is negative each of the negative 60 over 100. And so, using this expression, we find that the mean time to failure beauty. It&#x27;s the integral from zero to infinity of Are of tea DT. As we calculated above this is the integral from zero to infinity of mhm eight e to the negative three t over 100 minus 12 e to the negative 40 over 100 plus six e to the negative five t over 100 minus e to the negative 60 over 100 GT and evaluating each of these terms separately we get this is equal to eight times a simple heuristic. For this, Integral is going to be one of the exponents. So we get or one of the opposite of the exponents. So 100 over three, minus 12 times 100 over four plus six times 100 over five, minus 100 over six. And evaluating this is equal to 70. And so u of t is 70 hours

A M · Answer

So in the code below, we have that seven round of numbers are generated and they&#x27;re all indexed one by one into the vector U. So you is indexed. It&#x27;s Ah, uh, vector of random numbers, which has seven rows and one column. And so we have one for each of the seven components. Now, the sequence of end or conjunctions, as we have and or here this match is a serious and parallel ties and the system designed. So once you&#x27;re run this matter at court, you&#x27;re gonna get a is equal to 8159 So probably of a hat should take the back answer. That should be probably probably Hat of A is equal to 8159 over 10,000 which is equal to 0.81 flight.

Gennady Notowidigdo · Answer

Okay, so this question will be divided up into five parts. So what we have is a random process ex of tea, uh, which will end up being a linear function specifically 80 plus b Uh, the coefficients A and B our uniform random variables themselves. So specifically, a is uniform in the interval, 0 to 6 and B is uniform Between modest 10 to 10 are fair minded birth A M B. Ah, continuous uniformed distributions. Just bear that in mind as we sort of progress off through his question. Okay, sir, we start with finding the ensemble off Exit E, which pretty much just beans describe it, so to speak. So not just describing a bit, sort of figuring out what possible, I guess. Sample functions you can obtain, sir. It&#x27;s pretty obvious that exit T is linear. Moreover, the linear coefficients, selenia and constant coefficient coefficients to be more precise. Ah, uniformly distributed. It&#x27;s really Maur at this point. Just describe what dysfunction in tails. So I&#x27;ll probably leave it there because the next full pots become much more interesting. Her case and now to get the main function off the random verbal sir, to get change back the collar. So, um, you sub x off T by definition is just expected value off exit T, which ends up being the expected value off a T plus Be okay. Now T is a deterministic quantity. So if we use the Lini aridity of expectations, this ends up B tee times the expected value of a plus the expected Valley off be, um So if I go back to the first page, So if you look at these two things ah, the expected value off the random variables ends up just being the average valley off these two things. So if we go back to the main function, But this ends up being tee times three zero plus six, divided by two is three plus 10. Mona&#x27;s standing fight about two ends up being zero and you get three tea. So that&#x27;s the main function of X city, which we will use quite suit. Okay, so the next part we want to find the altar curve variance function of X t. I&#x27;m gonna go back to this. What I&#x27;m gonna do is I&#x27;m going to look at the altar correlation function because once we get the altar correlation function. Then we can get the water. Barbarians function just by subtracting the products off the main functions at T N s. More on that that suit. Okay, so the water co correlation function, which is denoted by our sub execs off s This by definition, is just expect. Valley off ex of tea times X off s. So if we substitute all the irrelevant quantities, uh, there are there, so should be a t plus be times a s plus, Hey, then what we can do is expand the brackets that the expected Valley off a squared T s plus a B T plus a B s. So what I can do is I can do that. So just factoring out tea out of a B, it just makes it easier later on to deal with the, um, deal with the product. And the other thing we need is b squared. A careful got to point something out at the start, which will become crucial here, and that is that I want to get the black color. Um A and B are independent. Okay, this is crucial for we going to do, um, in probably two steps time. So it&#x27;s probably prudent for me to just get that it. All right, So what are we left with? So using live in the area of expectations so that the expected value off a squared times T s plus the expected value off a B Times T plus s plus the expected value off B squared. Okay. All right. So I&#x27;m gonna go back to the start again. So let me just recall a couple of things. I do it in here. So if X is uniformly distributed, say, between A and B, then the variants of X is really good. Just up beam. Honest, eh? All that squared, divided by 12. So I think this is in Sections three, chapter three off the book. In any case, it&#x27;s always best to Google the continuous uniform distribution or searching on Wikipedia, which has sort of a list off what the variables are. All right, so once I am here. Okay. So I&#x27;ve got this. Um, I won&#x27;t bother doing any extra supplication. So which is gonna use that quickly? So the expected value of a squared. So, by definition, the variance of a it&#x27;s the expected value of a squared minus the expected value all squared. So this just ends up being the variance off, eh? Which is this? Minus the figures. I ve all squared. So just need to add on the expected value off, eh? Oh, squid. Okay, um then what we can do? He&#x27;s suggesting now, because I am be our independence we can rewrite. This is the expected value of a times three expected value off B times T plus s and to finish off. I just need to write this in to start to expect about it be squared in the same way as I read the expected value of a squared, which is variants off B plus the expected value. Sorry Off, sir. I think it would be, um so I&#x27;m just trying. All right, The rigor. So expected value off be all square. All right, so let&#x27;s cancel a few things out. So first of all expected value be was zero. So this term will not exist sick, and this term here will not exist. Okay, So what&#x27;s next? What&#x27;s next? Is that do compute eso variants off, eh? So remember a was uniform between Syria and six. So the parents of a is six minus zero or squared. Divided by 12 6 square. This 36 divided by 12 is three. On the various of B will be 10 minus minus. Austin, which is really just 10 plus 10. All that squared, divided by 12 eso 20 squared is 400 divided by 12. And if you really want to simplify this, it ends up being 100 over three. Okay, you&#x27;ve got those two qualities. So what is the altar correlation function? So the auto correlation function from what we&#x27;ve left off with. So this is going to be three, sir. Three squared is nine. Uh, this waas three and last of all this it is 100 over three. Hold it like that. All right, sir, Plugging all of daddy three plus nine is 12 served 12 times t times s plus the 100 over three terms on. That&#x27;s pretty much it with the auto correlation function. So once we get this, um, getting the so getting the Walter Kerr variance function in Part three is very simple, sir. Us up where we are this time we see my bad c sub x x off t s. This is equal to the expected value off ex of T X s. By definition, I think this was also something you had to show in problem 12. And now we need to subtract. Um, use up ex off tee times, muse up, ex off s. Okay, Well, this is the first term here is purely derived from pot de. They&#x27;re getting the altar correlation function. Um, so the altar correlation function waas if we call 12 times t times s plus 100 over three. Now, Musa Becks of tea is three t times that with three s users nine s t 19 s. And when you do this attraction, you end up with three t s plus ah, 100 over three. And that is our altar curve arians function. So we got one part left to figure out which is part E, which is D variance function, which we denote by Sigma squared sub X off T. And this is simply just the variance function evaluated at t and T. So from this formula here, we just substitute as equal to t and we end up with three t squared plus 103. And that pretty much concludes the question. The problem. Full team. Thank you

Mihajlo Grcic · Answer

So first, you stupid finding values for A and B that they&#x27;re asking the problem on their correspondent. You&#x27;re the probability so feel zero and b o y t energy states. Generally speaking, be off some energy is because daytime&#x27;s exponential function often minus constant be That was the energy. So from the figure 10. Why do we can see that from that for you? Zero. The probability is actually 0.3 days. Five So is going to be 0.385 Because this is going to be zero and the exponential function. What? So from here we can Ryan divi od cost a day now from the figure insane figure we should So how find for the energy Wani now devalued the coasted To be so big is one a And then the Sequels to cost day times means to mine has be and they want Wani energy level So when we divide this weight okay we have dangled be want a money over a is actually explosion function to be time Swanee and this left hand side here&#x27;s an American body and this value is this is from the vigor and also all But we already know which is equal to zero point 665 No notice that. Yeah, you good. Both sides as a snagged events off a natural operative fire shoe. So just to both sides of the equation, But they obtain that basically minus me over one he is equal to I&#x27;m off 0.665 Therefore, you can say that be equal, since this is going to be negative. Value the equals Wani times 0.41 and this will be Carly will look and treat these constant in future. So, basically what field? Same to say from the city off A is 0.38 fire times expression function negative. And then 0.4 to 1. That energy over who want e image on using this and putting it to calculate through the property values were paying that, for example, for busier. We know that would be good for B one B. It would be 0.385 times he in here. Look, you have negative 0.4 to 1. Ronnie over. Want the which cancels out and this is equal to you&#x27;re pointing 56 on the same matter. Well, just here, for example, For Toby, this will be a zero point fire 385 times special fashion. Now we will hear my guess is your point. For too long times Joe, he&#x27;s and Lunke and this will be This will be just castling and this will be to secure here We were left with one and the Sequels to 0.17. And in the same manner, we just re evaluated for all energy states which will be for the third energy states for the 400 g ST then for for the 1500 state. Then for the sixth Interview state for the seventh and 48. Finally, 145. No, this water&#x27;s should be plotted, but unions match lab board Wolfram out for our business. But way to plug this you and the energy will go from zero to hey, reserve money to beat me and so on. Probably to expand this. So this will be seven on them. Eight. And here this will go from zero to zero corns. One. Yeah, And though with the July we will show approximately how the when we blot each value where the college would look like this? What? They&#x27;re the exact her blood off. These probability of audience will look like this here and then here. And the exponential function awfully in this sort of. We tried to fit in. So it will be I thought they were going to here. There will be almost the same. But then so near the point to they will start to split. Well, exactly like this. And so this shows that comparing this wall is only be zero and be one e matches the exact policies, But then results starts to go away as we increase the energy on this is the sketch of the percentage off. Mashing between the exponential function ad is probability distribution function. So it&#x27;s a good approximation for lower energies about not really good for higher energies, energy states. This is the end of the problem. I hope you find it&#x27;s helpful and they hope to see you in another lesson.

Joshua Sieverding · Answer

For the first part of this question, we want to find the probability that a randomly selected component will last at most 12 months. So at most 12 months means it can last. We want to find the probability that t it basically is not greater than 12. Such the probability that t would be between zero and 12. Let&#x27;s do that. We&#x27;re gonna set up, are integral from 0 12 one of two e two negative t over to duty, and we&#x27;re going to see that this is negative. You to the negative T 02 Evaluated from 0 12 playing in the upper bound, we would get negative e to the negative six playing in the lower bound who get plus one. So rewriting this a little cleaner would have one minus e to the negative six for the next part. We want to find the probability that it has a lifetime between 12 and 20 months. So it&#x27;s gonna be basically the same problem except just changing our bounds of integration. So we&#x27;re gonna get this exact thing here and our bounce than playing in the upper bound of 20. We would get negative e to the negative 10 and the lower bound of 12 we get plus e the negative six. And again, I would rewrite this to be e to the negative six minus e the negative and the last part. Well, I&#x27;m saying the last part is to find the cumulative distribution function for this random variable. So are already part of the way there. Just by doing this integration, right? If we want to find the CDF, we want to do that. Thea integration of the pdf with no bounds are integral. And so it&#x27;s gonna give us negative e to the negative t over to plus C and the value for this constant that we want, um is going to give us Oh, either a one when we plug in the right and point or a zero when we plug in the left endpoint. So let&#x27;s start by plugging in the left endpoint here. Um, if we plug in zero to our function, we&#x27;re going to get e to the zero negative negative e to the negative zero, which is just gonna be negative one, um plus c So get negative one plus e. And we said that we want at the left endpoint zero to get a value of actually zero. That means air Sea has to equal one. And so if we go ahead and plug in the sea into this function here, we would get the FFT is equal to one minus e the negative t over to and that&#x27;ll be our cumulative distribution function for the given probability density function. Uh, and we&#x27;re supposed to use our answer from part C find the probability that a random selected component last at most six months sub d Probably that it will last at most six months. Kind of similar to a Actually, it was saying, at most 12 months. Um, well, if we want, if we know that we have the cumulative distribution function and it tells us the probability up to a certain point and we&#x27;re trying to find that it&#x27;s at most six months, that&#x27;s all of the probability up to six months So we can just go ahead and plug six in to our CDF and we would get one minus e to the negative three. Does it be negative? Six Over to that would be our answer. And if you actually look at solving a we could use the same CDF we could just plug 12 in on. We would get the same answer we got using our integration technique. Appear we would get one minus e to the negative six. Um, So our final answers are gonna be one minus seat of the 96. Even the negative six. Minus you tonight. 10th 1 minus e the negative t over to and the last part is one minus e to the negative third.

Linh Vu · Answer

in this question were given first asked to you could do no 37 plus six times able minus upon three day the first time we need to compute the integral from 0 to 6. Because we have the six years here under 37 Press six Able Moniz upon Treaty De and then we get equal to that. This 73 until you lived in the 37 usually get encoded, oppressed six a. Berman is a Ponzi treaty the dividing by, uh, minus organiser tree and even lamented Ondas to six and again to juju. And for this one, I think you the man of the upon 03 Tame six. Okay, 10 6 year on Dan. Ah, we have. Could you tell her that we have a power off the answer and then Thames eight Tim six year. And then they running by Manas Upon zero three, they would get in cordial minus one, not 67.54 and it will Buddha&#x27;s Aaron. So we go under minus zero and then plus, does your inside Uganda six divided by Is there a 0.3? So, six, they&#x27;ve anybody&#x27;s their bonds over three inequity 200 here. So we get ankle June. Uh, there&#x27;s only be 200. Therefore doesn&#x27;t get Echo Judah 200 plus na on due to ju minus 167.54 So it could You know, Joe, if I for one night from six, therefore the average and we go to the Joe 54.9 on 60 70 by six, and then we can get this one. Will be 44 42. Phone, phone 91

Problem

A certain machine has the following hazard functi…

Problem 127 Easy Difficulty

Answer

(a) $[1 (1 R_1(t))(1 - R2(t))][1 - (1 - R_3(t))(1-R_4(t))][1- (1- R_5(t))(1- R_6(t))]$(b) 70 h

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(a) $[1 (1 R_1(t))(1 - R2(t))][1 - (1 - R_3(t))(1-R_4(t))][1- (1- R_5(t))(1- R_6(t))]$(b) 70 h

Probability with Applications in Engineering, Science, and Technology

Boris M.

Heather Z.

Lucas E.

Joseph L.

Data And Variables

Data And Variables - Example 1

Matthew A. Carlton • Jay L. DevoreProbability with Applications in Engineering, Science, and Technology

Boris M.

Heather Z.

Lucas E.

Joseph L.

Matthew A. Carlton • Jay L. Devore

Probability with Applications in Engineering, Science, and Technology