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Consider system comprising three components_ The system requires at least two out of the three components to working for the system to work: Components fail indepen...

Question

Consider system comprising three components_ The system requires at least two out of the three components to working for the system to work: Components fail independently and the time to failure for each component has an exponential distribution with mean of one year:Determine expression for the probability that the system is working at time Hence, give the probability density function for the time to failre of the system_ mark Determine the mean and variance of the time to failure for the syste

Consider system comprising three components_ The system requires at least two out of the three components to working for the system to work: Components fail independently and the time to failure for each component has an exponential distribution with mean of one year: Determine expression for the probability that the system is working at time Hence, give the probability density function for the time to failre of the system_ mark Determine the mean and variance of the time to failure for the system_ [2 marks] Determine the probability that component one in the system is still working at time given the system is working at time What is the limiting value as t_0? 2 marks] Suppose now that the system has one spare component which can be used to replace component in the failed system_ The time to failure ofthe spare cOm- ponent is also an exponential distribution with mean of one year._ Consider the following repair regime As soon as the system fails the one of the failed components in the system is replaced by the spare and the system continues to function as long as there are at least two components are working: Simu- late the processes in R or MATLAB You may wish to use the function sort in MATLAB or R Provide histogram of the failure times for the repaired system and estimate of the mean time to failure for the repaired system_ Supply VOuT code. marks



Answers

Did you ever buy an incandescent light bulb that failed (either burned out or did not work) the first time you turned the light switch on? When you put a new bulb into a light fixture, you expect it to light, and most of the time it does. Consider 8 -packs of 60 -watt bulbs and let $x$ be the number of bulbs in a pack that "fail" the first time they are used. If 0.02 of all bulbs of this type fail on their first use and each 8 -pack is considered a random sample, a. List the probability distribution and draw the histogram of $x.$ b. What is the probability that any one 8 -pack has no bulbs that fail on first use? c. What is the probability that any one 8 -pack has no more than one bulb that fails on first use? d. Find the mean and standard deviation of $x .$ e. What proportion of the distribution is between $\mu-\sigma$ and $\mu+\sigma ?$ f. What proportion of the distribution is between $\mu-2 \sigma$ and $\mu+2 \sigma ?$ g. How does this information relate to the empirical rule and Chebyshev's theorem? Explain. h. Use a computer to simulate testing 1008 -packs of bulbs and observing $x,$ the number of failures per 8 -pack. Describe how the information from the simulation compares with what was expected (answers to parts a-g describe the expected results). i. $\quad$ Repeat part h several times. Describe how these results compare with those of parts a-g and with part h.

And this example will be looking at a rectangular distribution, also called continuous uniform distribution. This is the situation where we're looking at a rectangle formed on a closed interval alpha beta with the selection inside there on a closed interval A and B. And our formula for the probability that X lies between A and B is B minus A, divided by B minus alpha. And in our particular problem, alpha and beta are determined to be -0.05 And 0.05. So in the first part, I want to know the probability that X is less than or equal to 0.03 microseconds. All right. So it can't get any smaller than negative 0.05. So that means they will be -0.05 and B will be the larger value. Now, I'm going to plug the information into my formula. So B minus a, divided by beta minus alpha. And the next example, I want to know the probability that I might have Oh more than so. It's not equal to just more than And recall my area can't get any bigger than beta. So that means in this case a will be negative 0.02 and be Will be 0.05. And to find that probability B minus a divided by peter minus alpha. And then I do the calculations. Then there you go. For C. I'm actually looking at a situation where I want to know the probability that X will be between These two, values of 0- .04 and .01. So that means they will be -0.04 and B will be 0.01. Again, just a matter of plugging the values in B minus A divided by beta minus alpha. And then lastly, I mask defying the mean and the standard deviation. So our formula to find the mean is alpha plus beta divided by two. So in our example negative 0.5 plus 0.5 divided by two. The numerator is zero, Which means the mean is zero. And then our standard deviation formula is beta minus alpha, divided by the square root of 12. So that would be 0.05 -0.05, Divided by Radical 12. Mhm. And that's approximately 0.289. And then we need to were asked if our measurements are unbiased and in fact we can be sure that they are because We have a mean of zero and that would indicate that our information, our measurements are actually unbiased. I'm sorry, I ran out of room.

Um were given a model for the distribution of the random variable. Why? Which is the time to pavement city? But told that x one time to failure between running. That's two times of failure you to trans verse scripting. We're told these two random variables are assumed to be independent. Then we're told us running variable. Why is the minimum of these two random variables That's when the next to so the probability of fears would be to either one of these. Those is assumed the an increasing function of Time team and this would seem to make sense right? The longer the pavements out, more spokesman the weather and you think the more likely it's going to fail? Sure, now mhm to making assumptions mhm form of the CDF. Each node was detained to be five of a plus B t over the square to C plus B T plus E. T squared the fires, the standard normal, chaotic distribution function now it's cold were given the values of the parameters. A, B, C D V cracking and also different days or running were asked, defying the probability of statement failure within five years and also within 10 years. Okay, so first I'll consider the case of five years. Well, let f one and F two the accumulate distribution functions of X one and X two, respectively. Probability of pavement failure within two years. Mhm probability. That's why his last century, quarter T this is equal to the same as a probability at the minimum of X one, and x two is less than or equal to t. And this, of course, is the same as one minus probability that the minimum of X one next to is greater than teeth. Now notice that the minimum of two numbers could be greater than a number if and only if both of those numbers are greater than that number. This is the same as one minus ability. The X one is greater than tea, and that's two is great beauty. And since we've assumed that X one next to our independent follows that this is equal to one minus the probability that X one is greater than tea times, the probability that X two is great in T, which is the same as one minus one, minus the probability that X one is less than or equal to t times one minus the probability that X two is less than a quarter T which we know. It used to be one minus one minus form of tea times one minus F two of teeth. You're using the king of the distribution functions that we were given all that. This is equal to one minus one minus. Bye. Um, and this is for X one. We have negative. 25.49 plus 1.15 TV over square root. Um, 4.45 minus 1.78 t plus 0.17 won t squared, Mrs. Simply coming from the values that we were given for the constants times one minus cap If I, uh Mrs for the random variable x two negative 21.27 plus 0.3 to 5 Team over the square root of 972 minus 0.2 18 +100022 T square. Now if we are to evaluate a T equals five, see the probability that why less than or equal to five, then we have the argument inside. The first five is imagining it. The argument inside first five correspondent. F one is imaginary. However, we haven't defined dysfunction for imaginary numbers. Now, if we evaluate AT T equals 10 then we have the probability that why is less than or equal to 10. This reduces toe one minus one, minus five. Approximately negative. 7.22 times one minus phi of approximately negative 21.14 Mhm incredibly small number. Both of these really are. So this is really approximately equal to one minus one times one, which is zero. But there's a zero chance that we cracked within 10 years, which, okay there within 10 years, which is clearly nonsense. Obviously, there should be a chance that will fail within 10 years.

Okay, so that's what P. But a probability of a flaw being detected equal. Reports a were asked the probability of it being detected by the second fixation. That's better. It being detected on the first or not being detected on the first and being detected on the seconds that's is equal to P plus not being detected is one night. It's p times it being detected on a second. Okay, now we're asked report, Be it being detected on and fixation. It is equal to one, minus it not being detected. End times. No, I asked where the whole body of a flow item past persons inspection after Drea fixes that's equal to one witness. Be to part three. It's because it wasn't infected three times. And the property. If our random item, it's flawed probability of a random Oh, I don't applaud his 00.1. What is it? The probability of it passing after doing inspections that's the same as war minus P. There are three times this rate. Three times never are being problem ot of the flaw passing, given that P is equal support. Five. So that's one when it's fortified to Part three, which is able to 0.1 to fight

We have a system composed of five components connected in Siris so that as soon as one component fails, the entire system has failed. Each of the components lifetimes is exponentially distributed and the components failures are independent from each other, and each of the exponential distributions for each component has a parameter lambda equals 0.1 event a sub. I is the event in which the ice component lasts at least t hours. So let X be the time at which the system fails. Then for part A. We're looking for the event excess at least t We're as to express this event in terms of the A one through a five events. So for the system lifetime to be greater than t each individual component must last at least t hours. So therefore, this is the intersection of a one through a two through to a five. These five events must happen now for Part B were asked to compute the probability that the system lost at least two hours. Remember, since their exponential random variables, each components lifetime. They're CDF zehr, given by one minus E to the negative 0.1 t so this is for one individual component. And so we could say that the probability that it's called component one or a component I is at least t hours is equal to one minus the CDF. So this is the probability of an individual component exceeding at being at least t hours in terms of lifetime. Now, since we need all five components to last at least t hours And since the component failures air independent, the probability of all five components succeeding T hours is simply the products of their individual probabilities of exceeding five hours. So it's simply this probability to the exponents five, and that equals zero point E to the negative 0.5 t. So this is the probability that the system lasts at least 80 hours. And then we were asked to find the CDF of the whole system, which would be one minus the probability that we just solved for one, minus the probability that the system lifetime is at least t hours. So that is one minus e to the negative 0.5 t. And we were also asked to find the PdF of X so we can find this by taking the derivative of the CDF, and this comes out to 0.5 times e to the minus 0.5 times T. And of course, he has to be at least zero. We can't have a negative lifetime for the system, and this tells us that X is exponentially distributed with parameter Lambda equals 0.5 So now for Part C, we're told that there are n components in each having an exponential lifetime with parameter lambda. So what type of distribution does this system have that we can? We can continue in a similar fashion to part A. We find the probability that the system lifetime exceeds or is at least t hours. That's the product of the probability is that each individual each of the end components is at least t hours in lifetime. So this right here would represent the probability that an individual component last at least three hours. We're looking for the probability product of end of them, so that comes out eat to e to the minus n times landed T and similar to part A. This means that the pdf of the system is one minus this. Therefore, the pdf is N. Times, Lambda Times E to the negative and times, Lambda Times T and we can see that this is an exponential distribution with parameter end times Lambda.


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