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Question I: (10 marks)diesel engine is very complex system and when it fails is mnoSt olten due t0 some of many components; S0 when it is tixed, its hazard rate the...

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Question I: (10 marks)diesel engine is very complex system and when it fails is mnoSt olten due t0 some of many components; S0 when it is tixed, its hazard rate the same as what just before failure Assume that the hazard rate of the engine ish(o)= 120 400Q 40uu(a) What is the (ailure distribution of the system and its parameters? (3 marks)(b) If there is no preventive maintenance of the engine what is the probability that the first failure will be between 3000 and 5000 hours? (3 marks)

Question I: (10 marks) diesel engine is very complex system and when it fails is mnoSt olten due t0 some of many components; S0 when it is tixed, its hazard rate the same as what just before failure Assume that the hazard rate of the engine is h(o)= 120 400Q 40uu (a) What is the (ailure distribution of the system and its parameters? (3 marks) (b) If there is no preventive maintenance of the engine what is the probability that the first failure will be between 3000 and 5000 hours? (3 marks)



Answers

An automobile manufacturer is concerned about a fault in the braking mechanism of a particular model. The fault can, on rare occasions, cause a catastrophe at high speed. The distribution of the number of cars per year that will experience the catastrophe is a Poisson random variable with $\lambda=5$. (a) What is the probability that at most 3 cars per year will experience a catastrophe? (b) What is the probability that more than 1 car per year will experience a catastrophe?

Here were given a possum process, which is vehicles arriving at an inspection station at a rate of 10 per hour. So we have put some process and the rate is 10 per hour. But we're also told one other thing. We're told that the probability of a vehicle having no equipment violations is half, and so the concept that this question touches on is that of decomposition of porcelain processes. So if you have a personal process and some of the events could be characterized in a certain way, while the other events may not be characterized that way, then you can decompose the process into two separate personal processes. So in this example, we have a personal process, which is the arrival of cars at the inspection station, and half of them could be characterized Probabilistic Lee. Half of them could be characterized as having no equipment violations, so we can actually split this into two separate separate portion processes. So let's call one X sub n where N stands for no equipment violations. The rate for this process, then, is P Times lambda, which is equal to 0.5 times 10 or five, and so the second process is the events that involved vehicles that do have equipment violation. So let's call that XV for violation. And similarly, Lambda Savi is going to be five in this case as well, because it will be it's defined as one minus p times the overall rates. So in this case that comes out to five as well. Now, for part A were asked what the probability of exactly 10 cars arriving in one hour. So these rates are already per hour, so this rate is 10 per hour. So then are two processes are five per hour, whereas the probability that exactly 10 cars arrive in an hour and exactly 10 of them have no equipment violations. So this could be characterized as air expressed as the probability that, except N in one hour equals 10 intersection with X. Avi in one hour is equal to zero. So that's the probability of getting 10 arrivals of cars that have no violations and zero arrivals of cars that have violations in one hour. Now. One more thing to be said about, for some processes that air decomposed in this manner into to like taking one overall process and decomposing into two is that these two events are independent. These two processes are independent, so this probability, therefore it could be written as follow can be written as the product of these two probabilities, and this can be calculated as follows and it comes out to about zero point 000 12 So the probability of 10 cars arriving in the hour and all 10 having no equipment violations is 0.12 So now for part B for any were asked that for any fixed y being at least a bigas 10. What is the probability that exactly why vehicles arrived during the hour, of which 10 have no violations. So we're looking for the process of no violations in one hour to provide 10 events and the process of violations in one hour to provide why minus 10 events. So that way, the total number of events between these two processes well, some tea, why and exactly 10 of them will have no equipment violations. So we're looking further probability x of n in one hour was 10 and that XlV in one hour is equal to y minus 10. And because of the decomposition. These are independent processes, so we can rewrite the probability as to product as the product. So it's a probability except and at one it's equal to 10 times the probability that XlV in one hour is equal to you. Why minus 10 And we can write this out as follows and this expression can be simplified a bit to eat the minus 10 times five to the exponents. Why over 10 factorial times? Why minus 10 factorial and then for part C. We're asked, what is the probability that there are 10 no violation cars arriving in the next hour? So that is simply the probability that the possum process for no violation cars results in 10 events in the next hour. And that's calculated as e to the minus five on five to explain in 10 over 10 factorial and that comes out to about 0.181 And this works because this process is independent of the possum process pertaining to the arrival of vehicles with violations. So we simply have to find the probability of 10 no violation cars arriving irrespective of the number of vehicles with violations arriving

This problem, we're told that a buyer will only buy an engine if it starts successfully 10 consecutive times. I'm told the probability of the successful starting 0.98 and Soapy is 0.98 now on a We want to find the probability that the engine is accepted after only 10 starts. Now, this just means that it's successful 10 times in a row and so that probability is going to be a 0.98 to the 10th power in the four significant digits, that would be 40.8171 on B. We want to find the probability 12 attempted starts are made. Now, first, if we focus on the 1st 11 now, since the 10th one is going to tell us to stop, excuse me, we need 10 consecutive 10 consecutive. So rather than focusing on the 1st 11, if we're dealing with 12, that means that it had to fail the first two times. But then it was successful the next 10 because it needs to be 10 consecutive, not just 10 in general, but 10 consecutive. And so the only way for that to happen in 12 starts is for it to fail the first two and then make it the next 10 to grow. And so this probability this 0.0 3268 Uh huh.

We're told that the lifetime and thousands of hours of the motor of a brand of kitchen blender has a Waibel distribution with Alfa equals to invade a equals one. In part, they were asked to determine the reliability function of a motor and to graph it so we know from Chapter three that's the Waibel CDF is going to be function ffx equals one minus you to be negative x over beta to the alfa. So it follows that the reliability function is our tea, which is one minus F 50 which is e to the negative t over beta to the alfa. In plugging in the parameters, this is going to be eat to the negative t squared. So the graph this I'm only graphing for t greater than or equal to zero. Well, we have that zero r t is going to be one, and we have that around one r t r of one is approximately you. The negative first, which is about 0.4. We have a T equals two. Well, our of two is equal to eat the negative fourth, which is about 0.5 It's a very small, and it keeps getting smaller and goes to zero as T approaches infinity Now for the other values for T less than two. We have that. It's sort of a sigmoid all form, so starts a little bit flat and slopes down and slips back out like that. Next in Part B were asked to find the probability that a motor will last more than 1500 hours. Well, at TBD lifetime of a motor in thousands of hours, we were looking for the probability that T is greater than 1.5. So lifetime is greater than 1500 hours. This is the same as are of 1.5, which by part A. This is each of the negative 1.5 squared. Evaluating this is approximately 0.1054 in part C. We're asked to determine the hazard function of this motor into graphic. Well, now we're looking for the Waibel Probability density function, So the Bible probability density function is f f T equals to tee times e to the negative t squared, especially because we have the Alfa. So you go to and beta is equal to one mhm. So it follows that the hazard function is h of t and this is FFT over our tea And so we have to t e to the negative t squared over either the negative t squared just simply to t so it follows that the model has a linearly increasing hazard rate to graph h of t on the horizontal axis values of T on the vertical axis LF values of h of t so we have that t equals zero h of t is also zero t equals one h of t is going to be too You can dry straight line between these two points to obtain the graph of h of t finally in part D, whereas to find the mean time to failure of this kind of motor are using the data from this section we have mean time to failure is given by U T, which is the integral from zero deposit infinity. This is what t is to find on of r t d t and substituting. We get integral from zero to infinity e to the negative T Square DT and we know that looking in using you substitution u equals T squared D is equal toe to t d t so d t is equal to do you over to tea and we have also the tea. Since it's positive 20 equal to the positive squared of you. So this becomes you have zero is still zero and you have infinity is still infinity of e to the negative You and then GT becomes one over to tea. Do you or one over to route you? Do you, which can be written as one half times the integral from zero to infinity of eat to the negative You times you to the negative one half, do you? And now applying the gamma integral formula in this case with Alfa equal to one half in beta equal toe one this integral is going to be while mu of t is going to be one half gamma of one half times one to the one half and we know that gamma of one half is simply the square to pie. So this is going to be one half times route pie, which is pi over Route Pi over two, which is approximately 0.886 because T was in thousands of hours. It seems we have approximately 0.88 6000 hours for the mean time to failure or 886 hours. Now recall from section 3.5. The expected value of a Waibel distribution for Alfa equals two and B equals one. This is supposed to be one times gamma of one plus one half, which is the same as Gamma of three halves, which is the same as one half gamma of one half, which is the same as what we have above root pi over two. So this confirms.

Yeah. In this exercise, the random variable Y. Is the number of components not staying longer for $1000. So, and it calls for & P. Eco's .8. Using the formula of the phenomenal distribution, we can write the distribution of this random variable, prepared a. We want to find the probability that Because they actually two of the four components last longer than 1000 overs. So this is just a p. two and it echoes point 153, 6 four point B. We want to find out the probability that the sub system operates longer than 1000 Oliver's. So this is just the sum of the probability of p. two, p. 3 and painful. And it echoes point 9 7- eight.


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