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\fUsing the estimates of the parameters of the Weibull model obtained in Exercise 1, estimate the probability that this kind of diaphragm valve will perform satisfactorily for at least 150 hours. Exercise 1 A sample of 60 diaphragm valves, used in the control system of a chemical process, are placed on life test without replacement. The first 9 failures are observed after 3.6 6.9 9.5 15.7 27.3 41.2 81.7 178.3 227.1 hours. Using the Weibull model, estimate the mean life of this valve. How does this value compare with life that would have been obtained under the exponential assumption?As has been indicated in the text, one often distinguishes between initial failures, random failures during the useful life of the product, and wear-out failures. Thus, suppose that for a given product the probability of an initial failure (a failure prior to time t = a) is 01, the probability of a wear-out failure (a failure beyond time { = 8) is 82, and that for the interval o s t's / the failure-time density is given by f0) = 1 - 81 - 82 8-0 (a) Find an expression for F(f) for the interval astso. (b) Show that for the interval o s t s 8, the failure rate is given by 1 - 01 - #2 Z0)= (8- a)(1-81)- (1-91 -02)(1-0) (c) Suppose that the failure of a digital television set is considered to be an initial failure if it occurs during the first 100 hours of usage and a wear-out failure if it occurs after 15,000 hours. Assuming that the model given in this exercise holds and that 01 and 62 equal 0.05 and 0.75, respectively, sketch the graph of the failure-rate function from t = 100 to t = 15,000 hours.Show that the reliability function associated with the Weibull failure-time distribution is given byAn integrated-circuit chip has a constant failure rate of 0.02 per thousand hours. (a) What is the probability that it will operate satisfactorily for at least 20,000 hours? (b) What is the 5,000-hour reliability of a component consisting of 4 such chips connected in series?A system consists of 7 identical components connected in parallel. What must be the reliability of each component if the overall reliability of the system is to be 0.90?After burn-in, the lifetime of a solar cell is modeled as an exponential distribution with failure rate o = 0.0005 failures per day. (a) What is the probability that the cell will fail within the first 365 days that it is in operation? (b) What is the probability that two such cells, operating independently, will both survive the first 365 days they are in operation?If a component has the Weibull failure-time distribution with the parameters o = 0.005 per hour and 8 = 0.80, find the probability that it will operate successfully for at least 5,000 hours.Suppose that 50 units are put on life test, each unit that fails is immediately replaced, and the test is discontinued after 8 units have failed. If the eighth failure occurred at 760 hours, assuming an exponential model, (a) construct a 95% confidence interval for the mean life of such units; (b) test at the 0.05 level of significance whether or not the mean life is less than 10,000 hours.In a nonreplacement life test, 35 space heaters were put into continuous operation, and the first 5 failures occurred after 250, 380, 610, 980, and 1,250 hours. (a) Assuming the exponential model, construct a 99% confidence interval for the mean life of this kind of space heater. (b) To check the manufacturer's claim that the mean life of these heaters is greater than 5,000 hours, test the null hypothesis / = 5,000 against an appropriate alternative, so that the burden of proof is put on the manufacturer. Use a = 0.05.With reference to the data in Exercise 1, make a total time on test plot. Exercise 1 In a nonreplacement life test, 35 space heaters were put into continuous operation, and the first 5 failures occurred after 250, 380, 610, 980, and 1,250 hours. (a) Assuming the exponential model, construct a 99% confidence interval for the mean life of this kind of space heater. (b) To check the manufacturer's claim that the mean life of these heaters is greater than 5,000 hours, test the null hypothesis / = 5,000 against an appropriate alternative, so that the burden of proof is put on the manufacturer. Use a = 0.05
