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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.

(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

Intro Stats / AP Statistics

Chapter 4

Joint Probability Distributions and Their Applications

Section 1

Jointly Distributed Random Variables

Probability Topics

The Normal Distribution

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were given a six component system with component reliability functions are one through our six, and we'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'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'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'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

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