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The article "Response of $\mathrm{SiC}_{\mathrm{f}} / \mathrm{Si}_{3} \mathrm{N}_{4}$ Composites Under Static and Cyclic Loading- An Experimental and Statistical Analysis" (J. Engr. Materials Tech., $186-193$ ) suggests that tensile strength (MPa) of composites under specifions can be modeled by a Weibull distribution with $\alpha=9$ and $\beta=180 .$(a) Sketch a graph of the density function.(b) What is the probability that the strength of a randomly selected specimen will exceed 175$?$Will be between 150 and 175$?$(c) If two randomly selected specimens are chosen and their strengths are independent of each other, what is the probability that at least one has strength between 150 and 175$?$(d) What strength value separates the weakest 10$\%$ of all specimens from the remaining 90$\% ?$

(b) .4602, .3636 (c) .5950 (d) 140.178 MPa

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

Missouri State University

Piedmont College

Cairn University

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question we're dealing but the random variable representing strength of composites I denoted as Capital X here it has very able distribution with off nine and bit of 180. And in part, one of the question we're asked this Catch a graph of the pdf function. I already have the PD a function of label distribution here denoted with lower case F. It's a function of X variable and the two parameters Alfa and better to start sketching the graph. We should remember that off the shape parameter and better is this cave parameter. So it's kind of controlling the scale of the graph. And from the definition, we can see that for negative X, we don't have anything the pdf graph. So it's all zeros here and how offer controls the shape. For example, if offer equals one, the function is simplified to an exponential function like this, which is not the case we're looking for in this question. So what? We're more interest that in as rather off being significantly greater than one, which gives us this, uh, bill shape graph and you should also remember from the textbook that for that equals one the maximum off. The graph happens around one, and this is how the scale parameter works. For example, if we consider better equals two, we'll have this kind of graph where the maximum happens around too and so forth for larger betters like this. Fun for PETA equals tree. It has the maximum at tree, so better kind of stretches the graft words X axis, as you see for better cycles to is kind of the same as the graph for better I cause one, but stretch towards the X axes with the factor too. This way in this question are alpha is nine, which is significantly grated and one so we have this bell shape. Pdf there. We already know that the maximum happens somewhere close to better, which is 180 in our case. And we're left with only calculating the height a pdf at the maximum. So to do this, we simply you can plug in better to the pdf function here. And so, by replacing better with expert better, we can calculate this height. So what? Plugging in bed time instead of X to the pdf function. We're calculating f better, which replaces X barometers off on better again. So this would be over. Better to the power off. And here I replace X with bit to the power off minus one and eat the power minus again and replace X wit better here. Over. Better to the power Onda I canned is really simplify. This result writing Alfa. I can separate these two powers here off on minus one. So I have all four multiplied Better the power off multiplied. Better to the power minus one over it. Out of the power off. And here lots of life I e to the power minus one. So bit our better one to the power. And I can easily simplify better to the power Alfa from this, uh, result. And I also want to the power Alfa is one. So I can simplify this as a off multiplied bit of the power minus pond and into the power minus one, which can simply be written as off over that. And this is an interesting result. This is, uh, the height of the graph or the maximum point here off over better he And finally, we're ready to sketch the graph by calculating the maximum point on the graph having our Alfa and better So I am off on nine and better 180 don't supplied by E And for this one we have to use our calculator and the result would be 0.2 So this is, uh, the grab you were looking for. We can replace bid up by 180 here and the high it is zero pine zero to in part b of the question we have been asked calculate the probability of our random variable being greater than 175. And also we want the probability of X being between the two while use 150 175 here. We should use see the F function or accumulated density function because that is the one having relation direct relation to the probability values. As you should know, the CDF function, usually denoted with capital F, has defined this way F a B equals integration minus infinity to be to the point b of the function. If of x t x were F of X is the pdf function. And, uh, we already know that too propositions saying probability of X being greater than a constant value like B is equal to one minus CDF function or EHF other point B and also a probability of ex being between the two values A and B is equal to capital F B minus capital F A. So simply by comparing the propositions with values were looking for. You can rewrites the values as one minus F 175 for the 1st 1 And for the 2nd 1 we would have half a 175 minus f 150. So you see, we only need to values of F 175 F 152 cabinet, uh, answers. So, fortunately, we are given the CD a function for were able distribution in the textbook, so zero for negative excess and one minus e to the power minus X over. Better to the power off for positive accents. So by plugging in the 175 and 152 X here, we can simply calculate if 175 and 150 here we get one minus e to the power X is 175 better. We have it here 180 and off. We have nine. So this is something that we have to use the calculator to. Cantlie this term one minus e to the power minus 175. Overrun eighties Power nine. Using your computer, you can should be getting your point 54 disparate. And after 150 you can calculate the same way just replacing 175 with 150 you will be getting 0.18 So, having these two quantities, we can replace in the answers we already found. So one minus ever 175 will be one minus zero point 54 which we will be getting 0.46 and the 2nd 1 So 0.54 minus 0.18. The answer is 0.36. So this is the answer for the first question, and this is the answer to the second. He still okay. We have this result from part B at the question probability of ex. Being between these two constants is 0.36. And let's call this criteria the pass event that is X between 151 175. Let's call it pass. So probability of pass. You already have it as 0.36 and having probably have passed, we can calculate, probably fail as well. So failed the compliment event of past and its probability should be one minus probability of past, which is 0.36. So operability have failed. Would be 0.64. And here, in part C of the question, we were asked to take two samples of X. That is, for example, if considering considering this sample space, you want to take out one sample X and see if it passes or fails, that is, if it is in this period between 151 175 and then we take another sample and examine it to that criteria like the finish in of the question. And Parsi is to take to Sam pose of eggs, and then we want to compute the probability off at least one pass. So we want to know what is the probability that either or both of our samples pass so we can simply compliment that event. So at this one past, a complement of that would be no pass. So if none of our exes happened to be between the two values we have two fails. So the probability of at least one pass is equal to one minus probability off no pass or both failed. And because these two samplings are completely independent from one another, we can write the probability of both fail to the probability of fail for the first sample multiplied by the probability of failed for the second sample. This is the property off independent samplings. And we already had counseling to the probably of fail here 0.64 so we can simply plug in points. 64 year calculating one minus 0.64 multiplied by points 64. And that gives us point 69 which is the final answer toe the question which was probability of at least one pass of the question. We want to find a value that separates the weakest 10% of the samples from the wrist. To do that, let's call the value we're looking for X of s. So we're trying to find X of s in a way that 10% of the samples are weaker than than that and we care samples. They should have lower values, so thes points can represent weaker samples and, uh, to find the frequency or the number of the vicar samples. We need the information from the pdf. We already thought that the pdf on the part on part A of the question. It's something like this. And we already know that the value of pdf represents the frequency of each value. For example, the frequency of the value X s in our samples Race is proportional to this height. That is the value of pdf at X of s and the same for other points. So conclusion is, Dad. The area under the pdf function up to x of s represents the frequency or the number of ah, weaker samples in the whole example. Space death one, we want to be 10% and the rest of the area under pdf function should represent the rest 90% of the samples. So now to calculate the red area the area of the samples smaller than exit s, we should calculate minus infinity integral from minus infinity to the point except ass of the pdf that is f of X. And we already know this is the definition for capital F or a cumulative density at the point x of s. We want this value to be 10% off the whole number of values in sample space. And because the area under pdf function is 1 10% of one will be 0.1. And, uh, by solving this equation for X s begin, find the value the separating value we were looking for. So off this equation for exit as we already have the F which is C the F function. Ah, label distribution that is one minus me to the power minus X that we replace it with X of s over. Better to the power off and this should be equal toe 0.1. So by reorganizing this equation, we can get toe you to the power minus except s over. Better to the power off equals 0.9 on again By taking That's right. I look a rich him from both sides. We can get minus except s to better to the power off equals now and a 0.9 and continuing. Here we have x of s over better we take the negative to the other side so minus on and 0.9 and we also take the all for route from both sides. We have this one. And at last we have ex service equal toe better multiply by off road of minus L and, uh, point night. So now we can plug in offer and better for calculating the final ex service that it is the separation point between 10% and 90%. Uh, the weakest and stronger samples. So better is 180 off is nine and minus. L and appoint night. You can use calculator to get this value, and that will be 100 40 point 18. This is the separating point.

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