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Let $X$ denote the time to failure (in years) of a hydraulic component. Suppose the pdf of $X$ is $f(x)$ $=32 /(x+4)^{3}$ for $x>0$(a) Verify that $f(x)$ is a legitimate pdf.(b) Determine the cdf.(c) Use the result of part (b) to calculate the probability that time to failure is between 2 and 5 years.(d) What is the expected time to failure?(e) If the component has a salvage value equal to 100$/(4+x)$ when its time to failure is $x,$ what is the expected salvage value?

B.) $1$ $\\$ C.) $0.247$$\\$ D.)$24$ $\\$ E.)$4.35$

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

Aishwarya 1.

November 14, 2020

Andy E.

December 5, 2021

answers wrong for d & e? d should be 4 I think and 100/28 is definately not 4.35 even if you use the expected value of 24

Temple University

Missouri State University

Piedmont College

University of St. Thomas

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Okay, so we've been given a pdf for a function that is equal to 32. So we're gonna say f of X, which is the pdf, is equal to 32 divided by X plus four. It's cute for all X greater than zero. So this is our function right here. Our problem state right here. And the first thing I'm gonna do is we're gonna verify that this pdf this this f of X here is a legitimate and what that means is that the integral from negative infinity to infinity of this pdf is equal to one. However, given this boundary condition here, um, we're going to instead swap the lower bound for zero instead of negative finished. We're gonna take this integral, you know, everything. Like it out a little bit there. You of f of x, from zero to infinity with DX. Which will we evaluate with this expression here, Um, well, give us an integral which I will save some time here by not performing all the little steps, but we're gonna get negative 16 divided by X plus four. It's quick, and this is our integral right there. And now we're gonna evaluated at our bounds, which are zero and infinity. And we put this in, we could think about celibate. So when we put in the infinity here and we take the limit, right, if you're being mathematically accurate, so we're gonna take the limit as experts affinity. Well, this terms all gonna go to zero. So the first term of this substitution of our bounds, um, there's gonna be here, right? And now for the second term, when we put in this zero here, we're going to see that this x zero such as 16 we'll make it a 16/15. So that's going to negative 16 divided by 16 which, when you evaluate this whole expression here, uh, this is gonna be equal to one, which is what we're looking for, right? So what this tells us is that this pdf is a legitimate pdf. But how we can just think about it a little bit is that when we added up every possible value of X that this pdf can take, they all added to one or they'll let it to 100% This another way to look at it. So we know that all the probabilities had some to 100% which is just good. So we're looking for, um from here. The second thing we need to do because we're not quite done verifying that this is legitimate is read to verify that f of X is greater than zero for all X, which in this case is for all X greater than zero. So we're gonna check this function right here. I'm just gonna look at it for a second. Sorry, I could have at the top. We're gonna look at it for a second. We're going to say OK, so for any value of X, I pick, you know, what's this gonna look like? And what we see is that Okay, So for all positive X, which is the range that were for the domain that were restricted to, um for all positive X This this function is gonna be positive. So that satisfies their second check. So we can say that this pdf is alleged in pdf. From there, we can move on to the rest of the problems. So we're going to say that the CDF is the integral of the pdf, and it's kind of handy that we already have that written right here. Except we have some different bounce Serve with CDF. We're gonna integrate from zero but two X because we don't know the value that we want to take this accumulation up to because we're adding all the probabilities up to some value. Right? So we're gonna take it from zero x of the pdf, which is F of X. That's a little F uh, the axe, which is handy because we happened already. Have this that this exact calculation, The solution to this right here. So it's going to call the negative 16 over X plus four squared. And that is the CDF, also sometimes written as big f of X. So the CDF is big enough and the PDFs little F Okay, so this is our star CDF right here. Now we're gonna do is we're gonna take this cumulative distribution function, and we're going to find the probability that the time to failure is between two and five. So what a way to do this, as you probably remember from your calculus class, is that if we have a function that is the integral of some other function in this case that we have our big ffx to our little ffx, which is a pdf. Um, we could say that Well, the the second function the big F evaluated at the bounds. It's just the same thing is the integral. It's the same thing we're doing when we set up our bounds. Appear right So we could say that f of five years five minus f of two years, um is equal to the probability that the time to failure is somewhere between five and two years. There are 215 years because the bounds were flipped. So this evaluates to So we have effort five, which is equal to going take a 16 ounce. So is going to make of 16 times one over. So f of five is gonna be negative. 16/5 plus four squared. So it's 81 and then we're gonna minus, and we're going to say that this is number 16 over half of two, So two plus four, which is six square, which is 36. We'll evaluate this function or this expression here, and we're gonna find out that the probability that the time to failure is between two and five years is between isn option is exactly given this pdf 2.47 So there's a 24.7% chance that the pdf that that the machine will fail between two and five years. Now, what we're gonna do is we're gonna find the expectation, which is a different process, although it's very similar. So if with expectation of a function, we're going to integrate pdf where we're gonna multiply by an extra extra here. So we're gonna give integral from zero to infinity. That's integral. It's loopy. Say about that, um, the X we're gonna integrate into infinity, and I'm going to write this out slightly expanded, but I'm also gonna leave most of the integration for you guys. I'm gonna set it up for how you would do it. So we're going to the integration from zero to infinity and we're gonna write X here because we're multiplying our function. Uh, appear this this ffx right here? Um, we're gonna multiply that by X. So it's gonna be 32. It's about 30 right here, 32 times X. Right. And on the bottom, you have, uh, ex plus for cubed the X And to make this integration. Little bit simpler for you, Substitution. We're going to add a plus four minus four here. So we're not We're not fundamentally affecting the value of this problem. That is a really stop before my bed. All right, so we're not gonna fundamentally affect the value of this problem if we essentially at zero. So what this has allowed us to do is allows us to use substitution for Extras four, in which we can evaluate this the U minus four over you cubed. And we can split the U and the A minus four into two separate terms and integrate those, respectively. When we do this, it allows us to arrive at our solution here, which is 32 when we take it up front, Um, for the first term. So this 32 times 1/4 plus 2 56 times 1 16 Right. So this is our innovation maker, and when we're gonna do is we're gonna add these up. We're gonna value with this function, and it's going to give us an answer of 20 four right there. So that is our expectation of X, which means that this is the most likely value expectations sometimes written like this, if sometimes written, is our most likely value for for this function of the pdf. So if we had to pick one value and just say, all right, this is our number. This is our guy. This is gonna be the most likely outcome. This is the one you want pick, because it is the most likely out of all the possibilities that will occur. So the last part of this problem says that we want to find the expected value of the salvage when it fails. So what we're gonna do is we're gonna take the expected time to failure here, which, in this case, this is a year. So it's 24 years. We're gonna take our value of our so they give us a value here. Okay, so they give us a function for the salvage value of 100 divided by four plus X. And we evaluate this function at the expectation of X, which is the most likely about you. You call the 24 evaluate at the expectation which is equal to 24. When we do this, we get 100 over four plus 24. We just 100 over 28 which is equal to 4.35 Right. So this is our expected salvage value. Now, if this is in dollars, tens of dollars, $1000 I don't know, or even if it's in some other currency. But we know that this is the expected value given this formula here for the salvage value. So I hope this was up and good luck with your studies.

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