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The article "Stochastic Modeling for Pavement Warranty cost Estimation" $(J .$ of Constr. Engr.and $M$ gmt. $2009 : 352-359$ ) proposes the following model for the distribution of $Y=$ time topavement failure. Let $X_{1}$ be the time to failure due to rutting, and $X_{2}$ be the time to failure due to transverse cracking; these two rvs are assumed independent. Then $Y=\min \left(X_{1}, X_{2}\right) .$ The probability of failure due to either one of these distress modes is assumed to be an increasingfunction of time $t .$ After making certain distributional assumptions, the following form of the cdf for each mode is obtained:$\Phi\left(\frac{a+b t}{\sqrt{c+d t+e t^{2}}}\right)$where $\Phi$ is the standard normal cdf. Values of the five parameters $a, b, c, d,$ and $e$ are $-25.49$ ,$1.15,4.45,-1.78,$ and. 171 for cracking and $-21.27, .0325,972,-.00028,$ and .00022 for rutting.Determine the probability of pavement failure within $t=5$ years and also $t=10$ years.

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Undefined, $\approx 0$

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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University of St. Thomas

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Lectures

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02:27

Based on an analysis of sa…

09:05

Let $X$ denote the time to…

07:28

As soon as one component f…

um were given a model for the distribution of the random variable. Why? Which is the time to pavement city? But told that x one time to failure between running. That's two times of failure you to trans verse scripting. We're told these two random variables are assumed to be independent. Then we're told us running variable. Why is the minimum of these two random variables That's when the next to so the probability of fears would be to either one of these. Those is assumed the an increasing function of Time team and this would seem to make sense right? The longer the pavements out, more spokesman the weather and you think the more likely it's going to fail? Sure, now mhm to making assumptions mhm form of the CDF. Each node was detained to be five of a plus B t over the square to C plus B T plus E. T squared the fires, the standard normal, chaotic distribution function now it's cold were given the values of the parameters. A, B, C D V cracking and also different days or running were asked, defying the probability of statement failure within five years and also within 10 years. Okay, so first I'll consider the case of five years. Well, let f one and F two the accumulate distribution functions of X one and X two, respectively. Probability of pavement failure within two years. Mhm probability. That's why his last century, quarter T this is equal to the same as a probability at the minimum of X one, and x two is less than or equal to t. And this, of course, is the same as one minus probability that the minimum of X one next to is greater than teeth. Now notice that the minimum of two numbers could be greater than a number if and only if both of those numbers are greater than that number. This is the same as one minus ability. The X one is greater than tea, and that's two is great beauty. And since we've assumed that X one next to our independent follows that this is equal to one minus the probability that X one is greater than tea times, the probability that X two is great in T, which is the same as one minus one, minus the probability that X one is less than or equal to t times one minus the probability that X two is less than a quarter T which we know. It used to be one minus one minus form of tea times one minus F two of teeth. You're using the king of the distribution functions that we were given all that. This is equal to one minus one minus. Bye. Um, and this is for X one. We have negative. 25.49 plus 1.15 TV over square root. Um, 4.45 minus 1.78 t plus 0.17 won t squared, Mrs. Simply coming from the values that we were given for the constants times one minus cap If I, uh Mrs for the random variable x two negative 21.27 plus 0.3 to 5 Team over the square root of 972 minus 0.2 18 +100022 T square. Now if we are to evaluate a T equals five, see the probability that why less than or equal to five, then we have the argument inside. The first five is imagining it. The argument inside first five correspondent. F one is imaginary. However, we haven't defined dysfunction for imaginary numbers. Now, if we evaluate AT T equals 10 then we have the probability that why is less than or equal to 10. This reduces toe one minus one, minus five. Approximately negative. 7.22 times one minus phi of approximately negative 21.14 Mhm incredibly small number. Both of these really are. So this is really approximately equal to one minus one times one, which is zero. But there's a zero chance that we cracked within 10 years, which, okay there within 10 years, which is clearly nonsense. Obviously, there should be a chance that will fail within 10 years.

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