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In cost estimation, the total cost of a project is the sum of component task costs. Each of thesecosts is a random variable with a probability distribution. It is customary to obtain informationabout the total cost distribution by adding together characteristics of the individual componentcost distributions - this is called the "roll-up" procedure. For example, $E\left(X_{1}+\cdots+X_{n}\right)=$$E\left(X_{1}\right)+\cdots+E\left(X_{n}\right),$ so the roll-up procedure is valid for mean cost. Suppose that there are twocomponent tasks and that $X_{1}$ and $X_{2}$ are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 $\mathrm{th}$ percentile? That is, is the 75 th percentile of thedistribution of $X_{1}+X_{2}$ the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sumof percentiles? For what percentiles is the roll-up procedure valid in this case?

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Not valid for 75th percentile, but valid for 50th percentile;sum of percentiles =$(\mu_1+z\sigma_1)+(\mu_2+z\sigma_2)=\mu+\mu_2+z(\sigma_1+\sigma_2),percentile of sums =(\mu_1+\mu_2)+z\sqrt{{\sigma_{1}^{2}+}\sigma_{2}^{2}}$

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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full of a total cost of a project is the sum of the component task costs. We're told that each of these component has costs as they were in a variable with a probability distribution and were asked to find information about the total cost distribution using a roll up procedure. So we're told that there are two component tasks on the X one and x two are independent, normally distributed random variables where asked, the royal procedure is valid for the 75th percentile. So is the 75th percentile of the distribution of excellent classics to the same as some of these 75 75% houses to individual distributions. He saw that it was it was valid for the mean cost. Well, we have at least some of the percentiles. This is going to be new one plus z times sigma one times or plus in you too. Yeah, plus Z times signature, which is the same as new one. Plus Mewtwo plus z times signal one plus sigma too. And we have that the percentile of sums. On the other hand, this is going to be new one, plus mewtwo, the mean of the some plus z times the standard deviation of this some which is the square root of sigma one squared plus sigma to square. So we see that in fact, some of the percentiles is going to be the same as the percentile of sums. We would have to have these two expressions equal. So these air equal if and only if we have that sigma one plus sigma too is equal to the square root of sigma one squared plus sigma two squared. So in other words, we have that or we have that Z is equal to zero mhm. In this first case, we get signal in squared plus two sigma one sigma two plus Sigma two squared equals signal and squared plus sigma two squared. So we have that two times Sigma one Sigma two equals zero. So, in other words, Sigma one asked equals zero or a sigma to has to be equal to zero. Either of them could be zero. So it's the role procedure is not valid or 75th percentile because in this case, Z is not equal to zero. Unless, of course, one of the staring deviations zero

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