Question

An environmental engineer measures the amount (by weight of particulate pollution in air samples of a certain volume collected over two smokestacks at a coal-operated power plant. One of the stacks is equipped with a cleaning device. Let Y1 denote the amount of pollutant per sample collected above the stack that has no cleaning device and let Y2 denote the amount of pollutant per sample collected above the stack that is equipped with the cleaning device. Suppose that the relative frequency behaviour of Y1 and Y2 can be modelled by I k, 0 < y1 < 2,0 < y2 < 1, 2y2 < y1 f(y1, y2) = { 0 elsewhere That is, Y and Y, are uniformly distributed over the region inside the triangle bounded by y1 = 2, y2 = 0, and 2y2 = 41. (a) Find the value of k that makes this function a probability density function. (b) Find P(Y, > 3Y). That is, find the probability that the cleaning device reduces the amount of pollutant by one-third or more. (C) If we consider the stack with the cleaner installed, find the probability that the amount of pollutant in a given sample will exceed 0.5. (d) Given that the amount of pollutant in a sample taken above the stack with the cleaner is observed to be 0.5, find the probability that the amount of pollutant exceeds 1.5 above the other stack (without the cleaner).

Answer 1

(a) Given The joint probability density function of Y. and Yis :05 y, S2,05 y, S1,2y, Sy . .:) ; otherwise We will just modify the limits so that it becomes easy for us to calculate integrals We have oss 1,05 y, sl.), Re-writing them, we get os y, si So the probability density function becomes 10.) - ;0s, susi 0 otherwise We know that "TY, and Y, are jointly continuous random variables with a joint density function given by f ), then y)20 for all y: 2 | 107)dy,dy; = 1

By (ii) of the above statement IL 109,bey,dy; = 1 = f kdly.dty; = 1 => k[%tva dy = 1 =*[*dy, = 1 => #L4L = 1 k=1 So k = 1 Thus we get f..:)=0 i ;05 y 5 2,05 y, 51,2y, sy otherwise

(b) We have PLY, 23Y)= P Y ) = [»deyil DI Thus we get PCV, 237.)

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