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Homework #4. Due: Friday, September 19, 2003. <KEY> IE 230 Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers, John Wiley & Sons, New York, 2003. Chapter 2, Sections 2.6–2.7. 1. (Montgomery and Runger, Problem 2–99) A new analytical method to detect pollutants in water in being tested. This new method of chemical analysis is important because, if adopted, if could be used to detect three different contaminants—organic pollutants, volatile solvents, and chlorinated compounds—instead of having to use a single test for each pollutant. The makers of the test claim that it can detect high levels of organic pollutants with 99.7% accuracy, volatile solvents with 99.95% accuracy, and chlorinated compounds with 89.7% accuracy. If a pollutant is not present, the test does not signal. Samples are prepared for the calibration of the test, with 60% of them contaminated with organic pollutants, 27% with volatile solvents, and 13% with traces of chlorinated compounds. A test sample is selected randomly. (a) What is the experiment? _____________________________________________________________ Select one "sample" of the prepared water and test it. _____________________________________________________________ (b) Define useful event notation. _____________________________________________________________ Let O = "water with organic pollutants is selected". Let V = "water with volatile solvents is selected". Let C = "water with chlorinated compounds is selected". Let D = "pollutant is detected". _____________________________________________________________ (c) Write the given information in your notation. _____________________________________________________________ P(O ) = 0.60 P(V ) = 0.27 P(C ) = 0.13 P(D | O ) = 0.997 P(D | V ) = 0.9995 P(D | C ) = 0.897 _____________________________________________________________ (d) In your notation, write the probability that the test will signal. _____________________________________________________________ P(D ) _____________________________________________________________ (e) Determine the probability from Part (d). _____________________________________________________________ P(D ) = = = = P(D | O )P(O ) + P(D | V )P(V ) + P(D | C )P(C ) (0.997)(0.60) + (0.9995)((0.27) + (0.897)(0.13) 0.5982 + 0.2699 + 0.1166 0.9853 ← Total Probability Substitute known values Simplify Simplify (Comment: Is it obvious that the answer must lie between 0.897 and 0.9995?) _____________________________________________________________ – 1 of 4 – Schmeiser Homework #4. Due: Friday, September 19, 2003. <KEY> IE 230 (f) In your notation, write the probability that chlorinated compounds are present given that a pollutant is detected. _____________________________________________________________ P(C | D ) _____________________________________________________________ (g) Determine the probability from Part (f). _____________________________________________________________ P(C | D ) P(D | C )P(C ) = hhhhhhhhhhhh P(D ) (0.897)(0.13) = hhhhhhhhhhhh 0.9853 = 0.1183 ← Baby Bayes’s Rule Substitute known values Simplify (Comment: Notice that the numerator is one of the terms in the denominator.) _____________________________________________________________ 2. (Montgomery and Runger, Problem 2–90) A circuit operates if and only if there is a path of function devices from left to right. Assume that each device operates (or not) independently of the other devices. There are six devices, arranged three in a top path and three in a bottom path. The operating probabilities of the top three devices are 0.9, 0.8, and 0.7, respectively. The operating probabilities of the bottom three devices are all 0.95. (a) Sketch the network of devices. _____________________________________________________________ The textbook’s figure is good. _____________________________________________________________ (b) Define useful event notation. _____________________________________________________________ Let Di = "device i operates for i = 1, 2,..., 6". Let W = "the circuit operates (i.e., works)". _____________________________________________________________ (c) Write the given information in terms of your notation. _____________________________________________________________ P(D 1) = 0.9 P(D 2) = 0.8 P(D 3) = 0.7 P(D 4) = 0.95 P(D 5) = 0.95 P(D 6) = 0.95 All events Di are independent of each other. _____________________________________________________________ (d) In your notation, write the event that there is flow through the top path. _____________________________________________________________ P(D 1 ∩ D 2 ∩ D 3) _____________________________________________________________ – 2 of 4 – Schmeiser Homework #4. Due: Friday, September 19, 2003. <KEY> IE 230 (e) Determine the probability that there is flow through the top path. _____________________________________________________________ P(D 1 ∩ D 2 ∩ D 3) = P(D 1)P(D 2)P(D 3) = (0.9)(0.8)(0.7) = 0.5040 ← Independence Substitute known values Simplify _____________________________________________________________ (f) In your notation, write the event that there is flow through the bottom path. _____________________________________________________________ P(D 4 ∩ D 5 ∩ D 6) _____________________________________________________________ (g) Determine the probability that there is flow through the bottom path. _____________________________________________________________ P(D 4 ∩ D 5 ∩ D 6) = P(D 4)P(D 5)P(D 6) = (0.95)(0.95)(0.95) = 0.8574 ← Independence Substitute known values Simplify _____________________________________________________________ (h) In your notation, write the event that there is flow through the network (i.e, either path). _____________________________________________________________ P[(D 1 ∩ D 2 ∩ D 3) ∪ (D 4 ∩ D 5 ∩ D 6)] _____________________________________________________________ (i) Determine the probability that there is flow through the network. _____________________________________________________________ P[(D 1 ∩ D 2 ∩ D 3) ∪ (D 4 ∩ D 5 ∩ D 6)] = P(D 1 ∩ D 2 ∩ D 3) + P(D 4 ∩ D 5 ∩ D 6) − P[(D 1 ∩ D 2 ∩ D 3) ∩ (D 4 ∩ D 5 ∩ D 6)] = 0.5040 + 0.8574 − P[(D 1 ∩ D 2 ∩ D 3) ∩ (D 4 ∩ D 5 ∩ D 6)] = 0.5040 + 0.8574 − P(D 1)P(D 2) . . . P(D 6) = 0.5040 + 0.8574 − (0.9)(0.8)(0.7)(0.95)(0.95)(0.95) = 0.5040 + 0.8574 − 0.4321 = 0.929 ← Always true Substitute known values Independent events Independent events Simplify Simplify (Comment: Is it obvious that the answer must be larger than both 0.504 and 0.8574?) _____________________________________________________________ – 3 of 4 – Schmeiser Homework #4. Due: Friday, September 19, 2003. <KEY> IE 230 3. Consider the Sexual Transmitted Disease case study. (a) Explain why P(Tij | Di ′) = 0 is reasonable. _____________________________________________________________ A partner without the STD cannot transmit. _____________________________________________________________ (b) Suppose that there are n = 5 partners. Explain in words the meaning of T 1 ∪ T 2. _____________________________________________________________ Either the first or the second partner transmitted the STD. _____________________________________________________________ (c) Explain the difference between the probabilities that are listed under "Model Parameters" and the probabilities that are under "Model Conclusions". _____________________________________________________________ Model parameters are specified by the user so that the results will pertain to a particular situation. Model conclusions are calculated from the model parameters and the model logic. _____________________________________________________________ (d) For a given set of model parameters, explain intuitively why P(∪i =1Ti ) ≤ P(∪i =1Ti ). _____________________________________________________________ 4 5 Contact with yet another partner cannot reduce the likelihood of transmitting the STD. _____________________________________________________________ (e) Suppose that P(∪i =1Ti ) = P(∪i =1Ti ). Explain what you know about Partner 5. _____________________________________________________________ 4 5 Partner 5 is assumed to be completely safe. That is, Partner 5 is assumed to have no STD, or that condoms are always used and are completely effective. _____________________________________________________________ (f) What is the sign of the first derivative of probability of having a STD with respect to each of the model parameters tc , tn , n , pi , ni , and ci ? _____________________________________________________________ All signs are positive. Technically, there is no derivative with respect to the integer n , but the first differences are positive. _____________________________________________________________ – 4 of 4 – Schmeiser