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MAE 3020, Information Processing for Automation Midterm Exam, Fall 1998 (1) In a certain assembly plant, three machines, B1, B2 and B3 makes 30%, 45% and 25%, respectively, of the products. It is known from past experience that 2%, 3% and 2% of the products made by each machine, respectively, are defective. Suppose that a finished product is randomly selected, what is the probability that it is defective? If the product is defective, what is the probability that it was made by machine B3? S: Let P(Bi) be the probability of a product being made by machine Bi, and P(F, Bi) be the probability of probability of a defective product made by machine Bi, then P(B1) = 0.3, P(B2) = 0.45, P(B3) = 0.25 P(F, B1) = 0.02, P(F, B2) = 0.03, P(F, B3) = 0.02 Consequently, the probability that a randomly selected product is defective is: P(F) = (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02) = 0.0245 If the product is defective, then using the Baysian formula, the probability that the product is made by machine B3 is: P(B3 / F) = P(B3, F) / P(F) = (0.25)(0.02) / 0.0245 = 0.204 (2) A lot contains seven components including 4 good components and 3 bad components. A sample of 3 components is taken by a quality control inspector. What is the probability that one of the components is a bad component? What is probability that at least one component is bad component? What is the expected number of bad components? S: Note that this is not a binomial distribution problem because the sample space is small (n = 7). The probability that one of three selected components is a bad component can be determined by counting: 4 3 2 1 18 p 35 7 3 The probability that at least one component is bad component is equal to: 4 3 3 0 4 31 p 1 1 35 35 7 3 Finally, it may be noted that the probability distribution is hypergeometry. Hence, the mean of the number of bad components is: kn 3 3 1.286 N 7 (3) An appliance dealer sells three different models of upright freezers that having 13.5, 15.9 and 19.1 cubic feet of storage space. Let X be the amount of storage space purchased by the next customer to buy a freezer. Suppose X has the probability distribution as follows: x p(x) 13.5 0.2 15.9 0.5 19.1 0.3 Furthermore, suppose the price of a freezer having capacity X cubic feet is 25X - 8.5. (a) What is the expected price paid by the next customer? (b) What is the variance of the price paid by the next customer? (c) Suppose that while the rated capacity of a freezer is X, the actual capacity is h(X) = X - 0.01X2. What is the expected actual capacity of the freezer purchased by the next customer? S: (a) The expected price paid by the next customer is: E(25X – 8.5) = 25E(X) – 8.5 = (25)(16.38) – 8.5 = 401 (b) The variance of the price paid by the next customer is: V(25X – 8.5) = (25)2V(X) = (625)(3.9936) = 2495.75 (c) The expected actual capacity of the freezer is: E(h(X)) = h(X) f(X) = [(13.5)-(0.01)(13.5)2](0.2) + [(15.9)-(0.01)(15.9)2](0.5) + [(19.1)-(0.01)(19.1)2](0.3) = 13.66 (4) The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 70 and standard deviation 3. (a) If a specimen is acceptable only if its hardness is between 67 and 75, what is the probability that a randomly chosen specimen has an acceptable hardness? (b) If the acceptable range of hardness was (70 - c, 70 + c), for what value of c would 95% of all specimen have acceptable hardness? (c) If the acceptable range is as in part (a) and the hardness of each ten randomly selected specimens is independently determined, what is the expected number of acceptable specimen among ten? (d) What is the probability that at most eight of ten independently selected specimens have a hardness of less than 73.84? S: (a) Let X be the hardness of the metal, then X ~ N(70, 3), and therefore: 75 70 67 70 Z P(67 X 75) = P = 0.95221 – 0.15866 = 0.79355 3 3 c z 0.05 2 , since z0.025 = 1.96, c = (1.96)(3) = 5.88 (b) (c) Because the sampling is independent, its distribution is Binomial with p = 0.79, therefore the mean is = (10)(0.79) 8 (d) First P(X 73.84) = 0.89973, then, use the binomial distribution: P(p = 0.9, n = 10, x = 9) = 0.38742 P(p = 0.9, n = 10, x = 10) = 0.34868 Thus, the probability that at most eight of ten specimens have a hardness of less than 73.84 is: 1 – 0.38742 – 0.34868 = 0.264 (5) Consider the following measurements of the heat producing capacity of the coal produced by two mines (in million of calories per ton); Mine 1: Mine 2: 8260 7950 8130 7890 8350 7900 8070 8140 8340 7920 7840 Assuming that the populations are normally distributed, can it be concluded that the two population variances are equal? S: From the samples, we have: n1 = 5, s12 = 15750 n2 = 6, s22 = 10920 Note that the ratio of the variances conforms the F-distribution, the F-value is: s2 F 12 1.44 s2 On the other hand, from the F-distribution table, it can be seen that: F0.1(n1, n2) = F0.1(5, 6) = 3.1075. In other words, if the variance is really different (with a probability of 1 – a = 90%), the F-value will be great than 3.01075. Since F = 1.44 < F0.05(n1, n2) = 3.1075, we can say the variances are the same with a 90% probability. In fact, F = 1.44 correspond to a probability of 66.8%. We really cannot say the variances are different. (6) An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 95% confidence interval for the population mean of all bulbs produced by the firm. S: Since n = 30, x =780, = 40, z0.025 = 1.96, and the confidence interval is determined by: x za 2 n x za 2 Thus: 7.65.686 794.314 n