SCHOOL OF ENGINEERING ENGINEERING MATHEMATICS 2B SCEE08010 Exam Date: 10/05/2019 From and To: 14:30-16:00 Exam Diet: May 2019 Please read full instructions before commencing writing Exam paper information This exam paper consists of TWO sections. SECTION A: TEN multiple choice questions. Candidates should attempt ALL TEN questions. Answers should be given on the multiple choice question sheet provided. SECTION B: TWO long form questions. Candidates should attempt BOTH questions. Special instructions Do not remove this exam paper from the exam hall. Your exam paper must be returned with your script book. This is an open book exam. Students are permitted to bring the following textbooks: Glyn James: Modern Engineering Mathematics and Glyn James: Advanced Modern Engineering Mathematics. Students are also permitted two A4 sides of their own notes. Students should assume reasonable values for any data not given in a question nor available on a datasheet, and should make any such assumptions clear on their script. Students in any doubt as to the interpretation of the wording of a question, should make their own decision, and should state it clearly on their script. Please write your name in the space indicated at the right hand side on the front cover of the answer book. Also enter your examination number in the appropriate space on the front cover. Write ONLY your examination number on any extra sheets or worksheets used and firmly attach these to the answer book(s). This examination will be marked anonymously. Special items MCQ sheet, pencil, rubber, Data Sheet (2 pages) Convenor of Board of Examiners: Mr S Warrington External Examiner: Professor M Cartmell Section A Answer all multiple-choice questions on the machine marking form provided. You may use the script book provided for any rough working. Question 1 The are 3 children in a family. Assuming that the probability of a child been born a boy or a girl is exactly the same, what is the probability that all children in this family are girls, given that the first born child is a girl A. 0.5 B. 0.75 C. 0.33 D. 0.25 E. 0.66 (5) Question 2 Let X be a continuous random variable with probability density function fX(x) = λe λx for x ≥ 0. If Y = X2 what is the probability density function fY (y) of Y A. fY (y) = 1 e λ √ y B. fY (y) = λ 2y 1 2 e λ √ y C. fY (y) = 1 e λy2 D. fY (y) = λ 2y 1 2 e λ √ y E. fY (y) = 1 λ2y 1 2 e λ √ y (5) Question 3 Let X and Y be random variables with means μX = μY = 0 and standard deviations σX = 2 and σY = 10. Define the random variable Z = 2X + Y . If the covariance between X and Z is Cov(X,Z) = 4, then what is the variance Var(X Y ) A. 128 B. 86 C. 116 D. 99 E. 22 (5) Question 4 Let X be a random variable with Poisson distribution with expectation λ and further let Y = (X + 2)(X + 1). What is the value of the expectation of Y A. λ2 + 3λ+ 1 B. λ2 + 3λ+ 2 C. λ2 + 4λ+ 2 D. 3λ2 + 3λ+ 2 E. 4λ2 + 4λ+ 2 SCEE08010 Engineering Mathematics 2B – May 2019 (5) Please turn over Question 5 Suppose you are taking a multiple-choice test with 5 choices for each question. In answering a question on this test, the probability that you know the answer is p. If you don’t know the answer, you can choose one at random. What is the probability that you knew the answer to a question, given that you answered it correctly (5) A. 5p4p+1 B. p+ 15(1 p) C. 15(1 p) D. p4p+1 E. p5p+1 Question 6 Suppose that the weight in pounds of a British adult can be represented by a normal random variable with mean 150 lbs and variance 900 lb2. A lift containing a sign “maximum capacity 12 people” can safely carry 2000 lbs. The probability that 12 British people will overload the lift is close to (5) A. 0.5499 B. 0.9729 C. 0.0271 D. less than 0.0001 E. more than 0.9999 Question 7 A machine is used to manufacture gears for the transmission system of a car. After pro- cessing, gears are classified based on their characteristics in “Excellent”, “Good” and “Ac- ceptable”. Once mounted, “Excellent” gears have a 1% probability to fail, “Good” and “Acceptable” gears fail in 5% and 10% of the cases respectively. An engineer runs a test and finds out that, out of 1000 gears, 300 were “Excellent”, 500 were “Good” and 200 were “Acceptable”. How many of the 1000 gears do you expect to fail A. 131 B. 48 C. 90 D. 11 E. 910 (5) Question 8 Nova Rockets Ltd, a company that specialises in re-usable space rockets, developed a rocket composed of 8 first-stage boosters. A booster is re-usable if, after launch, it lands intact on earth. After extensive tests, engineers at Nova Rockets estimate that each booster has a 60% probability of landing intact. The company is planning two sequential launches and has a total of 12 boosters. Given that (a) each launch requires 8 boosters and (b) there SCEE08010 Engineering Mathematics 2B – May 2019 Please turn over is no time to build new boosters for the second launch, you have been asked to calculate the probability that Nova Rocket will be able to launch the second rocket. Which of the following R statements gives you the correct answer A. > 1 – pbinom(4,8,0.6) [1] 0.4511456 B. > 1 – pbinom(3,8,0.6) [1] 0.8263296 C. > dbinom(4,8,0.6) [1] 0.2322432 D. > pbinom(4,8,0.6) [1] 0.9718061 E. > 1 – pbinom(4,8,0.6) [1] 0.02819386 (5) Question 9 As an engineer at McCarrie Ltd, a fast-food chain, you have been asked to check and report on the ice-cream dispensing machine in a restaurant in Glasgow. On site you obtained 15 cups of ice cream and weighed them. Their average weight was 70 g and the standard deviation of the samples was 8.6 g. Using Student t-distribution you found that: > qt(0.95, df = 13:16) [1] 1.770933 1.761310 1.753050 1.745884 > qt(0.975, df = 13:16) [1] 2.160369 2.144787 2.131450 2.119905 Which of the following is the 95% confidence interval for the weight of the ice cream cups you will report A. 65.12 < μ < 72.19 B. 66.54 < μ < 76.51 C. 65.24 < μ < 74.76 D. 64.09 < μ < 76.30 E. 66.54 < μ < 72.50 (5) Question 10 Edison, a car manufacturer specialised in electric vehicles, is testing a new prototype, Model Z. The car is brought to cruising speed and its position (in metres) is tracked at intervals of 1 s. The test is a success if the car reached a speed of, at least 110 km/h. You have been tasked to determining whether this is the case. To do so you collect the time vector t (in seconds) and the position d (in metres) of the car. As you know there is a linear relationship between the distance d and the time t, you set out to find the speed the car reached by running a linear regression (model d = a t+ b ) with the following code: > r<-lm(d~t) > summary(r) Call: SCEE08010 Engineering Mathematics 2B – May 2019 Please turn over lm(formula = d ~ t) Residuals: Min 1Q Median 3Q Max -0.225621 -0.082933 -0.003759 0.083387 0.287022 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.127e+02 2.072e-02 10263 <2e-16 *** t 3.330e+01 3.563e-04 93467 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 Residual standard error: 0.1028 on 98 degrees of freedom Multiple R-squared: 1,Adjusted R-squared: 1 F-statistic: 8.736e+09 on 1 and 98 DF, p-value: < 2.2e-16 You then obtained the following diagnostic plots 500 1000 1500 2000 2500 3000 3500 0. 2 0. 1 0. 0 0. 1 0. 2 0. 3 Fitted values R es id ua ls Residuals vs Fitted 70 7349 2 1 0 1 2 2 1 0 1 2 3 Theoretical Quantiles St an da rd ize d re sid ua ls Normal Q Q 70 73 49 500 1000 1500 2000 2500 3000 3500
0. 0 0. 5 1. 0 1. 5 Fitted values St an da rd ize d re sid ua ls Scale Location 70 7349 0.00 0.01 0.02 0.03 0.04 2 1 0 1 2 3 Leverage St an da rd ize d re sid ua ls Cook’s distance Residuals vs Leverage 70 12 73 Which of the following statements is true A. The diagnostic plots allow to rule out that the residuals are homoscedastic. Since there are no issues with the remaining plots and the regression is significant at the 0.1% level the regression line, which explains 87% of the variation in the data, is t = 207.2 + 356x. B. The diagnostic plots clearly show that the noise is not homoscedastic, therefore no additional conclusion can be drawn. C. The diagnostic plots seem to indicate that the noise is homoscedastic. Since there are no issues with the remaining plots and the regression is statistically significant SCEE08010 Engineering Mathematics 2B – May 2019 Please turn over (below the 0.1% level) the test was successful as the car achieved a speed of almost 120 km/h. D. The diagnostic plots seem to indicate that the noise is homoscedastic. However the “Residuals vs Leverage” plot shows that there is a major issue with the residuals and therefore no additional conclusion can be drawn. E. The diagnostic plots show that the residuals are heteroscedastic. Since there are no issues with the remaining plots and the regression is significant at the 0.1% level the regression line, which explains 87% of the variation in the data, is t = 8.7780 + 2.3501x. (5) SCEE08010 Engineering Mathematics 2B – May 2019 Please turn over Section B Answer all questions in the script book provided, marking clearly the number of the question. Question 1 (a) For f(x, y, z) = x2y + y2z + z2x find f at the point (3,2,0) and the derivative there along the direction of the vector u = i + 2j + 2k . (5) (b) Use Green’s theorem to calculate the work of F = y2i + x2j on a path formed by a triangle with vertices at (0, 0), (0, 1) and ( 1, 0) in the anticlockwise direction. (10) (c) Verify your answer to the above part by computing the line integral directly. (10) SCEE08010 Engineering Mathematics 2B – May 2019 Please turn over Question 2 (a) Show using vector calculus arguments that a conservative field that is also solenoidal has a harmonic potential field. (5) (b) Formulate an iterated surface integral for the flux of F(x, y, z) = x2i + xzj + 3zk out of the sphere x2 + y2 + z2 = 4, in spherical coordinates. You are not required to solve this integral. (10) (c) Using the divergence theorem compute the flux out of the sphere by calculating a volume integral. (10) END OF EXAM SCEE08010 Engineering Mathematics 2B – May 2019 11.2 REVIEW OF BASIC PROBABILITY THEORY 911 11.2.5 Sample measures It is conventional to denote a random variable by an upper-case letter (X, say), and an actual observed value of it by the corresponding lower-case letter (x, say). An observed value x will be one of the set of possible values (sample space) for the random variable, which for a discrete random variable may be written as a list of the form {v1, v2, v3, . . . }. z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 .1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 .2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 .3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 .4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 .5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 .6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 .7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 .8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 .9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974 2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986 3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990 3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993 3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997 3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998 z 1.282 1.645 1.960 2.326 2.576 3.090 3.291 3.891 4.417 Φ (z) .90 .95 .975 .99 .995 .999 .9995 .999 95 .999 995 2[1 Φ (z)] .20 .10 .05 .02 .01 .002 .001 .000 1 .000 01 Figure 11.2 Table of the standard normal cumulative distribution function Φ(z). 920 APPLIED PROBABILITY AND STATISTICS The measured lifetimes of a sample of 20 electronic components gave an average of 1250 h, with a sample standard deviation of 96 h. Assuming that the lifetime has a normal distribution, find a 95% confidence interval for the mean lifetime of the population, and test the hypothesis that the mean is 1300 h. Solution The appropriate figure from the t table is t0.025,19 = 2.093, so the 95% confidence interval is (1250 ± 2.093(96)/!20) = (1205, 1295) The claim that the mean lifetime is 1300 h is therefore rejected at the 5% significance level. The same conclusion is reached by evaluating which exceeds t0.025,19 in magnitude. Example 11.5 n α = 0.10 α = 0.05 α = 0.025 α = 0.01 α = 0.005 n 1 3.078 6.314 12.706 31.821 63.657 1 2 1.886 2.920 4.303 6.965 9.925 2 3 1.638 2.353 3.182 4.541 5.841 3 4 1.533 2.132 2.776 3.747 4.604 4 5 1.476 2.015 2.571 3.365 4.032 5 6 1.440 1.943 2.447 3.143 3.707 6 7 1.415 1.895 2.365 2.998 3.499 7 8 1.397 1.860 2.306 2.896 3.355 8 9 1.383 1.833 2.262 2.821 3.250 9 10 1.372 1.812 2.228 2.764 3.169 10 11 1.363 1.796 2.201 2.718 3.106 11 12 1.356 1.782 2.179 2.681 3.055 12 13 1.350 1.771 2.160 2.650 3.012 13 14 1.345 1.761 2.145 2.624 2.977 14 15 1.341 1.753 2.131 2.602 2.947 15 16 1.337 1.746 2.120 2.583 2.921 16 17 1.333 1.740 2.110 2.567 2.898 17 18 1.330 1.734 2.101 2.552 2.878 18 19 1.328 1.729 2.093 2.539 2.861 19 20 1.325 1.725 2.086 2.528 2.845 20 21 1.323 1.721 2.080 2.518 2.831 21 22 1.321 1.717 2.074 2.508 2.819 22 23 1.319 1.714 2.069 2.500 2.807 23 24 1.318 1.711 2.064 2.492 2.797 24 25 1.316 1.708 2.060 2.485 2.787 25 26 1.315 1.706 2.056 2.479 2.779 26 27 1.314 1.703 2.052 2.473 2.771 27 28 1.313 1.701 2.048 2.467 2.763 28 29 1.311 1.699 2.045 2.462 2.756 29 ∞ 1.282 1.645 1.960 2.326 2.576 ∞ Figure 11.8 Table of the t distribution tα,n. (Based on Table 12 of Biometrika Tables for Statisticians, Volume 1. Cambridge University Press, 1954. By permission of the Biometrika trustees.) Tn = 1250 1300– 96 /!20 —————————– = 2.33–