CE 93 Midterm 1 2012

CE 93 Midterm 1 2012 - University of Caiifomia at Berkeley...

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Unformatted text preview: University of Caiifomia at Berkeley Structural Engineering Mechanics & Materials Department of Civil 8:. Environmental Engineering instructor: Armen Der Kiureghian Student name: Student ID#: CE 93 —- Engineering Data Anaiysis First Midterm Examination Monday, February 27, 2012 11AM — 12PM, 390 Hearst Mining Buiiding Work on ail three problems. Write clearly and state any assumptions or approximations you make. The exam is closed books and notes, except for one twofisided sheet of notes. The problems have the i‘oiiowing weights: Probiem 1 (40 points) Problem 2 (30 points) Problem 3 (30 points) Exam grade (100 points) Student name: Problem 1. [40 points; Carefuliy examine the MATLAB session shown beEow, where the resutting figures are shown on the right. Then respond to the itemized questions iisted on the next page. The data describes the daily water suppiy and demand in miliions of cubic feet for a commu— nfiyhnCafifinnm. >>clear all >>x w load(waterdata.dat); mm >>Supply = x(:,l); WM >>Demand m x(:,2); ~mm >>whos i Name Size Bytes Class Attributes in“ Denwnd 40x1 320 doubk 3mm Suppiy 40x1 320 double @005 x 40x2 640 doubk é gumzs >>normfreq(Demand,12) umfl >>scatter(8upply, Demand) 05:0 100 150 ECG 250 >>Emin(0emand), max(0emand)] ans = 76.6542 271.8221 >> {mean(Demand),var(Demand}] 2% ans = 1.0e+003 * game a o o :0 0.1446 1.8440 3, 0‘: so C3100 G Q CG CG >> {mean(Supply),var(Supply)] Q 9% O 0 1m 0’ g) o 0 m 0 ans m a l.0e+003 * 50 . . t . 0.2171 2.6076 “n 2‘” 8533.», 3““ >>prctile(Demand,510 25 50 75 90]) ans = . 101.8677 106.9565 137.2040 168.9907 200.4040 >>moment<Demand,3) ans = 6.510Se+004 >>sum m 0; >>f0r i=1:1:40 sum = sum + Demand(i)*8upply(i); end >>sum sum 2 l.2752e+006 l-J Student name: Questions for Problem 1 (5+10+5+5+5+10 = 40 points): a) Determine the range and [QR of the Demand sampie. range £ 271.8221 "76.6542 =~‘« 195.1679 [QR = 168.9907—1069565 :3 62.0343 b) Make a box plot of the Demand data. Clearly indicate ali coordinates of the box and whiskers. Does the data have any outiiers? If so, identify their values. (For this ques- tion, you will have to make one approximate estimate) 276322 + 106.956 —- 1.5 X 62.034 2: 13.9 < 76.654 253 2:6 168991 + 1.5 x 62.034r :— 262.042 < 271.822 2,, ~33” T 2 0 [ Data point 271.822 is an outiier. m a; t ‘5 168.991 From the normalized frequency diagram and “in the scatter diagram, the next largest data value “‘0 137291; is estimated around 230. This is within the 126 1.5 X IAQ range, so it is not an outlier. 153 $5956 1 89 76.634 _L Coiumn Number Determine the c.e.v. of the Demand sample ”a“ = 0.297 4.45 C.O.V.=: 0) Determine the mean—square of the Demand sampie. Demand sampie mean square =2 Var[D] + EEID] 7»- 1844 + 144.62 2 22,753 d) Determine the skewness coefficient of the Demand sampie. 65195 skewness coefficient 3 2 0.854 (391m)” e) Determine the correlation coefficient between the Supely and Demand data. (far/[19,5] _.—_ mix—(1275200 — 40 x 164.6 x 217.1) = 499.836 40*1 499.836 1‘ - W 2 0.228 11 91844 x 2667.6 L2.) Student name: Problem 2. £15 + 15 = 30 points! Columns A and B of a building are subject to settlement, depending on the conditions of the soil underlying their foundations. The probability that column A will settle is esti— mated as 0.15, whereas the probability that column B will settle is estimated as 0.l0. However, it‘colurnn A settles, the probability that column B will also settle is 0.4. Determine: a) The probability that either column will settle. b) The probability that the bridge will experlence differential settlement, i.e., one col~ umn will settle and the other will not. (Differential settlement may cause severe stresses in the building.) Solution: Let A : column A settles, B = column B settles a) PrUl U B) : Pr(A) + Pr(B) ._ Pr (AB) PrCAB) 2: Pr(B|A) Pr(A) : 0.4% x 0.15 7—: 0.06 Pr(A U B) m 0.15 + 0.10 —- 0.06 x 0.19 b) Pr(A:£—’_i’ U AB) = PrUlE) + 13MB) since events are mutually exclusive PrC/fi) == P1‘(A) “- Pr(AB) since PrUi) = Pi‘UlB) 4- Pl‘ (:40) m 0.15 “- 0.06 S 0.09 Praia) : Pr(B) —~ PI‘MB) : 0.10 # 0.06 : 0.04 pI‘CAE 0 its) 2 0.09 + 0.04 z 0.13 U} Student name: Problem 3. (10 + 10+ 10 w 30 points! 85% of the gravel in a quarry is known to be of good (G) quality with the remaining be» ing of bad (B) quality. Samples from the quarry are subjected to an imperfect test. If the sample is truly of good quality, the test will indicate good quality (1G) with 80% proba~ bility and indicate bad quality (B) with 20% probability. On the other hand, if the sample is of bad quality, the test will indicate good quality (16) with 10% probability and indi- cate bad quality (EB) with 90% probability a) If we test a randomly selected sample, what is the probability that the test will indi— cate good quality? b) If the test indicates good quality, what is the probability that the sample is indeed of good quality? c) If we test 5 randomly selected samples, what is the probability that two or more tests will indicate bad quality? Soiution: a) WK?) 2 PtUG ; G)P1-(G) + 9:061 Melts) = (osoxoss) + (0.10)(0.15) = 0695 <.: b) Pr(GlIG)=mPr(G) PZ'UG) = 03903530978 c: 0.693 c) PrClB) = 1 - 0.695 = 0.305 Pi‘CZ or more tests out ofS indicate [8) = 1 —- PrCO) —— 131(1) 5i (O.305)°(O.695)5 ... E£3,(0.3osmosgs)‘t D!><S! 2 1 —- 0.162 — 0.356 = 0.482 :1— U: ...
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