Final Exam with Solutions - Probability and Statistics Engineering | STT 351, Exams of Probability and Statistics

Download Final Exam with Solutions - Probability and Statistics Engineering | STT 351 and more Exams Probability and Statistics in PDF only on Docsity! Clam Be Z| / | OTE. p te Cer en! Cs (once HO RRUATAS _) ie STT351 Final Exam Fall 2008 Tests of hypotheses, 1. For the of score x = skin thickness, a 95% confidence inter- val [0.822, 0.847] has been obtained for the population mean yu. It is desired to harness this CI to test Ho: = 0.85 versus H,: 2 < 0.85, la. Which action, reject the null hypothesis or fail to reject the null hypothgsis, is taken based on the test employing this CI? fe TAA 2 o18S i Th FSH CL fits entindy hte Pt fF ug eeds thertire , eject fh ib. Is this test one-sided or two-sided? One oo bane Hoi eos m lc. What is the probability @, of type one error, for this test? af PC tip Fem) = of . 0:98 © 0,028 id. Ideally, what would be the desired probability of rejecting Ah if pis 0.84? desinad probability = | a 2 | finalt2-7-08.nb 2. Plots of P(reject Ho | p) are shown for two tests having the same a, Ho, Hg. wes . ohgetl Te yest Poreject Hy 1p) ob KOAB A OBE OGL oaL o2b L L 06 08 10 2a. Determine Ho. 2b. Determine H,,. Ho! Popol? 2c. Determine a. / or me oo 2d. For the poorer test, determine pos). 6035) = 1-0.875 0.125 finali2-7-08.nb | 5. Probability. 5. Box I: {4R, 3G, 7 Y}, Box II: {8 R, 2B, 4Y}. Box Lis chosen with probability 0.8, otherwise Box IL Then bails are ~ selected from the chosen box with equal probability and without replacement. (= ¢-# PITS) = or 5a. P(R1 Y2|D= (ale ny 7 / vA : 5b. PRI)= os (#) + 0.2 ($) / 5c. PURI = plznen = ecyecend . @ NGA) Poe Pee “Per 9a( 410.08 ) ae Bs 5d. PRI U YD = Ple son -~plarnyy = Pen POR Pearse drier) Sonltsoa( «<4 Ge) coke (4) — — [olde Bye "G > fy e 0.8m) o2l% i aon (2). oal& (4) . 6. P(OIL) = 0.1, P(+ | OIL) = 0.8, P(+ | OILS) = 0.3. J 6a. P(OIL ~) = plow) etter) =a) t-08) = eNO?) glow ees oiey = ON e) ate a (NED eute CA) aw. = La) 6b. P(+) = @.) (0-9) +(0.9) (0.3) c 6c, P(OIL | +) = Pins? = econ / TRCN” ikea) fone) 6dAre events OIL, + independent? Why se \ opto) pen) y peur Jon) —> f Bou eaten a x osfet 7. The number X of road service calls in one day is approxi- mately Poisson distributed with mean.144. Sketch the normal approximation of the distribution of X. -<¢¢ = /4* she vin t1Q / 10 “aye iy 6 | finait2-7-08.nb Expectation, Var, sd. 8. Random variables X, Y have ExX=5 Var X=3 EY=9 Var Y =4 — cen PEt tile EX es gs) a> 8a. E(6X “TYLA So eg n))eg e -RReNS 2 aes -27 8b. If X,Y neve spenden E(XY) = € (EN = 6a) = 44 8c. If X, Y are independent Var(6X -7Y+11-X) = = Vor( Sx Ty tHe Soar et * 26(3) - Hau) a = 76-144 =-12} ~p(w) 0 09 20 0.1 9. The distribution of r.v. W is: 9a. EW ofo9)4 22 (ody 9b. Var W= [o *(0-9) + 20* {9 \ - [ews tLe tors — 9c. Let T denote the total of 100 independent plays of the lot- tery whose returns are distributed as W. Sketch the bell-approxi- mation of the distribution of T. . | S¥ al [eran easy \ fat 1) taopiy) Vie | | N peeToa) tae Gt) \ | 9d. BCatp= cy (oa) s oe (0) ¢ | seg a ob (000) J ed finalt2-7-08.nb | 7 Continuous models. 10. Lifetime T of an electronic i dis- tributed with mean «4 = E T = 8 years, és Ks fp 10a. Determine P(T > 8)s</ 1- Pah = 710b. Determine P(T > 304.T > 12) myer T> a: 2é' 11. Time T and wear W are jointly distributed with density: {(t, w) = (1/6) (t+ w), O<t<1, O<w<3 = 0 elsewhere. - lla. Verify that f is a joint probability density. bf( are me ie y prasad Vat = 4] gee ve Ze), ~¢[ 521 ° = 4178 = | 6 (= | tadara Poy yout Crs ool iby deasidy 11b. Determine the marginal density f(t). Rt) = = Pi elawide « zd S btw dwn Afeveae te tbed a \ fat ea — lic. Determine the conditional density fy jr(w | t). . She) = FR, “) is fae) ; re) ce dy (\ \ Teg ) 1 . 2