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1994_MarkingScheme

1994_MarkingScheme - RES I RIC I t WE‘DXi-‘F...

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Unformatted text preview: RES I RIC I t!) WE‘DXi-‘F Sifi’i‘fitfi SONG KONG EXAMINATIONS AUTHORITY —flhl$§i&?§£&1£aea‘ EONG KONG ADVANCED LEVEL EXAMINATION. 1994 smmxummsnmzfscfifi 5%) As MATHEMATICS & STATISTICS (A) z? a 8 95 MARKING SCHEME m;mssx4¢. HEDGE-£555. KfslsLH-fflfifitfim. This is a restricted document. It is meant for use by markers of this paper for marking purboses only. Reproduction in any form is strictly prohibited. It is highly undesirable that this marking scheme should fall into the hands of students. They are likely to regard it as -a set of model answers, which it certainly is not. Markers should therefore resist pleas from their students to have access to this document. Making it available to students vould constitute misconduct on the part of the marker. ©§i§5§5§§v EEELEfi! $1¥figfififiifififi§§, 3215763 Efi§5¥¢H magmas. mfissmmnxnm. Mfiflfifi 7451.51. wsnmmssnsxziass P93. Fflfifi'flflfil'r'fflll. Eong Kong Examinations Authority All Rights Reserved 1994 RESTRICTED 79%“???- _...._1 a LHIEIJ.) SolutionA fi._;- 1. (a) Sample space, 5 = (BB, BG, 98, CG) (b) (1) Sample space, s = (BB, 36, GB) (ii) The event that the family has two sons, E = (88 1 P(E) = 3 dX kt . —— - 0 2 dc 1 Oke ‘ 2 __dx -= lOOk’e" dt’ Hence 100k’e*‘-200ke“-300 e“ = o k’-2k-3 I O (k-3)(k¢1) . o k - -1 or 3 : k is negative . k = -l 3. (a) Number of shortest paths from A to B = Number of permutations of the letters R,R,U,U,F,F _ 6! _ 25212: = 90 (b) (i) Number of shortest paths from A to C = Number of permutations of the letters R,U,F = 3! = 6 (ii) Number of shortest paths from A to B through = 6x6 = 36 Probability that Jack is caught by the trap g 2.6. 90 a 2 5 94-ASL-HATHS & STAT-A 1A + 1 1A 1A 1A 1A IN 1A 1A 1A 1A 1A IN 1A “1 RESTRICTED may”: Remarks IA for m notation .1 l. Il‘l-._— Viv.- Remarks Solution (a) Histogram of the weights Frequency U C 1A For veniul men-mu 20 ‘ '1", I “ 1A For hodunul Inc-sum Weight (kg) (b) From Figure 2, lower quartile - 46 kg Ausm45J46 upper quartile - 55 kg AnemSSdij 1A For either 1A Aucpl 9-10 Interquartile range - (55 - 46) kg - 9 kg (c) From the histogram, mean weight = 42.5x20*47.5x36+52.5x1%;57.5x14*62.5x8#67.5x4 kg 1H 1 = 50.8 kg 1A 9 4-ASL-HATHS E STAT-A RESTRICTED Fiji-“5354* P. 2 K VIKI‘ 7 {£1 ‘ r KtblKILICU E‘mXH‘ ‘- Solution Remarks Solution Marks Remarks 1M 7. (a) R: red card is drawn 1‘: response ‘True' 8: black card is drawn F: response ‘Falee’ p: percentage of persons who are homosexual 5. Let u=c"l, then du=2tdt. A x . f3c(c1v1) 'dc In P T R 3 .1 x < = 3ft] ’du 1A T 1-9 F J In = U74‘C ‘ 3 l—p T I - B - c‘el)’.c < ( 1 1A p F Since x-lo when c-o', 10'(01¢1)1*C 1M ' c-9 1M + 1. .1 x-(C’*1)1+9 . 13 1A .4 Alternatively Let x, y be the no. of interviewees who are 6. (a) Since ex-loxo —x—I+£ for x=0, homosexual and not homosexual respectively, ' 2! 31 then .5 z z z 1 2 2 1. _L +_ _x_ a 3; _£_ J —x+—y=790 1A e s (2) 2(2)4-6(2) 1M 3 3 2 1 a I c __ .. 1A - I-XT'XT'% for xao. 1A 3""3W410 1 Solving the equations, we have .L _ . . fl 9 3 dx I x(1—£+x_‘—L‘)dx x_30I Y=1170 1H Forreducmgmw o a 2 8 48 oneunkmwn 1 . 3O 1 . = X_L‘x_5_ :6" 1M .. The percentage required = 1200 6 40 336 o = 2.5! in . 1-1..L_; 6 4O 36 - O 8554 1A (b) WRIT) P(T RLNR) 9(3) (b) From the normal distribution table, 1 x, _ (0.025) (3) 1 .— -—l'—fe “mt-0.3413 1A ' 79 “(+1 42? ° 120 1 Hence — x 0.8554 - 0.3413 1M «2? - 0.0127 (or 7—19) 1;. I 1! II % I 3.141 IA 3.140forusingcuclulue 2xo.34132 7 0(0) 7 94-ASL-HATHS E STAT-A P.3 94-ASL-MATHS E STAT-A RESTRICTED WEBBUIF RESTRICTED WEBB???- REST RIC I tU wean; Solution d (X'2)-(x*i) 8. (3) 3i: —__(x—2)’ _ 3 s it? s R E ”'2’: SONG KONG EXAMINATIONS AUTHORITY For npplying m: quuicm ml: (b) Since y-- as x-2, x-Z is the vertical asymptote. ~7L71I££§iEEEELZE§€ 1 HONG KONG ADVANCED never. EXAMINATION, 1994 since . 1‘7! .. 1‘0 . 1 a. x-.. ‘ y 1.3. 1-0 I x ‘ y-l is the horizontal asymptote. (c) aim for any :02. when x-O, y=-l . 2 Y When y-o , x--1 . For the inurtcpu Eéfifififififlfiflfiéfiz’rSCZE) AS MATHEMATICS E STATISTICS (B) F" 9“ "Yfl‘PW' Forth: mph a? i‘i E ’5‘ MARKING SCEEME museum. Rflflflfififififlfl, KEHEWREK‘AFU. This is a restricted document. It is meant for use by markers of this paper for marking purposes only. Reproduction in any form is _strictly prohibited. It is highly undesirable that harassmsmess. MEX: this marking scheme should [all into EMEE¥EE Mfiglifigfi. the hands of students. They are likely to regard it as a set of E§§$HEEIIHZFU¢Efl Mfi'EEE model ansvers, which it certainly is TELL MfifimmflEEfixfla‘E-Ei‘ N) Area 09 the bounded {391-011 not- me, Eflfififlflfifl?flll. = - ° xtldx . —x X'Z Markers -should therefore resist : _f° (1,, 3 )dx pleas from their students to have -x x-2 access to this document. Making it _ _ _ o available to students would [xdlnlx 2”" constitute misconduct on the part of ' ' [0¢31n2—(—1*31n3) ] the marker. = -1-31n2°31n3 =1 0.2164 m. 0.116 © E % 55‘ R E) GEEH‘JZIE Eon; Kong Examinations Authority All Rights Reserved 1994 94-ASL—MATHS E STAT-B P.l RESTRICTED WEBB‘CHE RESTRICTED P935324?- l I VU*E)AI Solution Marks Remarks Solution le‘ - 1431 plm, a 1430 ~. The greatest monthly profit will be obtained when t=26, , l i.e. in February, 1997. The machine will cease producing cloth when x=0 100 2-0121: _. 659'°‘°2‘ _ 35 = 0 Put y=e‘°'°”, 100y-55y’—35 = o 13y2-20yé7 . o (y—1)(13y-7) = o For checking P when ;= 25 27 The greatest monthly profit is US$14.31. 1A AccszPl, : ~14); . . , 7 y = 1 or — 13 . r.l. 1m -a.o:: = L e 1 or 13 (c) If P = 500, then | In; 500 50.000 -0 I71: 5 02: I ‘ O ' - _ t g o (rej_) or c g 13 e 32500e 17800 n -0.0l 5 - 5°0e-a.on _ 325e-o‘oz: _ 178 . O .. ‘ ' 61 9 39 n 62 Alternatively It will cease producing cloth in February, 2000 500): _ 300 I 500 (") Th ttl tflth d ddi th x=1.s 11. e o a amoun o c 0 pro uce ur ng e -o.01c _ -o.oz: _ lifespan of the machine 100a 658 35 ' 1'6 11‘ H.901 u SOOe'°'°” - 325e'°'°“ — 153 - o = j; XdC Accept 0 NC 5 . . -0.axc = J” ”‘(100 e'°'°” - 65 e‘°-°“ - 35) dc P“: y 9 ' a 325y3-500y‘183 -o = -10000 e—Lau , 55 e-o.nzz _ 35 rial-’0‘ (65y - 61) (5y - 3) = O 0.02 61 3 a: _. ¢ _ = 141 (km) y 65 r 5 e-cmne = _5___ or 3 (b) Let P be the monthly profit, then 65 5 P = 800:: - 300x - 300 1 61 1 3 t: = — _ - 500x-300 V ~o.o1l” 5 °r —o.011n5 = 500(1006'°‘°”-65e'°'°“-35)-300 = 6.35 or 51.08 111 = SOOOOe'°'°“-32500e'°'°2‘-17800 P is increasing when t = 6.35 1 1": = ~500e'°~°”¢-650e'°'“‘ (OR The machine has not yet reached its dc production climax when t = 6.35) The machine should be re laced wh t = 51 0 d? _ -o.az g -o.ai P en - 8! E — 0 when 6509 ‘ 500:31 ‘ i.e. in April, 1999. 1A 1 650 = t = —— -— 6.2364 ..26 or c ca where o 0.011n(5°o = 2 n I d P = (5e'°'°”'-13e”°‘°“)lg.g " -3.85 < 0 Forpmvingm-x. dc: re, . Hence P is maximum when crco Alternatively x = 100 e'°'°” - 65 e‘°'°” - 35 ix = _e-n.axz,1_3e-o.azc QC d—X = 0 when e’°'°" = 1.Be'°'°" dc _ 1111.3 _ or C=Ca where ta - ”fil— - 26.2364 .' P = 800x - 300x - 300 = 500x - 300 is maximum when x is maximum '2 2‘ = (male-M”—o.ozse-°-“=)|m = -o.oo77 < 0 oc’ a.“ . ' Hence P is maximum when c-to p.: 94-ASL—MATHS s STAT-B p_ 94-ASL-MATHS 6r STAT-B RESTRICTED WEEK“: « RESTRICTED WEBB??? Solution (n Amount of pollutant I - ];z(t)dt .3 2 I 294(unltl) (b) (i) (04(11'3269) .90] :vaz‘ Lnr - lna ~blnt For uking log-Mm Foe “In" in m unbl- Farlhxgnph From the graph, lna - 1.35 Accept 13-13 I - 3.9 or 17-4.! rupcliwly 4.50-2.40 __ _ .t- . b ' 2.06-0.69 ' 1 5 Amp” H 3 échsl-HATHS & STAT-3 RESTRICTED P93133544: , BE§TRICTED EEfiB‘iH: m " Solution Run-k- (b) (11) Ame-"1c of pollutant / - f: 3.9cl-‘d: 1H 3-9 .14. ' [2.5 c L 1” _3_._9 3.5 - 2.5x. - 282.4 (unlto) 1A Amount of pollutant - I: lt’tlc - [raw]: 5 bu ' bus whore IE (3.7, 4.1) b e (1.4. 1.6) s (226.7, 351.4; (c) Amount of pallut-nt 31cc: 3 month- - L'3.9c"'dc 1H - 1)! I 1A a - 371-20“ "hex-I at (3.7, 4.1) bE (1.4, 1.6) Thu lake will "die” “tar x monthl Lt 3-9 2.5 3-9 2.5 —2-5X - 1000 (or —-2.sx 2 1000) 1M 1 _ 2.510.000 3-,; x (—————3'9 ) l x . (W)?! wharc a s (3.7, 4.1) be (1.4, 1.6) xe (12, 15) P.4 94*ASL-HATHS : sn'r—a 9.5 RESTRICTED W%BSE# i, TRICTED "Efib‘U—‘F Let X be the number of dry day. Ln A week. x — Bin(7, 0.3) 1H t,(x) - (L)(0.3)'(o.7)”' {or x-0,1,2,...,7 IA The prob. of heang exactly 3 dry days in a week Le t,.3) - (1)10.1)l(0.7)' - 0.2269 13 VSOIutLon 11. (I) (i) (ii) Let Y be the no. of days elepeed until the lit humid day. Y - Geom(0.7) 1H 2(y) - 3&7 1A Hence the mean no. of dry day: before the next humid dly in 1 - - am 1...fi 1 0.029 in (iii) The prob. at having 2 or morn humid day- before the next dry dcy 1| 1 — 0.3 - (0.7)(0.3) 1A - 1 - 0.51 - 0.49 1h biggsngsixslx f: (o.3)(o.7)‘r 1a k-z - (0.3) (0.7)'[1oo.7o(0.7)’o...] (0.7)2 1-0.7 ._- 0.49 , . 1A = (0.3) (b) Let a dry day and a humid day be denoted by D and H respectively. (i) 19th-20th-213t : D-H-D p(x on 20th, 0 on 21-: l 0 on 19th) - (1—0.9)(1—0.e) 1n - 0.02 1A (11) 19th-20th-21It = 0-3-0 or 0-0-0 p(0 on 21-: l n on 19th) = 0.02 + (0.9)(o.9) 1x - 0.03 1a (111) p¢a on 20th 1 0 on 19th and 21.:) 0.02 2H 0.03 - 0.02410 1A 9¢'ASL~HATHS 6. STAT-B RESTRICTED ESEBSCE Remlrkl I for ruminant, l for denominator 12. Coll: IA for m in mind IM. [A for 1» 2nd :nmfl an: Expected Frequency 1A for Int 2 ton“! u“. ZACOI." )A {0' Any 2 :ornct u. 9‘341' (I) Note: Under Po(4.2), ttx)- xl , expected frequency - SO£(x). 3.340 (b) (1) P(X<1.S) s so p(z<l4%22) - 0.0663 - l;£:fl - —1_5 a (2) p(x(2,5) - £;§2§11;219 50 p(z<34%32) - 0.1507 _ 2.5-2 _ _1 . v Solving the equations, we have 0-4.5, 0-2, 01-4. (c) with reference to the histogram: (P.I.o.), the Pail-on distributton, Po(4.2), 15 more suitnble for fitting the oblerved datn. ’ -LI r d _ 9—_£:2_ 0.5045 ( ) (1) 1 2; x, - (11) p(x>3.5) - P(z>-0.5) - 0.6915 1 94-ASL-HATHS E STAT-E P.7 RESTRICTED WEBIJ‘F : STRICTED $553644: Solution 1+- Solution Remarks 2:" - 10 13. (a) P(Z< CS 4 ) = 0.95 in Observed ' . __ -10 P041550“ _ Forlubcln C8 4 - 1.645 1A Accepllié-LLGS V -\° mal For uting hi‘uognm ' 10 - E For-ccuncy-nd nppmpriauneu C1 " 10'658 1R Accept 10.65640.“ f -12.3 Ea ' (b) P‘Z‘C’o—s’ ‘ °'°1 1“ 9 _ . . E '. C:'12.3 ‘ _ I - ‘ IO- . * Ea 0.6 = 2 327 1" “3 132 3 _ '_' E ______ _ " c, - 10.9038 1A 10.9ozuxo.9os ___.. duced under the favourable 7 5—.— (c) Given the batch in pro "ME ‘ condition, the required probability is 9’; ”g g P(cx<xand x<c,) : ----- . . =u=—= = 5 —- % . p(1o.ssa<x<1o.9o:a) 1H 5% E §§§=—§” g p(10.553-10 <z< 10.9038-10) m Efigggfl 0.4 0.4 5 EMEEE ___— - P(1.645<z<2.2595) Ea“ _"“ - 0.4881 - 0.45 1H Forwbmcfion EEEEE; —"—-- . 0.0381 n I ======_==——='——-" EEEEE Emfi_.— (d) P(X<c, where 0-0.4, p-lO) - P(X2c, where 0-0.6, 11-12. ) fl_ ===— =_—a=_==____m c _10 C _12_3 3— ""‘~~‘-==E=E=_= Le. _( J ) a _1— 1H + 1 g ===a=_== __... 0.4 0.6 ___._. _ EEEEE==W== .-:::1 ::J = 10.92 in 2 EEE==—L==:===5 E §E=====FEE= I: E—E=:E===E (e) The probability would be minimized if p is in the EEE§===E=E middle of the 2 limits, = ___— gqgugg§_=%_ i e = 10.3.9.4 m ====———E=== , _ p 2 . E§=====m_=_ ===========—__=_ s 10.1 m 0 1 2 3 4 5 5 7 8 E E P.8 RESTRICTED W§BIH= RESTRICTED maxim: t-ASL-HATHS G STAT-B ...
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