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1999_MarkingScheme

# 1999_MarkingScheme - Rﬁﬁﬁﬁﬁiéﬁﬂﬁ FOR...

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Unformatted text preview: Rﬁﬁﬁﬁﬁiéﬁﬂﬁ FOR TEACHERS' USE ONLY gmﬂgfﬁagsga FOR TEACHERS’ USE ONLY Solution Remarks \ 2‘ 3 (a) I (a) When x = l . 9 =? = 4 1A fora box-and-whiskw t=ln4 (or y=2|n2J Buys diagram IA for acorrect bu - d- (or _y =1.3863 ) 0—1 for r.t. 1.386 thiskcr dianganwith b .. _ in“ l)‘ ' H e— 2+1 FIFiﬁt—P—k-++iiti+~+—+—1+—— x 4 6 8 10 12 14 16 18 20 22 Reaction time IA for all being correct 1 IV = In x(x+ 1) taking log on both sides (0" sec.) ' 1: +1 (one side must correct) QB .t')‘=In.1.'-ij3Irt(:t+])-‘lt1(:tt2 + I} xﬂghvgl‘ 3' _ 21' thorpmductrule d: : x+l 1:2 +1 [M for differentiating log . When 1: d—y+ln4=l+1—l ‘ dx d ‘ lag-inti- (or 0.1137) [A a—I for r.t.0.ll-¢ (it: 2 . . (:1 +1)[(I+1)3 +3(:+I)1:}—z(z+1)3(2:) 1M for differentiating z" (x2 +1): IMHMHA [M for producti'quotient rule y la {5) 2. (a) 9-1: = l—2x+(_21)2 +041}: 4.... Reactiuntimc . 2! 3‘. (0,] sec.) 3 = 1—21 +2x: 74%....“ (b) Their chances afhaving a reaction time shorter than 1.1 seconds are equal. 1A This is because (i) the median reaction times for both the boys and girls are 1.1 sec, l I .‘ t i I t ; l 1 l l 3 or (ii) bath the probabilities ofhaving a reaction time being less than an 0‘”): =h[3}H[-51§IEWI} + 3? 5 5—! 3—2 I + foranyStenns 1.15m: forthebuyandthegtrlare 0.5. _l_ . . (5) x x2 :t:1 = +-———+— 2 8 6 I 5 . _ _ 1M farintegration l ‘ 4 (a) 1“ = J18” +11!‘)d: ' IMHA pp—l for missing Nor d! (1+1!)2 x x2 x3 2 4x3 . . ‘ n = ___ ._ 1- 2 __ a itn theresuittn a , — an ‘H 1 3 +15 + 2'“ x 3 + pp)! 3 () = 5:14.515 +c‘ forsnme constant c. 1A pp—i formissingc 3 7 2 1 3 N=100 whent=i = pagans—x Hid—8x + 100=6+6+c :9 c=83 1A . _ . i H ‘ FF" 2:313: +"' Le. Nnﬁt’ +6” +88 The expansion is validfor |.t'|<l . (or —1<.I:<1 ) 1 ﬂ - (b) When :=54, N = 6(64)3 +6(64)“ '33 = 13912 LJAW -- (51 --.'\S-M 8L 5—17 ‘ 99-AS-M {Q S-IS H'BRﬁﬂﬁigﬁﬁ FOR TEACHERS’ USE' ONLY ' Rmﬁgngrtg FOR TEACHERS’ USE ONLY HBEﬁfiEFﬁéiéliﬁ FOR TEACHERS’ USE ONLY Rmﬁgmgea FOR TEACHERS’ USE ONLY Solution Remarks 5, Let X be the no. of passengers using Octopus m a compartment. 7‘ Let Ex be the event that cable X is operative, (a) P“: 5) = C;”(0.5)’(i 705), M E, bethe event lhal cable 1’ is operative. : 0 ”00658 Ez be the event that cable 2 15 OPCI'EIIVC. and a 03007 (PI) IA a- I for u OlOl F be the evemthat A and B are not able to make contact. (h) 2m=np= I0x0V6 =6 1A.”. (a) (i) HE,r n52 )nwioisxuosm The mean number of passengers using Octopus = 0-00045 (P1) ﬂ-l for rt, 0.0005 in a compartment is 6 . (method musl be shown} {5] The probability that the third compartment is the ﬁrst une to have ., ' ' ' exactly 5 passengers using Octopus i“) P( E“. n Er 1'" 5.: )=(0‘015X01035)(0-030) z (1— {1200653)2 (0.200653) 1M (1 — p. )2 p1 = 0.0000: ‘25 (P2) ”" f‘” ”- 0'00”” a 0.1232 _m_ a—l for m. 0.123 (”"10“ "1"“ '3‘ WW“) (6) . . . . . . . - (iii) P(.F‘)=P(£x nEz)+P(E,. neg—Prey. nEynEz) ' ' = 0.00045 + (0.025)(0,030} — 0.00001125 PI + (QDZSKUMOJ v p, 6. (a) The number ofdif‘i‘erenl groups can be formed 0 00! '33.” ' _ In 5 I6 5 HI ‘ - _ Sx4xC‘m (or F2 me) 1M+IA leor P: or Cm ‘ t0.00]l39 (p3) r.t.0.001|89‘ = 160160 2—] forr.r.0.0012 (b) The possible numbers ofboys are 10, H, [2‘ ii. 14‘ 15. “5- - 1A (method must be shown) (c) Using (b) and the method of“rry and error": (b) P( Fl Ex ) = P( 52 ) Number of boys Probability of having a time keeping group = 0'030 ( p.) or 0'03 men the students with all the time - - bein_ be e "3 C" . P F E Ci = _....‘ [M for--‘% (c) max urn—HE“ ( I 3‘) cllg zoos Cm P(F) _ (U.015)(0.030) (0.0I5)p4 000113815 p] 2 0.3785489 M = 0.3735 n. 0.3735 There are 12 boys among the studean. _1L (6) 99-AS-M 3! 5—19 99-AS-M 3:. 3-20 Hmﬁﬁiﬁgﬁﬁ FOR TEACHERS’ USE ONLY Rmﬁaﬁgﬁﬁ FOR TEACHERS’ USE ONLY FOR TEACHERS’ USE ONLY Cliﬂﬁiﬁﬁﬁﬁ Solution 256 1 3 471 8. S =__ _ __.c (a) ‘ 9625(3 4' Hm] dSA 256 1 47 =——r ---r—r 120 dr 9525[ 2 + ] 123 =wr-716 21—15 9625( K ) >0 when Ogre; ﬂ =0 when (=13 d: 2 <0 when gasllﬁ . . 15 .4 attains its top speed at r: ? (or 7.5) 3 256 1 15 41 15 15 f Tos edofA=——— ——-— p 1” 9625[3[2] [2] 120(2)] M :100987 nv's (b) s,_%m"" \$3.3. = ﬂg'”(|_h) dr 50 Ic>ﬂ >0 when Osmi d3 —‘ =0 when (=1 d! k <0 when 1‘»1 k B nttainsitstop speed at r: i. l _15 From a — — (1 11—2 k—i (m01333) 15 ' SB 6.55626 9.39553 10.091329 9.64766 38.64106 6.5563 9.3955 10.0933 9.6477 6411 The distance covered by B in 12.5 seconds 11.: = S d! l. a m a 3530 +8.64106+ 2(6.55525+9.39553 + 10.09sz9+9.54755)} m :8 100.0457 m 99-AS-M 0'. 5—11 FOR TEACHERS’ Marks Remarks IA NW [A (1—1 for r.t. 10.099 pp—l for missing unit [A for solving & = 0 d: [M orsub.i= E into ESE-=0 2 d1 1A 9—! for r.t. 0.133 1M correct to 4 d.p. 1M 1A accept 100.045? to 100.0699 2—! for r.t. 3 d.p. USE ONLY ﬁning? \$5,, 183 2 k 2 d =—t “1;. (J m2 50 E ( 1:] 2! = 113225215045) <0 Solution FOR TEACHERS’ USE ONLY 183 4, (or —1ce 61—2 50 ( )) for 05.15115 The graph of SE is concave downward for 0 S t S 12.5 . i.e., The estimated distance covered by B in (c) is underestimated. Hence B CUVers more than 100 rn in 12.5 seconds. 8 ﬁnishes the race ahead of A . 113 Si)[ln{r +2)— 1n 2! (e) j" m—HZ d: 2S[|n E: = LUST: = 98.1092 C covers only 98.1092 m but both A and B ﬁnish the race in I2.5 seconds. C is the last one to ﬁnish the race among the three athlete X (4-2]: It 2 W31. 0 (+2 . 50(1n(r+2)*ln2]dr= ”HI 15+}. 2 If 25(ln 2 J =100 then lnr+2=2 2 Inc 12.78 C needs 12.78 seconds to ﬁnish the ace but both A and B ﬁnish the race within 12.5 seconds. C is the last one to ﬁnish I25 :11 (or50j[ 2 (or 50|:w4n21n(r+2)] ) ln(r+2) 1:12 — d r+2 H2] r) 12! 0 the race amon_ the three athletes. 09-AS-M 3: 5—13 913035100 5?: 5,00 1M Remarks FOR TEACHERS' USE ONLY Marks Remarks Solution Remark, Solution - . 3000 3000 9. (a) m N(t)= _M a .—-I=ae’”‘ (b) m N(r)= 3000 W 3000 1+0!“ H 49.4!" 3‘ IA Pp-l for ibrlne-Llnu VG): JOUOUbe'J" [N1 (£14124!)2 (ii) = 3000(49.4)(0.3)e W 44450e"”' [A accept a e [47.0510] and ‘ — (or —, I In[ 3000 — I] 2.40 0.90 —0.60 .2119 (I + 49-42“ )2 (l + 49-45” "T 300W 6 ”2300467101 \ NU) (2-4} (0-9) (—0.6) [—2.| 1A Correct to I tip. N'(t)> 0 for all r NU) is increasing | I t ' = _ (ll) 3' N (I) 100 NH) aoooabe'“ 1 3000 W=W1+aﬂ 1M ael470.5l.9],b=0.3 * E-m gm ael47.0.5l.9], 0:03 a _ t= o—g]n[a(1000 —l)] QB 44460;” _ _1_ _ 3000 (1+ 49.4241" )2 100 1+ 49.4!Mr 1A the line must pass .01, _0 3. through all the 4 points ”326 =1+49.4€ tau 24.2242 1A re [24.0581. 24.3687] 3000 MilnlaUOOb — 1)] ) = -——|————- = 2900 WI 0.3 m-mpmah—n]: ___ ___ l+ae "‘3 r QB 3000 14042242) = W = 2900 IM The greatest number ofmigrzmts found at Mai P0 is 2900. M (iii) Suppose all the migrants leave Mai P0 in 1 days. Then L'soJEu: = 2900 IM I 3 I for integration ‘05 1 = 390° ”‘ (including limits) 0 .t' 9: [7.3870 The number ofdays in which we can see the migrants is 24.22424 17.3870 9: 42 IA r.t. 42 From the graph, In :1 =19 . accept 3.35 — 3.95 a a: 49.4 [A accept 47.0 — 51.9 b t _ﬂflﬂ a, 0,3 t A 20— 5 W-AS-M Sr. 5—23 99-AE-M «3'. 5—2-1 .. Ramadan FOR TEACHERS’ USE ONLY Solution 10 Let .i' be the score on the questionnaire. (a) (i) Ptclassify as non-PD | PD) P(X< 75 |X~ MED. 51)) 7' _ P(Z-c s so] P(Z<—l) k 0.5 —0.3413 0153? (ii) P(classify as PD | non-PD) - P(X> 1'5 lX-«N(65. 51)) = ”25%) = r7234) =5 0.5 - 0.4772 = 0.0223 (b) The probability that out of IO PDs. not more than 2 will be misclassiﬁed e (1 — 0.1537)” + C;°(o. |587)(| — 0.1587)“ + Cll.°(0.1587)1(l—0.1537)' a 0.7911 [M for 2“ or 3" term lM for all {c} Let :0 be the required crilical level of score. P()(< :0 IX— N(80, 9)) = 0.0: — 80 P(Z < ‘9 ) = 0.01 1:0 —80 accept —2.325 to —2.33. I 43267 for the case 'Z< only :0 a: 63.3665 accept 68.35 to 63.375 (:1) [fa teenager is classiﬁed by the sociologist, then P(classify as PI) | non-FD} ‘ = PUU' 68.3665 |X~ M65. 52)) accept 68.35 to 68.375 = F(Z )- 0.6733) accept 0.67 to 0.675 3 0.5 — 0.2496 = 0.2504 accept 0.2498 to 0.2514 P(rnisclassiﬁed)n= (0.0l)(0.l) + (O.2504)(0.9) fer either in 0.2264 accept 0.2258 to 0.2273 lfa teenager is classiﬁed by the criminologist, then P(misclassiﬁed)= (0.1587){0.l) + (0.0228)(0.9) . 0.0364 0.2264 b 0.0364 The probability of teenagers miscalssiﬁed by the sociologist is greater than that by the criminologist. 99-AS-M & 5—15 FQWEiiﬁﬁéﬁﬂﬁ FOR TEACHERS’ USE ONLY ll (a) fix) (b) " (C) (d) WAS-M Se 5-26 t‘ 21:: FOR TEACHERS’ USE ONLY Solution Remarks = W = 3 > o for x i 2 I (2 - X) ' (2 - X): lim in) = lim ”‘4 = no and lim rm = lim 6"4 = —m 141' 3-.2' 2—x x—ol' x—n' 2—): x = 2 is a verlical asymptote to Cl . 1A 6 _ 3‘, lim fix) - lim ‘7 = —a x—um r—em i 7] X y = n6 is a horizontal asymptote to C1 . 1A fI-Zl=3(-2} and fil)=g(|) o _ —4 = E 1 +b 9-: J _ 2 = a[ ' I} b E a _ 69 _ e’ —1 1mm in = —¢| 1' 1A points of intersection 1A intercepts ‘Hﬁﬁﬂﬁﬁééliei 1A shape and asymptotes FOR TEACHERS’ USE ONLY 1» Uﬁenﬁf FOR TEACHERS’ USE ONLY El l: '1: .53“ Solution Remarks . Solution FOR TEACHERS’ USE ONLY Qﬁﬁﬂﬁﬁéﬁlﬁ FOR TEACHERS’ USE ONLY 12. Let N be the number ofcomplaims received on a given day and WAS-M 8L 5-28 Hﬂﬁﬁﬁﬁﬁﬁﬂ 23 _ a: X be the number ofcomplainls involving the lime schedule. , 6:: 9'2" —(e" —|)e'\, 65 1 ’ ‘ (a) g (x): 3 I 1‘ j = IE |.‘v1 muslhe simpllﬁed time schedule *rcsolvcd L - K ' r. — ’ ' o_4 not resolved g'U) 2- 0 and 11:31:: g(x) is (strictly) increasing for all values of x . IM The 2 menu“! marks can be _ awarded only when all . For x < _3 . ﬁx) 2 —6 but gm < —b. calculations and arguments manner of drlvem* resolved For x>8. flx)<—5 bm g{x}>-6. m comctemeptmt 03 not resolved Thus CL and C; has no poinrofintersection constantain g(x). routes resolved beyond the range 73 5x58 . 1 03 not resolved (0 Area ol‘lhe region bounded by Cl and C2 uLher things *mmlved I = I_2(g(X}—f{x))dx 1M 05 notresolved ”2 P(mannerofdrivc|;s|not resolved) IAfor p. . 1A for p; _ J.|[ 6e [9, Al] 4 6x74]dx 0351.08 PI p _ V _ x _ _ __ =—"----—-———-- — IM+1A+1A __l 1 9‘“ 9 2 " 0.43504+0.35x0.3+0.13K0.3+0.l2x0.5 (p2) “m” p} ' a 0.5[95 lA -l f 1.0.5 9 =?e_[2(2')5xe—e' —J;ld.r—I[—G+a]dx ‘1 m l e' —l 1 2"" IA for {(22 —z"‘)dx 10’ 40 6 -. l ' = s e = ‘e [82“! 172+[2x}"1+8[ln(2—x)]l_2 IAHA 5x_4 (b) (I) PW 5) 5' [A e' _| IA for I2 d: ' a i2.94312254 + 5 — l L090354s¢ "‘ z 0-037“ (P!) m a 7.8528 [A 51-] for 11!, 7.353 .. . 1019 'm s 3 z (u) P(N=5 and A=3J = (CAD-1) (0.5) } 1M p3(C3(0.4) (0.5) ) 95 0.008? M a—l for Lt. 0.009 (c) "29. (orP(N= Hand X=9)=0forn<9) 1M n -lD P(N=nand x=9) = '50 c: (0.4) (0.5")'° 1A 5 lo II :2 . m X _ 9 L x_ X m (l) Zk-9(k-9)! _’ + I! + 2! + 15+ 2 3 xx9(+—+£-+x— ) IA 1! 2! 3' = xqe' 1 (ii) P(X:9) = z: F(N =5 and X=9) n _m = Z:_qw+cg w. 4)“ (n. 5)” lM nu 10"e'm 9 = 04 0. 6 " EM 5215-3295 )" ( ) 2"“ I: (0. 4)° IA 9!(0 6) "'9 ("6 9)’ lo a 0. d . =;’~6 “ e‘ (bylbxo) 9![0 l5)° 495-l = 9' (or 0.0l32) [A 0-1 for r.LCI.0l3 99-AS-M & 5-27 FOR TEACHERS’ USE ONLY '. D If 31‘ :55” FOR TEACHERS’ USE ONLY Solution . I3. (0) Sample mean =3 (b) Number of 3'laai:[.'x:t:ted ﬁeqm—cy' ﬂtm dicinal herbs _3 7 3 29. 38 ”.73 (0:11.37) (or 5.23 ) (c)’ The maximum error for P00) is less than 1 (ll-1.94 — 14| :0.“ )while the maximum errors for B('.', 3/7) and the normal distributions are all greater than 1. The Poisson distribution as the best (d) (i) Let p be the probability that there is no medicinal herb in the tea. then p = e" (a: 0.0493) The required probability = (1o)J (I — p) = (e")’tI—e") x: 0.00m (ii) Let r,- bethe probability that acup oftea contains exactly 3 kinds of medicinal herbs, 33 0'3 3! = 0.22404 ( or 0.2240) then :1 = 111: required probability = Ho —q)'” +C.”’q(1 —q)"] . 0.6924 (or 0.5923 ) 99-A5—M «Rt 3—29 Qmﬂtoéﬁﬁﬁ FOR TEACHERS’ IA rnorlt-s iii (5) can be awarded independent of(a) IA *1 A for 29.38 and 22.03 IA+1A for 10.08 and 5.04 [AHA fat-[1.73 and 5.32 IA 1 provided that entries in the Poisson column are correct 1M for PUB) only IM IA «1—1 for 11.0.0001 W! [M 1A a -— l for H, 0.692 ...
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