2001_MarkingScheme

2001_MarkingScheme - FOR TEACHERS’ USE ONLY Rmfl%%fi...

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Unformatted text preview: FOR TEACHERS’ USE ONLY Rmfl%%fi Solution P(A u B) = P(A) + P(B) — P(A n B) and P(Anfl) = P(A)P(B) lM 0.7 = 0.4 + P(B)—0.4 P(B) 1A P(B) = 0.5 IA ------ “(4) - 2: d“ 2x 2, (a) Smce u=e , —— = 22 . 1A (11: ‘1’ = ELL" =( lM+lA [M for (1—2021: I! (b) Using (a), y = [(2—42“)1: = ZJr—e‘Jr +c for some constant c . Putting x=0 and y: l , wehave c=2. y = 2x—e"r + 2 IA -—-~-4s) lA for anyone correct _ _ = 7 = = = lA+1A 3 (a) a—8,b—6.c 5 8 b 7 c lAfora” (award 1A for 11:18, #36571, 565) (b) Before replacement mm mum Ages IM any box-and-whisker diagram with correct scale IA all correct. same scale lo Before Afier “T um“) 2t)01-,\5.M & 5.”. replacement replacement Rflfigflfifiéeflfi FOR TEACHERS’ USE ONLY Solving REflfigfi Solution FOR TEACHERS’ USE ONLY 7' = ’+[_—](a")+’2‘[‘-I—l—l](ax)z lM+IA IM for anyZ tenns correct n n 2 =l_flx+(n+13a 2 ’1 271‘ E _ i and ("HMI 2 w h (1A for anyone iftlie first two " w 3 2n2 - ' e ave rows are omitted) 2 901+ = 64112 n+l=4 n=3 and (1:4 lA+lA . . . l l l (b) The expanston IS valld for —Z < x < I . < z lM -------- «(6) 5. (a) " . . ' an IM correct to 4 d.p. ER d! a LED; + 6.45l61+ 2(783177 + 7.54717 + 704846)] IM as 44.5543 1A The total bonus for the first 6 months is 44.5548 thousand dollars. (b) 1A 2 _ = 7200(1 50) 1A is concave downward in the interval 0 S I S 6 . The approximation in (a) is an underestimate. lM --~-—-—(6) leOI-AS-M & 8—17 1 Rflflflflifigflfi FOR TEACHERS USE ONLY 6. 7. (a) (b) (c) (a) 0)) FOR TEACHERS’ USE ONLY Marks Rlfiflflfliéflfi Solution The probability that the heaviest student is in the selection 1M denominator, maybe awarded in (b) or (c) below lM numerator 10 m denominator Em numerator Willi fraction subtractedb l 1A The probability that the heaviest one out of the 3 selected students is the 4th heaviest among the ten students 0? =ETO— IM numerator 3 311615 1 6 s = _ _ _. f _ ._ ._ '(1098 IE("1098 =§ IA The probability that the 2 heaviest student are not both selected _ CiI IM for numerator and probability ~ __ subtracted by l C,“ 8 7 6 3 2 8 7 = _. _ _ +C __ _ _ £01918] [(WIgIg) @Sumofthetwocases l4 0.9333 1A 0—] form. 0.933 = E ~—-—-<7) The required probability = 0.39x 0.58 IA numerator 0.48x0.65+0.39x0.58+0.l3x 0.5 lA denominator = 0.375 (p) IA The required probability 1M binomial,_for any p IM Cip’a-p)’, for pin (a) IA a—l for Lt. 0.343 ----—-~-(6) = C;(o.375)2(| ~o.375)’ = 0.3433 2(Kll-AS-M K: S~|8 giggggmgga FOR TEACHERS’ USE ONLY grammes Solution 8. (a) (i) Since G(0)=9, 2a—12+(a+l2)=9 a=3 (ii) G(x) = 6—126” +l52‘2'“ G’(x)= t2ke'“ —30ke‘““ = 6ke'l“ (2 —5e"“) FOR TEACHERS’ USE ONLY Marks IA ., G’(x)=0 when 12"" =5- or x=—ln-5- IA 5 2 <0 when OSx<ilni and G'(x) 1 k5 2 lM >0 when x>—ln— Ir 2 . . . 4“ 2 G(x) is minimum when 9 = E . G"(x) = 42/85“ mom-w Since G(x) has only one stationary point for x 2 0 , G(x) is minimum when 2"“ =% . (ii) G(x) = 6—12e"‘x +15e'2’“ =15(e'“‘ —%e‘h)+6 = I5(e"" —%)‘ +¥ [ME G(x) is minimum when e‘k" =§ . [fl 2 2 2 ' The minimum CDO = 6—l2(§]+15[§) mg/L = 3.6 mg/L M m. *—-‘-—(5) (b) (i) Solving G(x) =4.5 , we have 6422'“ +15e‘2’“ :45 1M 10(e"")2 —se““ +1 =0 2““ = 4 i ‘5 1A l0 1 4 i J6 x = ——In It l0 Hence ——I—|n4_‘/z+iln4+‘/g=2.85 lM+lA It l0 k 10 i In 4 + J6— : 2‘35 4 ~ J5 k =05 (I tip) IA ztm-AS-M & S—l‘) a Rlifizflfliligéflfi FOR TEACHERS USE ONLY Rpfigymgea FOR TEACHERS’ USE ONLY R Bflflgmgfla FOR TEACHERS. USE ONLY Solution Marks . Solution (ii) Gm = 62'0'5‘ —15e_" 1212‘“ —30ke‘m 9‘ (a) (i) In Pm = mm 0.104.711 I. _ —o.5 —. _ 2 4“ 2 -2Iur —a G 00- “3e I +159 r 12" e +6°k 3 IM From the graph, = 3E4”; (Se—0.5.x _1) —8—(—3.5) l l —k nam— , It 2: 0.25 IA 11-1 for more than 2 d.p. G"(x) = 0 when .\' = ———In — (z 3.2) _ 0.5 5 0.04ak l I l I In 1 z —3.5 , a = 0.7512 2: 0.75 [A (1—1 for more than 2 d.p. —a <0 when x>———ln— [x>——ln—] , 41.25: and Gum 0.5 5 | k l5 1 M P (1):: 0.03e 1 [ ] P(I) ~ —O.lZe'°'25' 3H: IM 0 5 Since P(0)=o.o9. 1A Hence P(!) e 4.122412” +0.21 m __ _. 3 .. 5 . G (x)— 0-3<0 (u) ,u=P(3)=s0.1533 IA ;:e[o.1530,o.15331 Since G'(x) has only one stationary point for x 2 0 . 1 (iii) Stabilized PPI in town A = lim P(t) = 0.21 lM+lA G'(x) is greatest when e'k“ =— . 1-H” 5 -—-—-—(8) 3.2 km downstream from the location of discharge of the waste, 1A (b) (i) Suppose b = 0.09 _ 3 (I) Q’(r) = 0.2401”)? (iii) Solving G(x) = 5.5 , we have , -. the rate of change of the CDO is greatest. 1 __ 303-. 442-.» +1 =0 Q(r) = —(0.24)(—2)(3t+4) 2 1A 3 2.05, = nit/n4 J 30 =—0.16(3t+4) 2+c x____l_1n 12:4114 Since Q(0)-=0.09, c=0.l7 ' . 30 x~ 0.6 or 6.2 1A If W) =# ~ 0.1533 The river will retum to be healthy 6.2 km downstream form the _1 location ofdischarge of waste. 1 ’ 0- “(31 +4) 1 +0-17 = (11533 lM I — 0.16 1 Since 1211,00) = 6—122'o'5‘ +15e" (3H4) ” 0.0167 .. The river will return to be healthy. Sm“ 3‘ + 4 > 0 Solving G(x) = 5.5 , we have x a: 0.6 or 6.2 . ’"29-3 . I E [28.2, 29.3] The river will retum to be healthy 6.2 km downstream form the the PM WIll reach the'value of . location ofdischare of waste. Since Q(0) = 0.09 , Q(l) = 0.17 and "Hum-(lo) Q is continuous and strictly increasing (Q’(1) > 0) , EA] Q can reach any value between 0.09 and 0.17 includin a 0.1533 . (n) Stabilized PM in town a = lim Q(l) =0.17 1-0:: The stabilized PM will be reduced by 0.04 . (ii) 0.05 < b (<1). Otherwise. Q'(1)s 0 and the PM will not increase. It follows that the epidemic will not break out. -------- «(7) gum—AS-M .sz s—zn mm M M & S_,l FOR TEACHERS’ USE ONLY FOR TEACHERS! USE ONLY Solution Marks 10. (a) rto)=g(o) 2 his-=15 IA -11 m, 9 I 3 (ii) Let 14:2 9 , then lnu=——9"—x and dx=—fi~—‘du. IM r(9)=g(9) => LE 2 0:2 IA " u a 12 11+!) _11 _______ __(2) I l5-2 9 dx ll 2‘(ul1l-I 9 l (b) Since lim rm = lim 5"":5 =40 and limf(x) = lim 5"”‘3'5 =+oo. ' Iran “(—57)” 'M Chang°°fvamble and'imi‘s x—rl' x—ul' x+ x—vJ' x—fl' x+ x=-3 isavenical asymptote. IA 135 2mm . 45 = ‘T—Z" 1M Ignore limits 5+" n Z—al‘l . . = . x = u a Sm“ ,.'LT.f<x> ELL 3 5- = g 2-12-34 1+; ln2 y=5 isahorizontal asymptote. 1A ‘ a _________.(2) = I35 '2'; I 2an A (C) y l I Let u=2 "x , then_lnu=—ln—2x and dx=~i--l—-du. @ 9 |n2 u x IlS-Z 9‘dx = IISu —i-l du ln2 u n+9 J' 15 2 9 (1:: [E H y=_5_ ____________________________________________________________________________________ _. [426,5 ( ) x E] n+9 —lx n+9 ~lxln1 L 15.29dx =15!“ N dx -9x15 ——I—:|n1 ("9 = ln2 e 9 E ignorélimits I1 9 IA asymptotes “135 -g, M y=f(x) IA shape andposition = Inz 2 [E ' 1A intercepts and intersections a _._-...-(3) . _ 133 [2%-2—9] |n2 9 (d) (i) Azj 5x+45dx ‘A 0 x+3 I35 -5. _ 9 9 30 — '2 El =1 (5+ )d‘ 2|n2 o x+3 = [A 'noelimits _“_ [5x+301n(x+3)13 'g r If ‘35 .2 9 =45+301n4.thcn = 45+30|n4 86.5888 IA 0—1 for r1. 868589 21n2 —% In 2 = ln[(45+30ln 02352] 0:1‘5253 IA a—l for r.t. L525 -------- "(3) l, _’77 mm'AS'M AS " 200l-AS-M&S~Zl Rpfigflgfigflfi FOR TEACHERS’ USE ONLY Engaging/E55 FOR TEACHERS, USE ONLY Rflfiflflfiéfifi FOR TEACHERS’ USE ONLY Solution l I. Let X4 and X5 be the numbers ofpersons entered the building using entrances A and B respectively within a 15-minute period. 0 -J.2 (a) (i) P(X1=o)=(3L)0f4—=e"-’ 0:.) 0 -1.7 01) P<X1=0) =Q7—l—E—= e‘“ (m) 0! (iii) P(XA +x,, 21) = l—P(XA =0and X, =0) = l—P(XA =0)r(x,, =0) = l—e‘” .9973 (iv) For) +x,, =2) l_e-3,2e—2.7 = P(XA =2) P(X,, =0)+P(XA =1) P(X,, =1)+12(x, =0) 1=(x,, =2) = (3.2)2e’3-2 in + 3.22"” 27;“ +9.1, I (2.7)212'2-7 21 l! l! 2! = 17.4052"9 (b) (i) Since It is the most probable number of persons entered the building within a 15-minute period, P(X=k—1)sP(X=k) and P(X=k+ I)SP(X=k) lk-le-I Aka-A. Hence (k _ I)! k! k s ,1 and Ame—11 S Aka-l (k +1)! 1:! 21 s k +l A -I S k (ii) From (b)(i). k = 5 . The probability required = C; {P(X = left: —P<X = knzmx = 101 4 (5.9)’e"” 1 (5.9)‘e’5-9 1 (5.9)‘e"-’ =9 ‘5!— “T T 30.0l83 2001 'AS-M & S-ZJ lA a—l for r.t. 0.04l 1A a—l for r.t. 0.067 W l'(P1)(P2) IA a—I for r.t. 0.997 [M for the 3 cases 1A 1A 11—] for Lt. 0.048 --—~—-(7) lM+lM IA 1M for binomial 1M for all 1A a—l for r.t_. 0.018 ——~~48) Rpflflflfliéflfi FOR TEACHERS’ USE ONLY 12. Let E‘- and [5,; be the lifetimes of brand X and brand Y CFLs respectively. Rflfifimgfia FOR TEACHERS’ USE ONLY (a) max <8200)=0.l 151 = P 5J< 8200‘” =0.0808 400 400 :> 8200—” =—1.4 ----- -- -1A for either 400 :> y = 3750 IA Pas, < 3200) = 0.1587 :9 pféfl < M] = 0.1537 0' 0' 3 8200—8800 “Loo 0' :> a = 600 1A a, =0.3811, a2 =0.0548 M b, =0.2120, bz =0.2586, b, =o.2120 lb. =o.2109, b1 =0.260.8, b, =0.2109 ] (b) The mean of the lifetimes of the 2 brands only differ a little but the standard deviation of the lifetimes of brand X CFLs is significantly smaller than that of brand Y. I shall choose brand X because the lifetimes of its CF Ls are more reliable. i shall choose brand Y because there will be a bigger chance of ettin a Ion life CFL. I shall choose brand Y because the mean lifetime is larger. IA b. =17, e [0.2101,0.21201 b; e [0.2586,0.2624] ------ ---(s) (c) (i) Let X” . X, and X, be the lifetimes oflamps a. b and c resp. (l) The required probability = P(X, > 8200)[1=(x, >8200 or X, > 8200)] IM = [1 — P(Ex < 8200)][1—[P(Ex < 3200)]1} 1M a (1 —0.osos)(1—o.ososz) IM 5: 2(0.9192)1(1—o.9192)+(o.9192)J [m [m I'm ] a 0.9132 lA (ll) The required probability = P(X,, < 8200) P(X, > 8200) I’(Xc > 8200) IM for numerator 1—0.9l32 IM for denominator _ 0.080130 4.013021)2 1 — 0.9132 = 0.7865 IA 200 | -AS-M & 3—25 Rflfiiflflifigflfi FOR TEACHERS’ USE ONLY greatness FOR TEACHERS’ USE ONLY Rpasggmggg FOR TEACHERS‘ USE ONLY Solution Solution Note that P(EX < 8200) = 0.0808 and P(Ey < 8200) :1 0.1578 . v Since a brand X CFL is less likely than a brand Y CFL to have [3. Let X be the number of Grade A potatoes in the 8 selected potatoes. (a) P(,\’ s | | p = 0.65) r: 0.0002 + 0.0033 IM a lifetime less than 8200 hours, and lamp :2 is the most critical lamp 5 0.003 5 0.0036 1 A for the lighting system to work (according to the result of(c)(i)(ll)), l _..-.-..-.(2) Lamp a should be a bland X CFL. Hence I will put the brand Y CFL as lamp b or c. 1A with explanation (b) (i) p(X 5 31 p = 0.65) = 00002 + 0.0033 + 0.0217 + 00808 [M Let X a and Ya be the lifetimes of lamp a when using brand X CFL a 0.1060 (q) M and brand Y CFL respectively. Similar notations are used for the other two lamps. (ii) P(X > 3| p = 0.2) lM P01, > 8200)[P(Xb > 8200 or X c > 8200)] = 0.0459 + 0.0092 + 0.0011 + 0.0001 + 0.0000 lM = (1 —0.1587)(l —0.08082) = 0.8358 ~0.0563 IA 5 P(X,, > 8200)[P(Y,, >8200 or X: > 3200)] """" "( ) = (l — 0.0808)[(l — 0.0808) + (l — 0.1587) - (l —0.0808)(l ——0.l587)] = 0.9074 Hence putting the brand l’ CFL as lamp b or c will yeild abetter s stem. (c) The required probability IM for the 2 cases lM for Ist term 1M for 2nd term IE with explanation = C231]2 (1 ' q) + C3113 z C;(0.1060)2(1-0.1060)+C3(0.1060)3 lM+lM+lM z 0.0313 0.0314 IA --~—--—(4) (d) “the ' ’ ‘ .Iz whet ._ Withfi ' .: There are the armer w1 ave a 1gger chance of . lM ' ------ —-(|) (e) P(X 5 2| p = 0.65) a 0.0252 P(X s 3 1 p = 0.65) a 0.0252+0.0sos a 0.1060 Since P(X 5 2| p = 0.65) < 0.05 < P(X s 31 p =0.65) k = 2 . }lM+lA lM for 0.05 as a value between IA independent -----—----(3) """"“'~' “ 5 1" 2001-As-M & 5—27 Rlfiflfifigflfi FOR TEACHERS’ USE ONLY Rpfigggmgga FOR TEACHERS’ USE ONLY ...
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2001_MarkingScheme - FOR TEACHERS’ USE ONLY Rmfl%%fi...

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