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1998_MarkingScheme

1998_MarkingScheme - | RBEﬂEﬁgE—EI FOR TEACHERS’...

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Unformatted text preview: | -* ' RBEﬂEﬁgE—EI FOR TEACHERS’ USE ONLY Rmﬂﬂﬁﬁﬁ FOR TEACHERS’ USE ONLY Solution 16l+162 3. (a) Median -= cm I - = 151.5 - - :13” = e‘ +nx(_ i2) IM for product rule cm Ignore um: I l 1 3: IM for chain ml: 0:) Th: pmbability that a student with a hcight greater than 170 cm is 5319,;ch _ _ 9 z ,s ”a: - — 0.22 x 40 ( 5) d‘y 1l 11 1l - (i) Probabili 939 _=___, _ x __ re nit-ed 1—— — d1: xiv-r9: +3331 ty ‘1 "( 40) (40) 1 258l19 = 1+3" "0-1047 (or 2560000 ) a—l for m. 0.“)! .. . . . 9 9 (u) PrDbBblll re uu-gd . C5 — 3 -_ 3 1.313 a ._1_§ 1, U a “40)" 40) dx: x x’ s —x¢ -00684 (or 700569 ) ' 2 . . 10240000 0-! for LL 0.068 = o d .v y accept FE:- 4. (a) x = Issoﬂmd: pp—l formisslngd: _ _ _ 4.0M! , _ 1 (I) (1+axl" a l+(—4)(¢r)+(;4¥(m)1 +ﬂl‘.y)(i(m)1 +... For 8113'3 terms ,, ”2500‘ +‘ pp—I for unsung: ‘ . _ l for miuin . l-4ax+loa1;2 40.53,! +... or “mama-mam : m“ x: 51000 Wm“ “'0 ppm- 3x = 219500 “50: — 3 ' c a: _2 3°“ ! .r = 219500 - 152500;?“ b =—4a= s ‘ c: 10a“ _ 40 I (a) If moo x 2 - 219500 — 162500¢'°-°‘“' mm It: I I"[219500-114000] (b) The expansion is valid when _ 0'0“ 162500 —1<a::<l (or 1013“. I-mkl) unbeomitled - cu IDS (0:1019913) LL [0% _] < —2x < 1 I the number of customers will be doubled in _ 1 l 1 i IDS days she:- the start ofthe campaign. Le. ——<.t<— (or |x|<-—.—O.5<x¢0.5) 2 2 2 5. (a) (i) No. of arrangemcnls - 10! 9 3628800 (ii) No. ufarmngemcnrs = 2! 9! M for 9! = 725760 (b) (a No. ofnrranﬂements - Ior - 3623300 (ii) No. ofarrangemcms . m9: - a!) [A for (9! — a!) s 645120 mum. No.0fammgemenu: cgdmsszt (or 32-3!) M for chum = 645120 98vAS-M&S—15 95-AS-M3LS—16 HBRﬂEfﬁéiﬁﬂﬁ FOR TEACHERS’ USE ONLY Hmﬂﬁﬁgﬁvﬁ FOR TEACHERS’ USE ONLY . 1‘ ‘ 910030004300 FOR TEACHERS’ USE ONLY Solution 6. Let F. be the event that an ice-cream bar is contaminated. 0.003 E nelu 3|» a"... w 3" “3 0.992 5' “002 E For the tree diagram or all parts in (a) being 0 998 correct (a) (0 P(A)P('E'JA)-r%xo.992 =0.124 (:0) 00 P057 = P0“) P(E'IA) + PG!) P11? 1 B) + P63) WE‘ I C) 2 s 2 5 =0.124+§x0.998+§x] forpl+ip1+ip3 = 0.9935 a—i for Lt. 0.999 P(1"\)l‘(['3| A) P A E = —— (b) ( t J mm leDOE 1x030. = a (‘3' 1—4-2— ) “0-9985 ExO.8%+ Exam =I 0.6667 01—] for Lt. 0.567 98-AS-MJLSHI’.‘ Rmﬁﬁiﬁéﬁﬁ FOR TEACHERS’ USE ONLY Hﬁﬂﬂﬁiﬁﬁﬁﬁ Solution 3 '1 7. (20 Under Poissonm. '00:" p.195 4 -J and 100.1 4; 3195 41 J -1. 4 -1. Therefore 1001 e _ 100.1 c at 4! .1104 Since ,1. is an integer, 2:4. (b) [1' 1-110. then 98-AS-M&S— 1 8 Aluminium. By calculating the .1 P” 4 = E = 0.03 Hﬂﬂﬂﬁﬁéﬁﬁﬁ expected frequencies under Pow.) when 2. - l. 2, 3, , FOR TEACHERS’ USE ONLY Remarks IA 1M can be omitted IA IM A Winter: ‘ ' P311) 2:112) 20123) E1115.) 3 6.l 18.0 22.4 19.5 4 1.5 9.0 16.8 19.5 5 0.3 3.6 10.1 15.6 From the table above. ,1 as 4. 2 1A forjust writing .1 = 4 1M , (5) FOR TEACHERS’ USE ONLY Hliﬁ‘ciiﬂiﬁﬁﬁﬂ FOR TEACHERS’ USE ONLY I , HBEﬂEiﬁﬁEﬁ .FOR TEACHERS’ USE ONLY Solution .S 9. (a) .r-J;l e"dx J 25 _ -x "i“ in: _ 3.4.444: . 0.5244 (0.524446) | | i (b) J’ = 09—: +513" y-intercept is —3 a a -3 y' - we" +be" —bxe" neglecting the value are = (-a + b - bx)?" y attains its maximum when x =% _ 50002”(At—1) 3 - ——':-2— .. er+ b -3b = 0 l l < 0 when 0 e t <-—- _.. . 1 3 2 b 0 =0 when ”Al (-43,34m) b="6 Hence y = r32“ + 6x9" >0 when %<t<120 (c) lfy=0, 3e"(2x-1)=o N attains It: minimum when ”43.3le M. 43 (The number of ﬁsh decreased to the minimum in about 1 43 days after the spread_ of the disease.) I = 5 The x-intercept of the curve is l . Is (13’ 2 (b) In Ki? 1‘" = 95-1 ‘61!" u 3 _L _. y' =—9£" —6e" + ﬁre” = _ 2a ._ In J; so (4 e )d’ - -15e" Him" 3 I I ,5 = Sax-5):" = —— —20 2° to 75 * 50[ E + E J ; <0 if lilec:i a 2 =0.l570 y' =0 if x:2 The increase in the mean weight of ﬁsh in the ﬁrst 15 days is 0.1670 kg. g > 0 if .t :2- -2- ” I'dwmzns. a d; d . , a The point ofinﬁection is (-2.125) [ad-3.0.9850” then si[—2o£'ﬁ+ioe'ﬁ] =05 o ‘ ‘ a (d) (1) “mum-Im- .. -1 25 . mm—“m lGe_ﬁ ~202 1° e_-to 0-5 - 3 Jo - T[0.303265w0.205212+2(o.3tt'ttwst+0.334595+tt.2'n:r61'n] 1 -3. _- 1 40.5137 (0.5mm) 20 _ 20 _ = [e ] 2[e ]+ 5 o .40 e -3x0.524446+6x0.6l3742 .. 2.1091 (2.109114) :- 2° = 0.087! or 1.9129 _ (ii) The argument is not correct because the trapezoidal rule was used to a 4: 43.8073 or 42.9721 lreJ.) approximate the value of J only. [g 3"“ about 49 days for [he mean weight of the ﬁsh The convexity of the function re" should be considered instead of I A for elther reason to Increase 0.5 kg from the Recovery Day. the ﬁtnction —3e“ +6xe" . 1 for both 98-AS-M4945—20 98-A3-MJtS— l9 Rﬁﬁﬂgﬁggﬁﬁ FOR TEACHERS’ USE ONLY Rmﬂﬁiﬁaﬁﬁh FOR TEACHERS’ USE ONLY ; Hmﬂﬁﬁ%ﬁﬁ FDR TEACHERS’ USE ONLY 3 amongst FOR TEACHERS’ USE ONLY Solution Solution Remarks [0. (a) (i) (I): Inrtt)ulna+ﬂlnt (ll): [nr(r)=lny+,u Correct to 1 ctp. Correct to I d.p. for any 2 pointt being correct for all the 6 points being COHEN for any 2 points being C0111“ for all the 6 points being correct D l 4 2 From the graphs. equation (I) would be a better model and In a z 0.3 Accept 0.3 - 0.4 a, 33:. L3 Accept 1.3- 1.5 4.62—1.86 e A: 2.2 —. ,6 l.95~0.69 Accept 2.0 2.4 ‘ Acce t are [L3 [.5] Jr P t (b) I; at dr where an: 1.3, 13*: 2.2 I? E {10'1“ a 5,. M L} 11 ll = — 1 .... | p+a 1" ‘3 3.2l' 1° ) i n: [839 1889 hundred oftrecs would be destroyed in the ﬁrst I4 days. Accept M98 - 3015 k Consider In m‘dt = t889x2 ﬁb’l]; = 3775 k“ w 9299.59 lug!” 69 k a; A: a: [7.3339 The total number oftrees destroyed will be doubled in 4 days more. 98-AS-M&5—22 anagrams FOR TEACHERS’ USE ONLY HBEﬂBﬁiﬁrtﬁ FOR TEACHERS’ USE ONLY 93-AS-Mk5—2] 2 ' Hmﬂﬁﬂigﬁﬁ FOR TEACHERS’ USE ONLY : Emaiﬁlﬁgﬁﬁ FDR TEACHERS’ USE ONLY Remarks or objection as sample } mean and sample var-inn“; are not equal H_..__ For any entry being 12. (a) Sample mean a- 1.2921 Sample variance - 1.2914 Since the sample mean is approximately equal to the sample variance. the results do not point to any objections to the use nfa Poisson model. I I. (a) Let X be the no. ofpn‘nting mistake: on P23. then X— Po (0.2). ch= 0): a“ . 0.8183 (b) (i) Let p be the probability that there are printing mistakes on a page. then p = l— 2'“ Hence N - Geometric (p) and P(N53J- Puvs 11+ P(N=2) + P(N= 3) e PH?“-.t=)’rr?(l-r02 - i—(l—p)’ eorreerin the f: column = l—e'“ . - 0.45m . i For any entry rein; .. l correctinthe — (II) MeanofN=F 3] 14-: -S.Sl67 column to (6| -9 . 1-1, 2-02 F 11 ' be' Variance ofN= —: = ——_-5.—. -24.9l68 on. cum“ "‘3 p (1-: .2)? correct (c) M— Binomial (200,32) wherep = l—n'lu . Mean am: up - 200(1 43'”) =- 35.2533 Variance ofM: npﬂ — p) ... 2002'” (1—2“) a 29.682] As the maximum absolute diserepmey is 0.70 which is less. than 1, the Poisson model is acceptable. necepl contrary ‘1‘ conclusion due to wrong i m i' th tabl (e) (i) 9(2 emspeedingandbmlnrepﬁvatecan) m a n e c | | . . . l ‘ I (d) (l) l’- Binomial (40, —) . . . (0,4) ' 200 f _ 0.]6 J (ii) P(3 can speeding and 2 ofthem are private cars) = dream ~04) ' = 0.288 I 40 (ii) P(l’=0)=[l—ﬁ] 2:13.313} (d) Let X be the number ofprivale can speeding and l" he the [Dial number of cars speeding in an hour. (i) P(X= 2 and r: 2) = P(x=2] Y=2)P(i’-2) . 0.15 3: 0.22930: . 0.0361 (0.035683) (ii) ﬁx. 2 and Y: 3) = PM =2! r=3)P(r=3) as 0.283 x 0.098753 a: 0.0234 (0.023442) accept 0.0285 (iii) P(x=2lr<4) = Par-2 and Yeti) PO" 4; 4) P(X=2 and Y=2)+P(X=2 and l‘=3) ‘ M < 4) .. 0.036688+0.028441 0.957691 = 0.0680 accept 0.0673 — 0.06M 98-AS -M& 5—24 Rmﬂﬁﬁﬁﬁa FOR TEACHERS’ USE ONLY anaemia FOR TEACHERS’ USE ONLY 93-AS-M&. 5—23 u ' Hmﬁﬁfﬁgﬁﬁ FOR TEACHERS’ USE ONLY Solution 'Muks Remarks 1!. Let X, l’ b: :11: weight: ofrh: mndomly selected boxes in parts 1 and 2 015's test respectively. (a) P(X< 490 or X> 510) _ 490 -500 5 I0 - 500 deduct 1 mark once {or 01: _ 1 _ .— P( 5 z s ) 1A whole qumion for any =l-P(-2:§Z\$2) wrung' uali sign as l — 2 x 0.4772 - 0.0456 ' IA (1:) P0190 \$X<492)+P{508 <Xs SID) 1A 490—500 492—500 508-500 SID-50“) =P( SZ<—5——-—)+P( 5 «:25 5 ) 1A uP(—252¢—l.6)+?(l.6<2\$2) .. (0.4112 — 0.4452} x 2 20.0640 IA AW» pcx < 492) + P(X> 508) — P(a black signal is generated in th: rust part) < 492—500)+P(z ) 508—500)_o_0m = Pmlnck in pan l) + ”black in part 2) 3: 0.0456 + 0.0540 x 0.0456 a 0.0485 (d) P(508<XSSIOa.ndSOB-CYSSIOI490\$X<4920r508¢X551m P(508<X5510)P(505<Y5510) N490: x <492)+P(508 <X5510) - 0.0310x 0.0320 0.0320+0.0320 an 0.0!50 le)‘ P(red|pan 2) = P(503<Xﬁsloand508<}'\$5l0I49BSX€492M503<X\$510) +P(490£X £492 md49OSX£492|4905X<492or508<15510) :- 2 x 0.0II50 a 0.0320 (0 P010) = P(rcd | pm 2) H0011 2) a: 0.0320 x 0.0040 a: 0.0020 Annmatinlx. ﬁred)- P(503<X5510 and 503<Y5510) + P(4905X (492 and 4905 1’ <492) = 0.0320’ :2 . 0,0020 1M 1A 9B-AS-MELS—25 Hmﬂﬁﬁﬁﬁﬁ FOR TEACHERS’ USE ONLY ...
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1998_MarkingScheme - | RBEﬂEﬁgE—EI FOR TEACHERS’...

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