1995_MarkingScheme

1995_MarkingScheme - Rfififlfifiééfé ‘ - Solution...

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Unformatted text preview: Rfififlfifiééfé ‘ - Solution .__...._ l (a) St m in 10 ea in l O 5 7 8 1 0 1 2 2 5 8 B 2 0 1 2 3 5 5 5 3 0 2 4 5 0 (b) Mode - 18 Median - 19 Interquartile range - 25 — 12 n 13 2. (a) o‘ée’-xy el¢erd7 .y.x% dy . e'-y a; X-“ | l (b) y. (x—2)"(x:3)1 (x+1)’ lny - 11,.ln(x-2)#;1n(x*3)-.§ln(x+1) 1 dy I 1 , 1 _ 3 737‘ 20"!) IIXQ) 21"“) 2x = r 1 . 1 - 3 ) dx 7 x—Z x*§ x+I (or 1 l(x-2)(x+3) ( 1 § 1 _ 2(x*I) x+I x-2 1H5 y(19-x’) or 2("'2)(1¢*3)(M*I) ) 9S-AS—M&S-3 FOR TEACHERS’ USE ONLY mm mm 3 x41 Remarks 1“ + 1 Awud IM if order or not more dun J gnu-in are wrong. 1A 1A 1A For either (6) IA + 1 1A 1“ + 1 1M for ukjn log, onbodl aidu and app in: lnab - na~lnb nlna - lna' 1A ilny - dz ydx (5) 3. (a) (b) RBESEEFH$E l-DK ItACHtKS’ um: UNLY Solution Marks Remarks No. of ways of grouping the students . C‘uxc‘n 1A - 17153136 1A Alteggatlvelx, 18! No. of ways m 1A - 17153136 1A No. of ways of grouping the students so that each group contains one girl - c,“xc,”x3! 1M + 1 leorJ! - 4540536 te a VG , No. of ways —--—l5I x3! 1M 4- 1 leorJ! 5! S! 51 - 4540536 4540536 Required probability m 1n = i 34 (or 0.2647) 1A (5) 95-AS—MkS-4 4‘ Rmfifi$fi ‘.~. Solution uln- dt 3t‘4 4. (4:) Hm -I[ 1 4:925)- .1 = %ln(3t~4)*2(t¢25)1*c .‘ rum-3.1 c = 3.1—%1n4-10 ‘ = -—§1n2-6.9 (0: -7.3621) Put) - %1n(3t*4)*2/C*55-%1n2-6.9 ( or %1n(3c+4)«2/—2'5:+ -7.3521 ) The value of the house, in million dollars, t years after the end of 1994 is %ln(3t+4) +2/_I§c+ -%ln2-6 .9. (b) The rise in the value of the house, in million dollar , between the end of 1994 and the end of 2000 is "(5)‘”(0) %ln22*2,/3'1'——§-ln2-6.9-3 . 1 §1nl2$+2mqo (or 1.7038) Alternativelx, M(6)-M(O) - Emmy“ 92m“ 0 = %ln22+2,/3T-%1n2-10 = %ln}2—1+2/3T-10 (0: 1.7038) 9iAS$¢kSS IN IA (7) FOR TEACHERS’ USE ONLY Remarks pp-l for mining dilferenml [A for inugrltiug l tum kifl§$flamassa run ltACHERS’ USE ONLY Solution Remarks 5. (a) Let A’ be the event of having no accident within a year. A 0.7 L 0.99 A' A 1A 0.88 A’ W. pa.) - 0.7, Ni!) - 0.3 P(A’|L) - 0.99, pm’lu) - 0.88 1A P(A/ H) P(H A’) - —— I P(A’) _ P(A’ H)P(H) P(A'|H)P(H)+P(A’|L)P(L) I 1H+1H* IM forlhenumemor O.88x0.3*0.99x0.7 1M for the denominator - 0.2759 (or 75;) 1A (b) P(L|A’) - 1 - P(H|A') - 1 — 0.2759 in = 0.7241 (0: 21) 1A '2"; Mix, P(A’ L)P(L) m. m - __——l—-—— ' P(A’{H)P(H)‘P(A’|L)P(L) 0.99xo.7 . ._________ 1H 0.68x0.3*0.99x0.7 21 . . 1A 0 7241 (or T9.) (7) 95-As—Mm (b) RPRWEW§W I'UK Solution 2"“4 - 5(2') (2">‘-5(2')*4 - o (2‘-4) (2"-1) - o 2"-4o:1 xs2or0 The intersection points are (0,5) and (2,20) If 2" - e“ for all values of x: then a = ln2 Area - [0’ [5(21) - 2" — 4]dx . f1 [Sexlnz _ exlnl _ dldx o I 5 xln22 _ 1 xlmz _ 1 1n2[’5 1° 1M 9 1° “x1” _.1_ 1112 5 =15( fi)—e 1 - (or 2.8202) 21n2 ll 9 5 -AS-M&S-7 l tACHtKS’ USE ONLY (b) Hbfififlfligfi FOR TEACHERS’ USE ONLY Solution i mammalian mi 1.00504 1.02062 1.04828 1.09109 1.15470 Il - o.1[%(101.1547o) * (1.00504 ‘1.02062 +1.04828‘1.09109) ) =- 0.5242 X (n) f’(X) - -—; (l-x‘)‘ 2211*1 til (x) . ___! (1-29)1 (111) By (nun). t”(X) >0 for 05x51} I {(x) is concave upward (or convex) on [0, é) Hence I1 is an over-estimate of I. -_1 (:1) f(x) - (1-x‘) ‘ 1 1 1 (-—)(——-1)(-——2) 4, .2—.2_.—2(-x3)J o 31 1 2 3 A i t o l. . 1+3x‘8x+16x#... for 5x52 -. p.(x) - hale-g-x‘oTss-x‘ u NIH 9 95-AS-M&S-8 Remarks Using conful formuln mun be limplified Argument for convexity IM (or binomial series (ii) 9$ASJH&S& Rfiflfigfi Solution 1 1 1 - (-~)('—-1) (-—-z*n . 2 2 2 f(x) - p(X) + I, (-x') 1 l l (—) (—‘ll . (—‘t-l) 2 2 2 - mm + n x“ > p(x) to: 0<xs%. ‘ Hence I > I, i.o. II in an under-estimate of I. 1 mark for the following argument in b(ii) Sum to infinity p(x) is just a truncation Hence underestimate withhold 1 mark 952; for incorrect degree of accuracy. FOR TEACHERS’ USE ONLY Remark: 8- (a) (i) 9$ASM&&IO REfifi$fi Solution N(x) - ae“‘ in N(x) a: lna 4>ln e‘ - in a — bx A: :=_—5'===fl=3:= "‘33::- D. m _---‘m—Illum-I-l “III-"Ill..."- I.‘ “mlfl-m-I “III III—III I _ n-ul-L‘IIl-- “mun-mum.‘ Ila-.- m._" -I_ --—---=-::m::... I... umfl=ull==I==-IIIII== :=-—=.=:..==:=:=---- .- .- m-I g§-__.:=:====a.-==:: ::I:_Iln-II —I-IIl—I I..._I_-I-=IaB-I II:- a—"L-mamaum ...__.......-.....=E==E;.;'=EEE:::E-."-'- .... I‘m-Illumaflgm I. I luuull—III— IIE: From the graph, in a a 6.9 a - 992.27 4.91-6.11 -_EU:IU__ b a 0.04 -2): FOR TEACHERS’ USE ONLY Al Inn 1 d.p. [A for the point: [A for the line Accept 6,85 - 6.95 Accqu 94138 - 1043.15 no mark for b-slope or Dal-0.04 in my calculation. Hmfififigfi FOR TEACHERS’ USE ONLY Rfififififi$fi FOR TEACHERS’ USE ONLY 1 E Solut ion Remarks Solution .27 -o.ou _ (b) 992 e 400 m 9. (a) (i) Cfleo) . EMS—0’. . 3,200 400 . 100-80 -0.04x I in w” ' The cost required is 3,200 thousand dollars. 2: - 22.7 = 2000 1 ae'°~°” = 400 where a e (943.86,1o43.15) 3 x - 71— (or 71.4286) -0.04x= 1:11-29- 7 Hence 71-571‘ (or 71.4286‘) of the loudness of ~ x E (21.5.24.0) - ,. .. v .. 03 o lo *X the noise released from the Stadium can be The price of each CD should be $22.7. 1A Accept $215-$240 reduced, (c) (i) G(x) - 992.27 (x-1o)e-Mu m (b) (1) Cum) = W ‘ (loo-x)2 ' 60000 C(x) - a(x—10)e’°~°“ (mo-x)“ where a 6 (943.68,1043.15) 1A cm, = w (loo-x)I (ii) G’(x) - 992.27[e'°~°“.(-o.04)(x-1o)e‘°-°“] 1M - 992.27 'M“ 1. - .o 9 t 4 ° 4"] 1" (n) c,”(x)>o for OSx<100 The curve y - C‘ (x) is concave upward in (or convex). : ae-°~°4x[1_4—o,o4x] 13 The vertical asymptote is x=100. where a E (943.88,1043.15) Y G’(x) =0 when x=35 and 1M roro’uc) =0md solving , )0 if x<35 G (X) {<0 if x>35 1“ G”(x) = 1.59xe‘°'°"‘-95.18e'°'°“‘ For not crossing the nympwla G”(:§s) = —9.750 M - For the curve Therefore C(x) is maximum when x335. For maximum profit, the selling price for each CD should be 535. 1A 95-AS-M&S-l l 95vAS-M&S-12 Rpmq-xptuvlfij run IEAbHEKD' Ubt UNLY Solution (iii) y ' C1(x) and y - C,(x) interect at (75, 2400). 9$A$M&543 The local engineering company is more cost effective. Remarks For the graph ofy - C10) (aw-rd 1A if: cmdidnu correctly skctchl the graph of y I cor) only). 9153313ng I-UK ItACHtKS’ U51: UNLY Solution Exwxud‘ Frequency Exyxud‘ Frequency [M for Expect-ed Frequency 1M for Probability 1A+1Afor the 7. pain of am [A for all being comet ‘ Correct to 1 decimal place. (ii) Po(3) fits the data better. Let x be the no. of bicycles sold, then X ~ Po(3). P(X=0) = 0.0498 (b) (i) (ii) Let Y be the no. of days that no bicycles will be sold, then Y - B(7,0.0498) pass) a c;(o.o493)3(1-o.o498)‘ - 0.0035 (c) 9(3 items sold) P(3 bicycles sold) + P(2 bicycles and l tricycle so d) + P(1 bicycles and 2 tricycles sold) + P(3 tricycl s sold in For Addition 3 0.2240X0.1353+ 0.224OX0.2707+ 0.1494X0.2707 + 0.049BXO.1804 1M forth: 1st&4lhtennl 1M for the 2nd 81. 3rd term: - 0. 1404 954ELM&SJ4 11.‘(;3 Probability of rejection, (b) Let X be the number of cartons inspected by Madam Wong in a day, then x ~ Geom(p,). ’. mean ‘ ——- (C) (i) (ii) (5‘) 95d¥$bfifi$45 Probability of acceptance, p,-(1-0.02P Solution '0.9039 pr'l—pa I0.0961 1 h P: - 10.4 Prob. that Madam Wong can achieve her target, p, - P(All cartons are acceptable) + P(exactiy 1 carton is not acceptable) + P(exactly 2 cartons are not acceptable) . (p.)za,(212)pt(p.)z1 1(222) (p‘)3(p‘)zo I 0 . 5445 Accept 0.6444 - 0.6445 -lter ative, , l - P(the 1st 20 cartons are accepted) + P(1 is rejected in the 1st 20 cartons and the 218t carton is accepted) + P(2 is rejected in the 1st 21 cartons and the 22nd carton is accepted) = (p‘) ’°+( 21°)p,(p‘) ’°+( 221) ([3,) 3 (1).)" b = 0 . 6445 Accept 0.6444 - 0.6445 If Madam Wong can achieve her target, the prob. that she needs to inspect 20 cartons only ‘ - (1).)” P1 - 0.2058 Accept0.2057 (l-rt)s z 0.95 r% 5 0.010206 r 6 1.0206 The greatest acceptable value of r is 1.0206 . Rmflfifi$fifl FOR TEACHERS’ USE ONLY Solution 12. Let x denote the test score and D the event that a person has the disease. (a) P(x>63.2|D’) - 0.33 P(Z>§2j%;fl) = 0.33 From the normal distribution table, §3i§LE - 0,44 s. u - 61 66-70) 5 - P(z>-o.8) - 0.7881 P(X>66|D) - P(Z> (b) (1) 66-61) and P(X>66|D’) - P(Z> s - 9(z>1) - 0.1587 P(the person will be classified as having the disease) - 0.15x0.7881+(1-0.15)x0.1587 - 0.2531 (ii) 0.7881 >66 566 0.1587 >66 :66 P(X566|D) - 1 — 0.7881 ' = 0.2119 3. P(the person will be misclassified) - 0.15XO.2119+(1-0.15)x0.1587 - 0.1667 95—AS-M&S- [6 ...
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This note was uploaded on 08/17/2011 for the course MATH 0001 taught by Professor Unknown during the Spring '11 term at CUHK.

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1995_MarkingScheme - Rfififlfifiééfé ‘ - Solution...

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