1997_MarkingScheme

1997_MarkingScheme - E1N§$&Efiééfifi FOR...

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Unformatted text preview: E1N§$&Efiééfifi FOR TEACHERS’ USE ONLY Soluuon chlurks 1 _ I I V l r,( r _1) _ 1Aforfl1clsl&2ud term 1 _ _ r 3 1 l. (a) 0+“) -_1+ 3‘" + 24 (ax) +--- JANA lAfurlhc'firdlerrn z =l+-!ax—a—xz+... 1" 9 1 H a: ~fi=v1 ) 9 ‘ (1:3 or ~3 —r-—-1A ‘- (4) 2. (a) Range=(26— l8)°C=8°C [A (b) (i) Median =2|.s °c =:%(1m+32] “F = 70.7 “F [A . 9 9 lnlerquarule range = §(22.S)+32 — ?(20) +32 “F [A for 215—20 = 4.5 “F 1A (ii) Mcan=[%(22)+32|“F=716°F [A Standard deviation 2 gr 2 "F =- 16 “F 1A (6) 7 3‘ 1A for lb: 2 vertical asymlmcs IA! 1A! IA for each part of the cunt-s 1A for file local minimum {5) 97-AS-Mk5—3 HWfl%%fi mREmwm’mEmu Rmflfihéfifih FOR TEACHERS’ USE ONLY Hmntnere FOR TEACHERS’ USE ONLY “, Solution Remarks bolution Remarks _..__-_ G. (a) Let X be the number ol'ears passing through the auto-toll in a iiiinute. 4. (a) v0; = [20004301 IA then X~ 130(5). 7—100:2 -30001+c IA Pu”) , __. vwrzoooo. .'.c=20000 IA = 1—2” 5 9 Hence V(n=ioo:1—3000r+20000 for Usrsk. “0 X! as 0.3840 a 71 for r t. O 384 (b) Wk): 0 I (Is) Out of the next 4 minutes. let Y be the number of minutes in which look —3000k +20000 = 0 IM more than 5 cars will pass through the auto-toll, then }'—- Bt-l. 0.3840) k2—30k+200=0 P(Y=3) (k — 20M — 10) =-0 as (gttuswfu — 03840) For binomial formula ii :3 or 20 (reacted) 1“ = 01395 (or 01396) a~l for m. 0.140 tc) W5) h W0) 'M = 100(5)I —3000(5) + 20000—20000 7. LC! A. be the event that the original motor breaks down, = 42500 A: be the event that the backup motor breaks dovm and W be the even that the machine is working. (3} Wt :42) = 0.15 x 0.24 = 0.036 (b) PM): :— P(A,A2) : l — 0.036 = 0.964 The total depmiation in the first 5 years is $12 500. Altgrn-gtjvely. 5. (a) Nugget of “ans In which the 10 students can Lake I IE 5122 pup): p(,41)+ PHIAZ) : ‘ , =0.35 +0.I5XO.76 'M ll 41 4! : 0.96-1- = 3150 The probability that the machine is operated by the original motor (I!) Number oi'ways in which the 10 students can take the seaLs with the 2 students from school A are next 10 each other 7 9! ' Cu 4! = ,M =63“ =‘ 0-33” IA 3—! font (1832 The thhhtttty that the 2 5tUdCht5 from “hoot A We 11'3"! to each other (C) The prob. thal the 131 break down of the machine occurs on the 101]] day : ii = (ammo—003mm" N I = 0-0259 1A n—l l'or rt. 0.020 ' 5 (7) Ahemnu‘velv. The probability that the 2 students from school A are next to each other _ l l l 2 9'1! — 2 -fi-3 + B'fi'h (0' WI” ) 2A Marks can be awarded 1 independent of part (a), = IA (5) 97-AS—MScS-4 9745M“ ‘5 Rmflhfiglfi FOR TEACHERS’ USE ONLY matinee FOR TEACHERS” USE ONLY mama-mg FOR TEACHERS’ USE ONLY grammar FOR TEACHERS’ USE ONLY Remarks Solulion Solution N(0) = l6 9_ (a) b .3 349—195 _49_ = 16 8—14 l H“? = —0.1 .- b = 1.5 Sub. (8. 7.49) ink) lnN(x) = —0Alx+lna . 7.49 n: Ina —0.8 NW) 7 17.4 a a: 40m) l ‘ -4) 1 =17"; - 1': -ol [+1.5e" (b) (I) Mr: =ac‘=4oom ‘ Daily profit (in dollars) ofsclling Nor) clams: 13(1) = N(r) - r - (2 fo) +5000) = (x— 2) N(x) - 5000 a 400m —2)e °-“ —5000 e Jr _ _L[flLl} _ 15 m for2N(x)+5000 r= = 002 (m P11) = amrx~2)(—o.1e*'-”)+e “*1 40 4|] = 4002'M(12—x) (b) “0) = (or *7 7 A ) “be” “1.59—0.92: :0 if 04.1112 4 _b —fl' u 4mm P ' = ' = N'(r)=————0{ m I) (or W) (x) o If,» 12 (1+be’”) (1+1_<e‘*"“2')2 <0 if n12 = “braifl or Ila-0702! Par) attains its maximum when x = 12. (I “’94); ( _(1+ 156. mar—)2 ) Hence the scllirrg prior: Breach clam = $12 > 0 the number ofclams sold per day = M12) NU) is increasing. x 4m4‘1‘l2) as 1205 (c) lim 6"" r 0 - Hm (C) The difference bem‘een the numbers of clams sold on the n-th N = “m 40 (or um 40 ) and (rt-1H}! days after the launch of the promotion program: a 1-H" l+bc'” r m l+152'°°2' = MU?) ‘ Ml” “ 1) v 40 = [1500+1000(l—e‘°"")]—[l$00+1000(l-e'°"”'”)] : imw—o.ln+f43.in_eo.r) (d) (i) N "0) = [maln(eo,1_l) ; mil};- M“ )(12) 7 mm" (2;(15;}(1+ 1.5e'°°" )(—o_02)u'°f'_1f “+ 0.01:)4 Ir M(n) — M(n — m 15 _ OIOIZF-fl.alt(3€-002r _2) 4"" H “Hi-0.02:): then e -=C —lm8h| _n r: >- “3.475 > 0 whcn r < r0 The promotion program: should run for 10 days (ii) From (i). N"(r) :0 when r:- 1“ <0 when (>10 71* 2 where r0 = mm: in; 9: 20.273] The rntc ofincrcase Is lhe greatcsl when r: to 9: 20.2733 For Solving N "(n = o For checking maximum N '(20) r: 0.199999 N '(2I) = 0.199959 The company should slam to advertise on the 20m day after the first week 97-A5-M&5—6 97-AS-M&S—7 Hmflfiifiéfififi FOR TEACHERS’ USE ONLY Rmfigmgga FOR TEACHERS’ USE ONLY nmfln%n FOR TEACHERS’ USE ONLY - " Solution —u—"—‘_ _ 10. (a) y = I" In y = x111 x 1A i E l I x V (11' _ + n lawn“) 1A (b) 12.! = x’i(l+lnx)+(l+lnr)£:' 2 dx dx = x‘-%+(l+in:)r’(l+inx) 1A = I“ + x‘(l+ [:1 Jr)2 3‘ 0 for I S I S 2 1A y is concave upward (or convex) for 15 r 5 2 I won“ be overestimated if the trapezoidal rule is used to cslirnzllc I. l 1 (c) I+J =11 x"(l+|nx)d.: 1A = [r]2 tlv (a) 1 . = 3 l 97-M-M-Esws HEfifi$fi Remarks FOR TEACHERS’ USE ONLY Rflfififiifififlfi FOR TEACHERS’ USE ONLY Solution Remarks (d) (i) I'm-II “ IA Jo = 3931277259 0 2((12269! 4.053393 myormnmnn] IN! = 0.9535 ]A (ii) y 3.5 3.0 2.0 1,5 73 1.0 0,5 1A+1M l 1.2 1.4 1.6 [.8 2 From the plotted graph, y = r‘ In 1' Is concave upward (or convex} for l S x S 2 . Jn is an overestimate of J . [M (iii) The estimation can be improved by increasing the number of sub-inicn-als 1 (iv) 1'1. is an underestimate of I because the value 3 for I + J is exact and Jo is an uvcrcstimnlc of J I 91ASMkfi—9 RBEflEn%I}fi FOR TEACHERS’ USE ONLY HWflfi$fi Let .\' be the number of FIGS per day, then X- Po(-|). 0 -4 I'(.t‘=0) :4 P (1:) Let J” be tlte number of Fle which are related to house fires in 5 FIGS, then Y— B(5_ll.l5). P022) =1 *P(Y=U)—P(}'=l) - = t—Cgm-i)S — C‘:(O.6)(0.-t]“ a. 0.9130 (C) Let I! and L be the events of “a FIG is related to a house fire" and “a FIG is large". Let A be the amount ofa FIG. (1) Pal II): PM > 20 000) = P” 2300000400000) ' 50000 = P(2 > 2) 9: 0.0123 PUJ fl) =P(.4 > 20 000) =Hz>§0000450000 20000 = P(Z> 2.5) 9: 0.0062 m.) = P(Ll may) + Pg] .3 )P( r? ) e 0.0223(06) + 0.0062(04; x 0.0162 I’ll-i H) PHI) Pu.) t 0.0223 x 0.0 0.015? a: 0.3444 (iii Pit-111.): (iii) P(5 FICs and at least 2 ol'them are large) = [’(2 or more out 01's FICs are large)P(X=5) -4 J = [t—(1—0.0l62)’ —S{t),tt162)(l-0.0162)“1%; r 9: 0.0004 97-AS-M&S~It] Rfiflfififi FOR TEACHERS’ USE ONLY Remarks FOR TEACHERS’ USE ONLY gramme/.500 Solution I2 (a)&(b) Note: Under B(5, 0.4), enpcclcdl‘req =6le C:(l).-I)‘(U,6)5“ , ‘ Concetta 2 decimal places, (b) L Net. a?) P(.t'< 9000) = fl 60 PU? a: 90m" V i: L00 (= 0.06667) 0' 60 w :3 —1 50 ............. ..(1) U 90;: 12000) g m < L300” 3; “‘5” ('0: 0.30333) U W =_0_50 .. .12) 0' Solwng (1) and (2). we have ,u: 1350f) _. a: 1000 (C) N(l350ll. 30002) is used to model the sales volumes. (i) The probability that a salesman will get a “timing = P050)“ + 313020)2 men + 3P(_t'=0)P(.t’=l )I + 101030)z mm) (-tzaflutelteufit (ii) The probability that a salesman got no merit points in {flfifiq " 60 ' 50 200203 at least 2 ol‘tl'tu previous 3 months I_ .1P(.\'=0;_P(._i'gn{ Ulllll'l fflpfl ' 00 60 00201 al- =lJ 4246 The number of sale5niun n he are expected to get no merit points in at least 2 ol'the previous 1 months z lino-1246 = 4 9T-AS-Mt‘kSvl I HEfifi$fi 4.00 ‘ 22.99 lAl'lA 1A+1A lA 1A IA IA IM lA IA lM IA [M IA FOR TEACHERS’ USE ONLY Reina rks For the 3rd column IA for any one being correct 1A for the remaining two For the #111 column 1A for any one being correct 1A for the remaining tum FOR TEACHERS’ USE ONLY Rmflfiifiéfififi FOR TEACHERS’ USE ONLY Solution 13, Let [.crn bclhc length 01111:: from portion oer. Wong's necktie. (3‘) H44 < I. '1: 45) 14 —4-1,6 ? 45 44.5 _ 1.2 fl 4 ‘ < 12 ) = 121—05 < z M13333) 9:11.19” 4 0.1293 (or 0.1915 +0.13%) For cithcr 20.3203 (0r 0.3221) _ (b) 111:! 1' be the number 011115115 that Mr. Wong gets the first perfi'cf lying. then 1‘ ~ Geomcu'icfio), where p 20.3203 (01 U312] ) 1 EU) = — P 33.1172 (01' 3.1046) (c) Pmut more than 3 trials) = P(1 trial) +P(2 triais) + P(] lrials) = p+p1|—p)+p(I-p1’ = 0.6867 ( 01' 0.6335 ) or 1—(1--p)3 (d) Lel T be the event 1.11211 Mr. Wong has to go to work by 111d. (1) Pm a 1 #116867 (or 1—0.6885) =o.3133 (or 03115) P(Icss than flout off: days) = (3015357)“ +cf(0.63671‘(o.3133) (or ($061135?+Cf(0.511115)5(0.31153 ) R: 0.3919 (or 0 3957} vl. (n) Puzsl 1);: W 20.2179 (or 0.2134) (1:11 Probability-required = 10313311106357)‘ (or 5(0.3|15)2(0.6385)‘ ) =0.l1191 (or 111090) 9108-1114154 2 HBRflaflififi FOR TEACHERS’ USE ONLY ...
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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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1997_MarkingScheme - E1N§$&Efiééfifi FOR...

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