MIPU3 H2 Mathematics Paper 2 2012 Prelim II Answer Key

PU3 H2 Mathematics Paper 2 2012 Prelim II 1(a)(i) 1(a)(ii) 4 2 5 1 4 2 5z     1(b)(i) 1(b)(ii) 12zi 2(i) x (3, 4) -1 5 x (1, 2) 2 3 1 y = -1 0

2(ii)       2 1 1 2 1 ln 1 2 ln 1 2 : ln 1 , 1 xye yx yx f x x x      2(iii) 2(iv) 2 2 1 1 0.797 0 x x ex ex x x     3(i)       2 2 2 2 1 2 1 2 2 1 22 dx tdt dy dt t dy tdx t dy dx tt         3(ii) 2 1 2 2 1 2 1 1 1 1t x t t yt             At 1,t 1 2 dy dx  Gradient of normal 2  1 2 2 25 yx yx     0 x = -1

3(iii)   211 22 32 12 2 4 3 5 0 1.158 1 2.158 t t t t t t t t            4(a)                1 2 11 62 6 6 4 4 1 2 1 4 1 1 2 1 12 1 2 11 n r nn rr nn n n rr rr n n n nn nn                   4(b)(i)         11 11 12 11 23 11 1 1 1 1 1 1 11 ... 1 nn rr nn n r r r r              4(b)(ii)    11 21 211rr r r r r    As 1 1, 1 1nn     1 2 21r rr     5(a) (i) Consider the green, yellow and purple tiles as 1 unit. Number of ways to arrange the green, yellow and purple tiles within the unit = 5! 3! Arranging the unit with the rest of the tiles = 7! 3!3! Thus number of possible arrangements for the tiles = 5! 7! 3! 3!3! =2800

(ii)Number of ways to arrange the 3 red, 3 green, 1 yellow and 1 purple tile = 8! 3!3! Number of ways to slot in the blue tiles = 9 3   Number of possible arrangements such that no blue tiles are placed next to another = 8! 3!3! 9 3   =94080 (iii) Number of possible arrangements such that a red tile at the beginning and another red tile at the end of the line = (3 3 3)! 3!3!  = 10080 5(b) (i) Number of ways =  10 1 ! 362880 (ii) Insertion method Number of ways   7 27 1 ! 2! 2! 60480C      (iii) Number of ways  9 1 ! 10 2! 806400     6(i) P(same colour) 45 22 9 2 4 9 CC C  Die is 1 Die is 1 Die is not 1 Die is not 1 Die is 1 5 6 Same colour Different colour Die is not 1 Die is not 1 Die is 1 1 6 4 9 5 9 1 6 1 6 1 6 5 6 5 6 5 6

P(wins exactly one DVD) 4 1 5 5 1 37 2 (0.228 (3 s.f.))9 6 9 6 6 162       6(ii) P(different colour|did not win any DVD) P(different colour and did not win any DVD) P(did not win any DVD) 5 5 5 25966 (0.510 (3 s.f.))4 5 5 5 5 49 9 6 9 6 6        6(iii) P(wins 2 DVDs| different colour) 1 1 1 6 6 36   6(iv) Let X be the number of DVDs John can win, out of 2 dice throws. Then X ~ B(2, 1 6 ). Mean number of DVDs = 112 63 7(i) 0.946r 7(ii) The scatter diagram shows that the relationship between t and x is non-linear. 7(iii) From the scatter diagram, we see that y decreases as x increases, which is the case for model A ( ,0bx a b t   ). Hence model A is appropriate. 7(iv) 0.75046151 11.0423697 11.04236970.75046151 a b x t    11.04236970.75046151 0.9958475033 0.99645x    This estimate is reliable since 45t is between the range of values of t from 5 to 80, and the product moment correlation coefficient between x and t is 0.9754 (3 s.f.), which suggests a strong positive linear correlation between x and t. 8(a) Let I be the r.v. ‘no. of packets of macademia nuts sold in a week’.  Po 10I    P 11 1 P 11 1 0.69677 0.30323II       Let W be the r.v. ‘no. of weeks, out of 52, of which more than 11 packets of macademia nuts

are sold per week’.     B 52, P 11 , i.e., ~ B 52, 0.30323 W I W   P 0.08Wk  P 0.92Wk Using GC,  P 20 0.869 0.92W    ,  P 21 0.9209 0.92W    Least value of k = 21. Let I be the r.v. ‘no. of packets of macademia nuts sold in a week’.  Po 10I    P 11 1 P 11 1 0.69677 0.30323II       Let W be the r.v. ‘no. of weeks, out of 52, of which more than 11 packets of macademia nuts are sold per week’.     B 52, P 11 , i.e., ~ B 52, 0.30323 W I W   P 0.08Wk  P 0.92Wk Using GC,  P 20 0.869 0.92W    ,  P 21 0.9209 0.92W    Least value of k = 21. Required probability = P(exactly 3 bonuses within the first 10 wks & bonus in the 11th wk) = P(exactly 3 bonuses within the first 10 wks) × P(bonus in the 11th wk)       3710 3 P 11 1 P 11 P 11 0.00668C I I I             Or, Let X = number of weeks (out of 10) where bonus is paid. X ~ B(10, P(I > 11)) P(exactly 3 bonuses within the first 10 wks & bonus in the 11th wk) = P(exactly 3 bonuses within the first 10 wks) × P(bonus in the 11th wk) = P(X = 3) × P(I > 11) = 0.00668 8b(i) Let B( , )X n p Given 3, (1 ) 2.85 60, 0.05np np p n p      For X  o 60 is large (>50) and is small , 3 5, P 3 approximately n p np X     Similarly for Y  o 80 is large (>50) and 0.02 is small , 1 .6 5, P 1.6 approximately n p np Y     

By additive property of Poisson distribution, hence  oP 4.6XY 8(b)(ii) From GC: P( 3) 0.163XY   (3 s.f.) 9(i) Let X and Y be random variables for the amount of time Singaporean youths and American youths spend at an ice skating rink per month respectively   2~ 10.1,3.2XN and   2~ 9.3,2.3YN ( 5) 0.0555PX  9(ii)   2 2 2 12 2 ~ 2 10.1 2 9.3, 2 3.2 2 2.3 (1.6, 41.64) X X Y N N          )2( 21 YXXP  )02( 21  YXXP =0.5979 =0.598 9(iii) Let W be the random variable for the cost, in $, spent by a Singaporean youth per month.   227 ~ 7 10.1,7 3.2 (70.7,501.76)W X N N    ( 120) 0.0139PW  10 For each of the classes from Arts and Business faculty, P(a student being selected) = 6 20 . For each of the classes for Science faculty, P(a student being selected) = 6 30 . Since the probabilities of selection is not common, the sample is not random. Line up the 240 students in some order (eg alphabetical order of name). From the first 240 460  students, select one randomly. Thereafter, select the next 4th student till 60 are selected. 11(i) Let X be the volume of coffee dispensed in a cup (in ml) with population mean  0 1 : 100 : 100 H H     Assumption:   2~,XN  Test Statistic: XT s n  Level of significance: 5%

Reject H0 if p-value < 0.05 Under H0, using GC, p-value = 0.0550537= 0.0551 (3 s.f) Since p-value=0.0551>0.05, we do not reject H0 and conclude that there is insufficient evidence at 5% level of significance that the machine is dispensing too much coffee To conclude that the machine is dispensing too much coffee i.e. 100p value  5.51 