MI H2 MATH P2 Answer

MI PE II 2015: PU3 H2 Paper 2: Marking scheme   1i 1ii 72 ( ) ( 52 ) 72 52 7By observation, to have only 1 root, 5 x x xe k x x ekx k         1iii 2a 22 2 20 dd22 0 dd d2 d 2 d2 d 20 20When = 5, 25 d2 ( 5 ) 2.32 cm/s (3 sf)d 2025 Rr RrRr tt rR R R tr t r Rr R r t                y  x y x

2bi 2 2 2 d 2d d2 d( 2 1 ) d1 d( 2 1 ) 4From given line, gradient of line . 9 d1 4We observe for all d( 2 1 ) 9 so it is shown there is no such tangent. x t y tt y xt y txt          2bii   2 Grad of normal 2 1 11At , 0, and grad = 422 1Eqn of normal is 4 . 2 When normal meets curve again: 11 4(2 1)21 2 91 or (NA, already used)16 2 91 7For , coordinates of point is , 816 8 t tx y yx tt t t           .  3i 45 15 2 1 5 5 and 10 AB k BC k k                         3ii Since both triangles share same base with points A, B, C, so ratio is 1 : 5. 3iii r 41 21 , 02               

3iv 1Since , 3 41 11 20 . 544 00 4 2 2 41 3 20 . 5 1 . 5 22 5 15 2 10 5So or 2 OP PB OP OB OQ PQ PQ OB OB                                                           

 25 2 1.5 8.5 52 5For , 4.5 and for , 14.522 52 5 Coordinates of are (1.5, -4.5, -5) and (-8.5, -14.5, -25). OQ OQ Q                           4i,ii Midpoint of A and B = 35 22 i 4iii h = 1 E

4iv 1 2From diagram, (tan 3) 3 12As is isos. , greatest arg = and 23 6 1smallest arg = 23 3 Range is . 36 PCD PDC z                  5i The group should decide on quota for each male and female stratum, eg 50 males and 50 females. They should then situate themselves at the exit. Select and survey people according to strata until respective quotas are reached. 5ii Not appropriate. The reason is since the period is June holidays, there is a high likelihood that many school children will visit. 6i Let X be the random variable for weight of a durian. Let T be the random variable for total weight of a basket of durians.    2 12345 1.52 N 5(2.36) 1.52, 5(0.07 ) N 13.32, 0.0245 P( 13.2) 0.222 (3 sf) XXXXX T T   

6ii 0.0245N 13.32, 10 P( 13.2) 0.992 (3 sf) X X   

7i Let X be the number of hours of sleep for each student in the recording. 0 1 H: 6 . 5 H: 6 . 5     Assuming H0 is true, since σ is unknown and n is small, we use t- test and assume X follows normal distribution. At 10% level, we reject H0 if p-value ≤ 0.1 Using GC, p-value = 0.4096 Since p-value = 0.4096 > 0.1, so we cannot reject H 0 and conclude, at 10% level, that there is insufficient evidence to support Mr Lee’s claim.

7ii 0 1 H: 6 . 5 H: 6 . 5     For α = 0.03, z–value = −1.8808 Let Y be the number of hours of sleep for each student in a sample of 60. Since sample size = 60 is large, by Central Limit Theorem, 2 N6 . 5 , 60 sY   approximately. For H0 to be rejected, 2 6.42 6.5 1.8808 60 0.109 (3 sf) s s    8i Let X be the number of defective pencils in a packet of 40 pencils. B(40, 0.013)X P( 1) 0.905X  (3 sf) 8ii Let Y1 be the number of defective pencils in a packet of 20 pencils. 1 B(20, 0.013)Y Let Y 2 be the number of defective pencils in a packet of 30 pens. 2 B(30, 0.015)Y 2 pencils, 0 pen: 12P( 2) P( 0) 0.01612YY  1 pencil, 1 pen: 12P( 1) P( 1) 0.05887YY  0 pencil, 2 pens: 12P( 0) P( 2) 0.04934YY  Required prob = 0.01612 + 0.05887 + 0.04934 = 0.124 (3 sf) 8iii Po[0.015( )]Tn approximately P( n – T = 50 ) ≤ 0.2 Using GC, the least number is 52. 9i Let X be the number of customers in a hour during the peak period. Po(15.6)X P( 12) 1 P( 11) = 0.851 (3 sf) XX   9ii Expected number

= 2(15.6) + 0.5(8.2) = 35.3 9iii Let A be the number of customers during the peak period. Po(31.2)A Since λ = 31.2 > 10, so N(31.2, 31.2)A approximately Let B be the number of customers during the off-peak period. Po(41.0)B Since λ = 41 > 10, so N(41, 41)B approximately For required condition, we have P( B > A) = P( B − A > 0) N(9.8, 72.2)BA P( B − A > 0) = P( B − A > 0.5) [Continuity correction] = 0.863 10i r = 0.970 (3sf) Since r is close to +1, it suggests a strong positive linear correlation between X and Y so a linear model is appropriate. 10ii 10iii Simplify 2 eqXYp to 2ln ( ) lnY q X p . Obtain r value = 0.977 (3 sf) The better model is 2 eqXYp as the r value is higher. 10iv 2ln 0.00646( ) 0.833YX Y = 6.85 (3 sf) Result is reliable as r is close to +1 and it is obtained through interpolation. 10v r value will not change. A change in the scale of the variables will not affect the scatter / cluster of the points which determines the r value. 11ai 18 2 22 20 4 0.0316CC C   11aii Males from basketball: 37 22CC Males from volleyball: 49 22CC Males from badminton: 29 22CC 12.0  5.68  8.71  14.1 X Y

Prob = 37 49 29 22 22 22 20 4 0.0650CC CC CC C   11aii i 461 0 12 1 41 6 13 P(3 females and 1 must be netballer)P(3 females | 1 netballer) = P(1 netballer) = = 0.268 CC C CC 11bi 5! × 4! × 3! × 2! = 34560 11bii (6 2 × 52 × 42 × 32 × 22 × 12) / 6 = 86400