2012 H1 A Level Solution (SAJC)

1 Let 2xeu  13 4( 1)uu  2 2 3 4(1 ) 3 4 4 0 uu uu     Solving, 2u or 2 3u 2 2xe  or 2 2 3 xe  (rejected) 22 ln 3x 12ln23x 2 3 20 100 4 2 80 2 40 (1) x y x y xy xy          ( 20)( ) 3(20 ) 60 (2)20 x y x xy x    Solving (1) and (2): 2 2 2 602 40 20 2 40 60 40 800 2 60 800 0 30 400 0 xx x x x x x xx xx             Solving, 40x or 10x (rejected) When 10x , 60(10) 2010 20y  Length of HF = 10 + 20 = 30 3(i) 2 2 2 22 3 4 1 4 1 2 k k x xk xk    2012 H1 A Level Solution

3(ii) 1 2 2 22 1 2 1 222 0 1 3 22 0 33 3 3() 4 12 4 12 43 112 8 24 1 6 k k k k k x k dx k x dx xkx kk k            4(i)a)  2ln 3 2 2 ln(3 2) 32 32 6 32 d xdx d xdx x x         4(i)b) 1 2 2 4 21 4(2 1) 4( 1)(2 1) 2 8 (2 1) d dx x d xdx x x           4(ii) 2 4 2 4 2 42 2 1 1( 2 ) 2 ln2 16 4 8 ln 4 4 ln 222 ln 4 ( 2 ln 2) 2 ln 4 ln 2 42 ln 2 2 ln 2 x dx x x dx x x xx                                  

5(i) Coordinates of axis intercepts = (-0.767, 0), (2, 0), (4, 0), and (0, 1). 5(ii) When 1.5x , gradient of C = -1.0395 5(iii) When 1.5x , 0.57843y 0.57843 1.0395(1.5) 2.1377 1.0395 2.1377 c c yx       5(iv) At A: 0, 2.1377xy At B: 1.0481xy Length of AB =     22 1.0481 0 2.1377 1.0481 1.51(3 . ) sf    6(i) It is a method to obtain a sample of size n from a population of size N via systematic sampling, where we line up the population and randomly select a first member out of the first k members, where k  . Then we automatically select every kth member following this first chosen member. 6(ii) Advantage: This method is cost effective as it takes less time and effort to carry out. Disadvantage: The sample may not be representative of the town people as it may not include adults from different strata in a correct proportion. E.g: Too many men. 6(iii) Instead of standing outside the main supermarket, he can obtain a list of adults who use computers and choose 100 people from the list to do his survey. 7(i) A and B independent => ( ) ( ) ( )P A B P A P B ( ) ( ) ( ) ( ) ( )P A P B P A B P A P B    25 9p p p   2 520 9pp   (ii) 5 (rejected since 0 1)3pp   or 1 3p 2 1() 9P A B p   n N

8(i) (vote for A and male) 0.5 0.6 0.3P    (ii) (female) 0.5(0.4) 0.35(0.6) 0.15(0.8) 0.53 P     (iii) (vote for C and male)(vote for C|male) (male) 0.15 0.2 1 0.53 0.0638 PP P    9(i) (ii) 0.984r This value indicates that there is a strong negative linear correlation between the age of ‘Pluto’ cars and advertised price. The advertised price decreases at a constant rate as age of the car increases. (iii) 19.21 183.12yx  (iv) (a) when 4, 19.213(4) 183.12 106.27 xy    Advertised price = $10 600 (corrected to 3 s.f)

(b) when 9, 19.213(9) 183.12 10.203 xy    Advertised price = $1 020 (corrected to 3 s.f) (v) For (a), since |r| is close to 1 and 4x is within the data range, the estimate is reliable. For (b), 9x is outside the data range, thus the estimate is not reliable. 10(i) Let S be the r.v. “no of Sunbrite plants that will flower in one tray” ~ (12,0.8)SB ( 10) 0.283PS  (ii) ( 8) ( 7) 0.0726P S P S    (iii) Let X be the r.v. “no of Sunbrite plants that will flower in 8 trays” ~ (96,0.8)XB Since 96 50 is large , 76.8 5, 19.2 5,n np nq      ~ (76.8,15.36) approximatelyXN ( ) 76.8 ( ) 15.36 EX VX   .( 75) ( 75.5) 0.62994 0.630ccP X P X      (iv) Let Y be the r.v. “no of gardeners out of 3 with more than 75 of their plants will flower” ~ (3,0.62994)YB ( 2) 1 ( 1) 0.691 P Y P Y     11(i) ( 300) 300100 60 300100 299.4 xx     2 2 1 ( 60)(1240 )100 1 100 12.162 12.2 s    (ii) Let X be the r.v. “length and string” and  be the population mean. 0 1 : 300 : 300 H H     Under 0H , since 100 50n is large, 12.162~ (300, ) approximately by CLT100XN p-value = 0.0427 < 0.05 Therefore, reject 0H and conclude that at 5% significance level, there is sufficient evidence that the average length of string in a ball is at least 300m. (iii) Under 0H ,

12.1~ (300, ) approxim ately by CLT100XN Do not reject 0H , p-value > sign level ( ) 0.10P X k Using GC, 299.554k  Hence, the least possible value of k = 299.56 12(i) Let A be the r.v. “mass, in kg, of grapefruit of type A” Let B be the r.v. “mass, in kg, of grapefruit of type B” 1 10 ~ (2.5,0.004)A A N 1 10( 2.4) 0.0569P A A    (ii) 1 6 1 5( ) ( ) ~ ( 0.25,0.0069)D A A B B N       ( 0.2 0.2) 0.274PD    (iii) Let W be the r.v. “amount that Mrs Woo pay” 1 2 3 1 2 31.5( ) 2.4( ) ~ (3.645,0.018252)W A A A B B B N      Let T be the r.v. “amount that Mr Tan pay” 1 101.5( ) ~ (3.75,0.009) ~ ( 0.105,0.027252) T A A N W T N     ( 0) 0.262P W T  