RI 8865 2022 Prelim Stats Solutions vetted

H1 Mathematics 2022 Year 6 Preliminary Examination (Solution) Page 1 of 7 Question 6 No. Solution (i) Let  and  denote the mean and standard deviation of X. P (X  0) = 0.06  P  Z  0 −   = 0.06 where ( )~ 0,1ZN  0 −   = −1.55477   − 1.55477 = 0 --- (1) P (X  15) = 0.029  P  Z  15 −   = 0.029 where ( )~ 0,1ZN  15 −   = 1.89570   + 1.89570 = 15 --- (2) Solving (1) and (2),  = 6.76 and  = 4.35 (3 sf ) (ii) 6.76 8 −8

H1 Mathematics 2022 Year 6 Preliminary Examination (Solution) Page 2 of 7 Question 7 No. Solution (i) P (AB) = 1 6  P (A  B) P (B) = 1 6  P (B) = 6 k P ( A  B ) = P ( A ) + P ( B ) − P ( A  B )  9 20 = 1 5 + 6 k − k  k = 1 20 (ii) P (A'  B) = P (A  B) − P ( A ) = 9 20 − 1 5 = 1 4 (iii) P (A' )  P ( B ) = 4 5  6 k = 6 25 Since P (A'  B)  P (A' )  P ( B ), the events A' and B are not independent. Question 8 No. Solution (i) Let X denote the number of defective lamps in a sample of 30 lamps X ~ B (30, 0.05) The expected number of defective lamps is E(X ) = 30  0.05 = 1.5 (ii) P (X  1) = 1 − P (X = 0) = 0.785 (3 sf ) (iii) Required probability = P (X  2X  1) = ( ) ( ) P 1 2 P1 X X   = P (X = 1) + P (X = 2) P (X  1) = 0.59754 0.78536 = 0.761 (3 sf )

H1 Mathematics 2022 Year 6 Preliminary Examination (Solution) Page 3 of 7 Question 9 No. Solution (i) (ii) From GC, r = 0.945 (3 sf ) r is close to 1. There is a strong positive linear correlation between the number of years, x, of experience of the prosecutors and the percentage, y, of their cases that ended in guilty pleas. As the number of years of experience of the prosecutors increases, the percentage of their cases that ended in guilty pleas increases. (iii) From GC, y = 1.77 x + 70.4 (3 sf ) The percentage of a prosecutor’s cases that ended in guilty pleas is expected to increase by 1.77 % when the number of years of experience of the prosecutor increases by 1. (iv) From GC, when x = 20, y = 106 (3 sf ) An estimate for the percentage of cases that ended in guilty pleas for a prosecutor with 20 years of experience is 106 %. The estimate is not reliable because (1) x = 20 is outside the data range of 1  x  15 and (2) the estimated percentage of 106 does not make sense since it exceeds 100 %. x y 15 1 70 95

H1 Mathematics 2022 Year 6 Preliminary Examination (Solution) Page 4 of 7 Question 10 No. Solution (i) Let S (in grams) denote the mass of fries in a small packet. S ~ N (80, 2 2) P (S  77) = 0.066807 Required probability = [ P (S  77) ] 2 = 0.00446 (3 sf ) (ii) Let L (in grams) denote the mass of fries in a large packet. L ~ N (150, 3 2) Let X = L 1 + L 2 ~ N (150  2, 32  2), i.e. N (300, 18) Let Y = S 1 +  + S 4 ~ N (80  4, 22  4), i.e. N (320, 16) Y − X ~ N (320 − 300, 18 + 16), i.e. N (20, 34) P (Y − X  15) = 0.80441 = 0.804 (3 sf ) (iii) 1.72S ~ N (80  1.72, 22  1.722), i.e. N (137.6, 11.8336) 0.86L ~ N (150  0.86, 32  0.862), i.e. N (129, 6.6564) 1.72S − 0.86L ~ N (137.6 − 129, 11.8336 + 6.6564), i.e. N (8.6, 18.49) P (1.72S − 0.86L  k)  0.7 From GC, k  6.35  The largest integer value of k is 6. Alternatively, when k = 5, probability = 0.79876  0.7, when k = 6, probability = 0.72729  0.7, when k = 7, probability = 0.64509  0.7.  The largest integer value of k is 6.

H1 Mathematics 2022 Year 6 Preliminary Examination (Solution) Page 5 of 7 Question 11 No. Solution (i) Number of ways = 5P4  8! = 4838400 (ii) Required probability = 4P3  9! 12! = 1 55 (iii) Required probability = 3!  2  9! 12! = 1 110 (iv) Number of ways for the team to consist of 2 pupils who play sports = 3C2  9C3 = 252 Number of ways for the team to consist of 3 pupils who play sports = 3C3  9C2 = 36  Required probability = 252 + 36 12C5 = 4 11 (v) Let A denote the event that the team consists of at least 2 pupils who play sports , and let B denote the event that the team consists of all boys. P (A) = 4 11 P (B) = 7C5 12C5 = 7 264 P (A  B) = 5C3  2C2 12C5 = 5 396  Required probability = P (A  B) = P (A) + P (B) − P (A  B) = 299 792

H1 Mathematics 2022 Year 6 Preliminary Examination (Solution) Page 6 of 7 Question 12 No. Solution (i) Unbiased estimate of the population mean, x _ = 3290 60 = 329 6 or 54.8 (3 sf ) Unbiased estimate of the population variance, s 2 = 1 59  201100 − (3290) 2 60 = 350.82 = 351 (3 sf ) (ii) A 1-tail test should be carried out as the investigation is to find out if the minimum recommended time is met, and thus we are only concerned if the mean time spent is less than 60 minutes. (iii) Let  denote the population mean time spent. Test H 0 :  = 60 against H 1 :   60 at the 5 % significance level Under H 0 , since n is large, by Central Limit Theorem, X _  N  60 350.82 60 approximately Using a 1-tail test, x _ = 3290 60 gives p-value = 0.0163 (3 sf ) Since p-value = 0.0 163  0.05, we reject H0 and conclude that there is sufficient evidence at the 5 % significance level that the minimum recommended time of 60 minutes is not met. (iv) Test H 0 :  = 60 against H 1 :   60 at the 5 % significance level Under H 0 , since n is large, by Central Limit Theorem, X _  N  60 152 60 approximately ( )2 60 ~ N 0,1 15 60 XZ −=

H1 Mathematics 2022 Year 6 Preliminary Examination (Solution) Page 7 of 7 2 60 15 60 cal kz −= Using a two tailed test, reject H 0 when 1.96 or 1.96cal calzz − − Since H 0 is rejected, 1.96 or 1.96cal calzz − − 22 60 60 1.96 1.96 15 15 60 60 kk or−− − − Hence k  56.2 or k  63.8 (3 sf )