Cambridge A Level Mathematics - Further 9231 — 2022 Oct/Nov Paper 4 · Variant 2

1 A basketball club has a large number of players. The heights, x m, of a random sample of 10 of these players are measured. A 90% confidence interval for the population mean height, n m, of players in this club is calculated. It is assumed that heights are normally distributed. The confidence interval is 1.78 G n G 2.02 . Find the values of / x and / x2 for this sample. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 2 M1 s 2.02 = x + 1.833 10 s 2 1.78 = x − 1.833 10 2.02 + 1.78 Add, x = = 1.90 Add. 2  x = 19 A1 2 M1 Allow 1.372, 1.383, 1.812 instead of 1.833. s Subtract: 2.02 − 1.78 = 2  1.833 10 s 2 = 0.042859 A1 May be implied. s = 0.20702 2 1 2 (  x ) 2 M1 But s = (  x − ) 9 10 2 2 19 2 A1 36.486  x = 9 s + = 36.5 10 Using 1.645: maximum M1A1 M0A0 M1A0 3/6 6

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2 In the colleges in three regions of a particular country, students are given individual targets to achieve. Their performance is measured against their individual target and graded as ‘above target’, ‘on target’ or ‘below target’. For a random sample of students from each of the three regions, the observed frequencies are summarised in the following table. Region A B C Total Above target 62 41 44 147 Performance On target 102 94 95 291 Below target 56 45 61 162 Total 220 180 200 600 Test, at the 10% significance level, whether performance is independent of region. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 H0: performance is independent of region B1 H1: performance is not independent of region M1 A1 Calculate expected frequencies, allow 1 error. 62 (53.9) 41 (44.1) 44 (49) 102 (106.7) 94 (87.3) 95 (97) 56 (59.4) 45 (48.6) 61 (54) Test stat = 1.217 + 0.218 + 0.510 + 0.207 + 0.514 + 0.041 + 0.195 M1 Calculate test statistic. + 0.267 + 0.907 4.08 A1 SCB2 for 4.08 if totally unsupported. SCM1A1 B1 for 4.08 supported only by expected frequencies. Tabular value, 4df = 7.779 M1 Compare with 7.779 and conclusion. 1.08 < 7.779 Accept H0 Insufficient evidence that performance is not independent of A1 Correct conclusion in context, following correct work, level of region uncertainty in language. ‘Prove’ gives A0. 7

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3 A scientist is investigating the masses of birds of a certain species in country X and country Y. She takes a random sample of 50 birds of this species from country X and a random sample of 80 birds of this species from country Y. She records their masses in kg, x and y, respectively. Her results are summarised as follows. / x = 75.5 / x 2 = 115.2 / y = 116.8 / y 2 = 172.6 n The population mean masses of these birds in countries X and Y are n x kg and y kg respectively. Test, at the 5% significance level, the null hypothesis n x = n y against the alternative hypothesis n 2 n x y . State your conclusion in the context of the question. 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Mark scheme: 3 2 1  75.52  M1 239 xs =  115.2 −  = 0.02439 49 50 9800   259 2 1  116.8 2  ys =  172.6 −  = 0.02623 9875 79  80  A1 Both correct 2 0.02439 0.02623 M1 A1 May be implied by 1.75 for z s = +  = 0.00081565 50 80 75.5 116.8 M1 − 50 80 z = s 1.75 A1 Compare with 1.645: 1.75 >1.645 M1 Reject null hypothesis Using areas, 0.04 < 0.05. Sufficient evidence to suggest that population mean in country X A1 Correct conclusion in context, following correct work, level of is greater than population mean in country Y uncertainty in language. ‘Prove’ gives A0. 8 Pooled variance: z = 1.74 M1A1 M0A0M0A0 M1A0 max 3/8

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4 The continuous random variable X has probability density function f given by k 0 G x 1 1, f ( x) = kx 1 G x G 2, * 0 otherwise, where k is a constant. (a) Show that k = 25 . 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(b) Find the interquartile range of X. 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(c) Find Var ( X) . 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Mark scheme: 4(a) 1 2 B1 Integration and k + 3k/2 = 1 oe seen.  k 2  3 2 x , k + k = 1, k = kxdx =  kx  +    kdx +  2 5  2  0 1 1 4(b)  kx 0 x  1, M1 CDF or integration with 1 x 2 + 1 seen. F ( x ) =  k 2 1 5 5 x + 1 x 2  2 5 k 2 1 M1 Method for UQ UQ: x + = 3, 2 5 4  2 3 2 11  1 A1 UQ x + 1 = 5  , x = x = 11    4 4  2 1 5 B1 LQ LQ: kx = , x = 4 8 1 5 A1 CAO IQR = 11 − = 1.03 ( 3 ) 2 8 5 4(c) 1 2 M1 With correct limits. 2  k 2   k 3  17     E(X) = kxdx + kx dx = x + x =   2   3  15 0 1 1 2 M1 With correct limits. 2 3  k 3   k 4  49     E(X2) = kx dx + kx dx = x + x =   3   4  30 0 1 2 M1 Using correct formula with numerical values. 49  17  Var(X) = −  30  15  157 A1 ( = 0.349 ) 450 4

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5 A 6‑sided dice, A, with faces numbered 1, 2, 3, 4, 5, 6 is biased so that the probability of throwing a 6 is 1.4 The random variable X is the number of 6s obtained when dice A is thrown twice. (a) Find the probability generating function of X. 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A second dice, B, with faces numbered 1, 2, 3, 4, 5, 6 is unbiased. The random variable Y is the number of 6s obtained when dice B is thrown twice. The random variable Z is the total number of 6s obtained when both dice are thrown twice. (b) Find the probability generating function of Z, expressing your answer as a polynomial. 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(c) Find Var ( Z ) . 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(d) Use the probability generating function of Z to find the most probable value of Z. 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Mark scheme: 5(a) 9 6 1 2 M1 A1 2 probabilities correct, in a polynomial GX (t) = + t + t 16 16 16 2 5(b)  9 6 1 2  25 10 1 2  M1 Second PGF correct and multiplied by part (a). GZ (t) =  + t + t  + t + t   16 16 16  36 36 36  1 2 3 4 M1 Obtains a quartic polynomial. 225 + 240t + 94t + 16t + t ( ) 576 A1 1 2 2 25 5 47 2 1 3 1 4 Note: ( t + 3 ) ( t + 5 ) scores M1M1A0. Or + t + t + t + t 576 64 12 288 36 576 3 5(c) 1 2 3 M1 Differentiate twice. GZ’ (t) = 240 + 188t + 48t + 4t ( ) 576 1 3 GZ’’ (t) = 188 + 96t + 12t ( ) 576 1 5 25 M1 Use correct formula. Var (Z) = (188 + 96 + 12 ) + − 576 6 36 47 A1 72 1 3 1 5 M1 One term correct. OR: Var(Z) = 2   + 2   M1 Two terms present and added. 4 4 6 6 47 A1 72 3 5(d) 1 B1 FT FT their power with greatest coefficient in part (b). 1

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6 The manager of a technology company A claims that his employees earn more per year than the employees at technology company B. The amounts earned per year, in hundreds of dollars, by a random sample of 12 employees from company A and an independent random sample of 12 employees from company B are shown below. Company A 461 482 374 512 415 452 502 427 398 545 612 359 Company B 454 506 491 384 361 443 401 472 414 342 355 437 (a) Carry out a Wilcoxon rank‑sum test at the 5% significance level to test whether the manager’s claim is supported by the data. 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(b) Explain whether a paired sample t‑test would be appropriate to test the manager’s claim if earnings are normally distributed. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 6(a) M1 Wrong test: max B1 for hypotheses, B1B1 for correct mean and 359 3 342 1 variance 374 5 355 2 398 7 361 4 Attempt at rankings (allow up to 4 errors) 415 10 384 6 427 11 401 8 452 14 414 9 461 16 437 12 482 18 443 13 502 20 454 15 512 22 472 17 545 23 491 19 612 24 506 21 Test statistic = 127 A1 Clearly identified H0: population medians are equal B1 H1: population median for X is greater than population median for Y 1 B1 Mean =  12  25 = 150 2 1 B1 Variance =  12  12  25 = 300 12 6(a) 127.5 − 150 M1 Allow incorrect or no continuity correction. 300 −1.299 A1 Compare with −1.645: −1.299 −1.645 , or 0.097 > 0.05 M1 Valid comparison with 1.645 or 0.05 and reach correct ft Accept H0 conclusion. Insufficient evidence to support manager’s claim A1 Correct conclusion in context, following correct work, level of uncertainty in language. ‘Prove’ is A0. 9 6(b) Not appropriate/no, not the same people B1 OE No and reason needed, e.g. individuals in the samples cannot be paired up. 1

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