Cambridge A Level Mathematics - Further 9231 — 2025 Oct/Nov Paper 4 · Variant 2

1 A large company claims that the median salary of its employees is $32 500. The salaries ($) of 15 randomly selected employees are listed below. 18 750 30 500 125 000 42 500 25 000 26 000 52 500 23 000 27 500 19 500 25 500 33 000 30 000 21 500 29 000 (a) Explain why a Wilcoxon signed-rank test may not be appropriate to test the company’s claim in this case. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Carry out a sign test at the 10% significance level to investigate the company’s claim. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) Data has outliers / is skewed / lacks symmetry, suggesting B1 Must be in context. population/distribution may not be symmetrical. Condone no explicit mention of population/distribution or a definite statement relating to population. 1 1(b) H0: population median is [$]32,500. B1 Allow m , but not . If m defined, must see population. H1: population median is not [$]32,500. Median not mean or average. Test statistic = 4 (or 11) B1 Can be implied by P( X ≤ 4) or P( X  4) . (For use of Normal approximation, the value of the test statistic can be implied by 4.5 or 3.5 for 4 and 10.5 or 11.5 for 11). [ X ~ B (15,0.5 ) ] B1 AWRT 0.0592. Might see 0.1184/0.1185/0.118/0.119 from 2P( X ≤ 4) or 2P( X ≥ 11) . P( X ≤ 4) = 0.0592 OR P( X ≥ 11) = 0.0592 Use of N(7.5,3.75) scores B1 for ±1.549. '0.0592'  0.05 , accept H0 / do not reject H0 / not significant. M1 Compare their 0.0592 (must be a tail probability) with 0.05 (or 0.1 if 1-tail test used) OE and appropriate result (may be in terms of H1). Result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Use of Normal: compare their ±1.549 with ±1.645, signs must be consistent (1.282 for 1-tail test). Insufficient evidence to reject [company’s] claim. A1 Correct conclusion from correct working ignoring hypotheses. OR In context. Insufficient evidence to suggest that median [salary] is not Level of uncertainty in language is used (for example, not ‘prove’). [$]32,500. Allow “Not enough evidence / no sufficient evidence that / to show / conclude that…”. Do not accept “No evidence…”. Do not accept statements such as “there is sufficient evidence to suggest…”. 5

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2 A factory produces packets of biscuits. The total mass of biscuits in a packet has a normal distribution with mean n. A random sample of 12 packets is taken and the mass of the contents of each packet, x g, is recorded. The results are summarised as follows. / x = 2390 / x 2 = 476117 (a) Find a 99% confidence interval for n. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ A test of the null hypothesis n = k is carried out on this sample using a 5% significance level. The test does not support the alternative hypothesis n 1 k . (b) Find the greatest possible value of k. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a)   = 32633 = 9.8788  M1 Correct expression, implied by AWRT 9.88. s 2 = 111 ( '476117'− '2390'12 2 ) 3.106 B1 3.106 or 3.11 seen. '2390' 12  '3.106' '9.8788'12 M1 Correct form, must be a t -value. 196, 202  A1 Accept with inequality signs or open brackets. Condone [202, 196]. Do not accept 199 .3 4 2(b) −1.796 B1 1.796 or  1.80 seen. '199.17'− k '199.17'− k M1 Correct form, must be a t-value, consistent signs. −' 1.796' or  '1.796' Condone non-strict inequality. '9.8788' '9.8788' 12 12 200.8 A1 Accept 200.7, 200 or 201 from correct working. Condone final answer of the form k < 200.8 OE. Condone non-strict inequalities. 3

More questions on Inference using normal and t-distributions

3 A traffic expert claims that the number of breakdowns occurring each day on a busy section of a motorway follows a Poisson distribution with mean 0.7. The number of breakdowns each day over a 200-day period was recorded. The following table contains the observed frequencies together with some of the expected frequencies using the expert’s distribution. Number of breakdowns per day 0 1 2 3 4 H 5 Observed frequency 88 73 26 7 3 3 Expected frequency 99.317 m 24.333 5.678 0.994 n (a) Find the value of m and the value of n, correct to 3 decimal places. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Carry out a goodness of fit test at the 5% significance level to investigate the expert’s claim. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) m = 69.522 B1 n = 0.156 or 0.157 B1 Subtraction gives 0.156. 2 3(b) H0: Po ( 0.7 ) fits the data. B1 Must mention distribution and data/number of breakdowns. May see H0: Po ( 0.7 ) is a satisfactory model for the data. H1: Po ( 0.7 ) does not fit the data. H1: Po ( 0.7 ) is not a satisfactory model for the data. Combining last 3 columns give 13 and 6.828 or 6.829. M1 Combine expected frequencies less than 5. ( 88 − 99.317 ) 2 ( 73 − 69.552 ) 2 (13 − 6.829 ) 2 M1 At least 2 expressions (not including 200) of correct form seen. + ++ May see individual terms not summed. 99.317 69.552 6.829 May be implied by 2 contributions correct to 3sf seen. [1.2896 + 0.1740 + 0.1142 + 5.5764] 6.828 gives 5.5790. OR 882 732 26 2 132 + + + − 200 99.317 69.552 24.333 6.829 7.15 or 7.16 A1 SC B1 for 7.15 or 7.16 following M0. '7.15'  7.815 , accept H0 / do not reject H0 / not significant. M1 Compare their 7.15 with 7.815 (9.488 if 2 columns combined, 11.07 if no columns combined) and appropriate result (may be in terms of H1). Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Insufficient evidence to reject expert’s claim. A1 Correct conclusion from correct working ignoring hypotheses. OR In context. Insufficient evidence to suggest that Po ( 0.7 ) is not a good Level of uncertainty in language is used (for example, not ‘prove’) Allow ‘Not enough evidence that / to show / conclude that…’. fit/model [for the data]. Do not accept statements such as ‘there is sufficient evidence to suggest…’ or ‘no evidence…’. 6

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4 A continuous random variable X has cumulative distribution function F given by Z ]0 x 1 1 , ] 1 x + a 1 G x 1 4, F ( x) = [ 51 2 x + b 4 G x G 6, 50 ]] 1 x 2 6, \ where a and b are constants. (a) Find the value of a and the value of b. 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(b) Find the probability density function of X. 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(c) Given that E ( X ) = 529 , find Var (X ). 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(d) Find the 10th and 90th percentiles of X. 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Mark scheme: 4(a) a = − 15 B1 b = 257 B1 2 ≤ x  4, M1 Attempt at differentiating both parts.4(b)  15 1  1 ≤ x ≤ 6, f ( x ) =  25 x 4 A1 All correct, fully defined including domain. 0 otherwise. Domain must cover all reals, must not overlap at boundaries.  2 4 64(c) M1 Attempt to integrate x 2f ( x ) with correct limits. E X 2 x 3 dx =  1 x 3  +  1 x 4  ( )  15  100  1 4 = 15 14 x 2 dx + 251 46 Can be implied by answer of 735 or 14.6. 73 529 2 M1 2 . Variance formula used with their E X ( ) Var ( X ) = 5 − ( 150 ) 2.16 A1 3 4(d) 10th percentile: 15 x − 15 = 101 M1 Award M1 for a correct method to find at least one of the 10th or 90th percentile. 1 2 7 9 90th percentile: 50 x + 25 = 10 FT their a and b. 10th percentile = 1.5 A1 90th percentile = 31 A1 AWRT 5.57. 3

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5 An engineer is comparing the tensile strengths of steel rods made from two machines, A and B. The engineer randomly selects 8 rods from machine A and 6 rods from machine B. The tensile strengths, in appropriate units, are given in the following table. Machine A 402 403 415 412 409 407 406 410 Machine B 401 398 395 397 410 405 You should assume that the two distributions are normal and have the same population variance. Use a t-test at the 5% significance level to test whether there is any difference in the mean tensile strengths of steel rods from the two machines. 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Mark scheme: 5 H 0:A = B H1:A  B B1 If in words, must have population means. Allow H0: =d 0 , H1: d 0 . '1331848'− '3264'8 2 [ = 136] M1 OE, may be implied by s A2 = 19.43 . Biased estimate of 17 earns M0 unless recovered. 31.6 . '964964'− '2406'6 2 [ = 158] M1 OE, may be implied by Bs2 = Biased estimate of 26 earns M0 unless recovered. 2 '136'+ '158' M1 Correct form using their values, may be embedded. ps = = 24.5 8 + 6 − 2 A1 OE, for example 492 . '3264' 8 − '2406'6 M1 Correct form using their values. t = = 2.62 ' s p ' 18 + 16 A1 AWRT 2.62. '2.62'  2.179, reject H0 / significant. M1 Compare their 2.62 with 2.179 and appropriate result (may be in terms of H1). Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Sufficient evidence to suggest that there is a difference in the A1 Correct conclusion from correct working ignoring hypotheses. mean tensile strengths of steel rods produced by the two In context. machines. Level of uncertainty in language is used (for example, not ‘prove’). Allow “(Some/enough) evidence that / to show / conclude that…”. Do not accept statements such as “there is insufficient evidence to suggest…”. 9

More questions on Inference using normal and t-distributions

6 The discrete random variable X has probability generating function GX ()t given by t G ()t = . X ( 3 - 2t) 2 (a) Find E ( X ) and Var (X ). 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The discrete random variable Y has probability generating function GY ()t given by t 2 G ()t = . Y ( 3 - 2t) 2 The random variable Z is the sum of the random variables X and Y. (b) Assuming X and Y are independent, find P ( Z 2 4 ) . 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Mark scheme: 6(a) X ( t ) . ' ( 3 − 2t ) 2 + 4 ( 3 − 2t ) t  3 + 2t  M1 Attempt to differentiate G G X ( t ) = 4 = 3  ' 1) . ( 3 − 2t )  ( 3 − 2t )  May be implied by expressions for G X ( ' A1 E ( X ) = G X (1) = 5 Attempt to differentiate GX' ( t ) . '' 2 ( 3 − 2t ) 3 + 6 ( 3 + 2t )( 3 − 2t ) 2  24 + 8t  M1 G X ( t ) = 6 = 4  '' G X (1) . ( 3 − 2t )  ( 3 − 2t )  May be implied by expressions for '' ' ' 2 M1 Use correct formula using their values. = 32 + 5 − 25 Var ( X ) = G X (1) + G X (1) − ( G X (1) )  12 A1 CWO  Var ( X ) = 5 6(b) t 3 1 3 2 M1 Multiply and attempt at binomial expansion. t ( 1 + ( −4 )( − 3 t ) +) G Z ( t ) = 4 = 81 ( 3 − 2t ) P ( Z = 4 ) = their coefficient of 4t . M1 Attempt at finding their coefficient of 3t or 4t . Either seen earns M1. OR P ( Z = 3 ) = their coefficient of 3t . 1 A1 Both values seen.  8 and   P ( Z = 4 ) = 243  P ( Z = 3) = 81 P ( Z  4 ) = 1 − ' 243 8 '− ' 811 ' M1 Attempt to find P ( Z  4 ) using their values. 232 A1 AWRT 0.955. P ( Z  4 ) = 243 Note: considering the expansions of the probability generating functions for X and Y separately, then finding P ( Z  4 ) = 1 − P ( Z ≤ 4 ) with P ( Z ≤ 4 ) = P ( X = 1, Y = 2 ) + P ( X = 1, Y = 3 ) + P ( X = 2, Y = 2 ) earns full credit. 5

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