Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 4 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/42 Paper 4 Further Probability & Statistics October/November 2024 1 hour 30 minutes You must answer on the question paper. You will need: List of formulae (MF19) INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● If additional space is needed, you should use the lined page at the end of this booklet; the question number or numbers must be clearly shown. ● You should use a calculator where appropriate. ● You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. INFORMATION ● The total mark for this paper is 50. ● The number of marks for each question or part question is shown in brackets [ ]. * 3 9 1 3 0 0 3 9 6 3 * DC (PQ) 337156/3 © UCLES 2024 ,










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3 9231/42/O/N/24 © UCLES 2024 [Turn over 1 A scientist is investigating the lengths of the leaves of a certain type of plant. The scientist assumes that the lengths of the leaves of this type of plant are normally distributed. He measures the lengths, x cm, of the leaves of a random sample of 8 plants of this type. His results are as follows. 3.5 4.2 3.8 5.2 2.9 3.7 4.1 3.2 Find a 90% confidence interval for the population mean length of leaves of this type of plant. [4] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000003 * ,










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4 9231/42/O/N/24 © UCLES 2024 2 The random variable X has probability generating function ( )t GX given by ( )t pt qt GX 5 1 2 = + + , where p and q are constants. (a) Given that ( ) . X 1 1 E = , find the numerical value of Var(X ). [4] … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000004 * ,










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5 9231/42/O/N/24 © UCLES 2024 [Turn over The random variable Y has probability generating function ( )t GY given by ( )t t t 1 GY 3 2 2 1 2 = + b l. The random variable Z is the sum of independent observations of X and Y. (b) Find the probability generating function of Z. [2] … … … … … … … … … … … … (c) Find ( ) Z 2 P 2 . [1] … … … … (d) State the most probable value of Z. [1] … … … … … … * 0000800000005 * ,










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6 9231/42/O/N/24 © UCLES 2024 3 Rosie sows 5 seeds in each of 150 plant pots. The number of seeds that germinate is recorded for each pot. The results are summarised in the following table. Number of seeds that germinate 0 1 2 3 4 5 Number of pots 12 40 43 35 16 4 Rosie suggests that the number of seeds that germinate follows the binomial distribution B(5, p). (a) Use Rosie’s results to show that . p 0 42 = . [1] … … … … … … … … … (b) Carry out a goodness of fit test, at the 10% significance level, to test whether the distribution B(5, 0.42) is a good fit for the data. [9] … … … … … … … … … … … * 0000800000006 * ,










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8 9231/42/O/N/24 © UCLES 2024 4 The random variable X has probability density function f given by ( ) ( ) , . x x x 1 0 2 5 f otherwise 21 1 2 G G = - * (a) Find the cumulative distribution function of X. [3] … … … … … … … … … … … … … The random variable Y is defined by ( ) Y X 1 4 = - . (b) Find the probability density function of Y. [3] … … … … … … … … … * 0000800000008 * ,










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9 9231/42/O/N/24 © UCLES 2024 [Turn over (c) Find the median value of Y. [2] … … … … … … … … … … … … … (d) Find E(Y ). [2] … … … … … … … … … … … … … * 0000800000009 * ,










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10 9231/42/O/N/24 © UCLES 2024 5 Dev owns a small company which produces bottles of juice. He uses two machines, X and Y, to fill empty bottles with juice. Dev is investigating the volumes of juice in the bottles. He chooses a random sample of 35 bottles filled by machine X and a random sample of 60 bottles filled by machine Y. The volumes of juice, x and y respectively, measured in suitable units, are summarised by . , . , . , . . x x y y 30 8 29 0 62 4 76 8 2 2 = = = = / / / / Dev claims that the mean volume of juice in bottles filled by machine Y is greater than the mean volume of juice in bottles filled by machine X. A test at the % a significance level suggests that there is sufficient evidence to support Dev’s claim. Find the set of possible values of a. [9] … … … … … … … … … … … … … … … … … … … … … … * 0000800000010 * ,










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12 9231/42/O/N/24 © UCLES 2024 6 A sports college keeps records of the times taken by students to run one lap of a running track. The population median time taken is 51.0 seconds. After a month of intensive training, a random sample of 22 new students run one lap of the track, giving times, in seconds, as follows. 51.3 52.0 53.4 49.2 49.3 51.1 52.2 47.2 53.0 48.5 49.4 50.3 50.8 51.6 49.1 52.3 51.8 52.4 47.9 48.9 50.6 51.9 It is claimed that the intensive training has led to a decrease in the median time taken to run one lap of the track. Carry out a Wilcoxon signed-rank test, at the 5% significance level, to test whether there is sufficient evidence to support the claim. [9] … … … … … … … … … … … … … … … … … … … … * 0000800000012 * ,









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14 9231/42/O/N/24 © UCLES 2024 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000014 * ,









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16 9231/42/O/N/24 © UCLES 2024 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. BLANK PAGE * 0000800000016 * ,









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This document consists of 13 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/42 Paper 4 Further Probability & Statistics October/November 2024 MARK SCHEME Maximum Mark: 50 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 13 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 13 Mathematics Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, non-integer answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5 Where a candidate has misread a number or sign in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 A or B mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 13 Mark Scheme Notes The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. DM or DB When a part of a question has two or more ‘method’ steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly, when there are several B marks allocated. The notation DM or DB is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. FT Implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. • A or B marks are given for correct work only (not for results obtained from incorrect working) unless follow through is allowed (see abbreviation FT above). • For a numerical answer, allow the A or B mark if the answer is correct to 3 significant figures or would be correct to 3 significant figures if rounded (1 decimal place for angles in degrees). • The total number of marks available for each question is shown at the bottom of the Marks column. • Wrong or missing units in an answer should not result in loss of marks unless the guidance indicates otherwise. • Square brackets [ ] around text or numbers show extra information not needed for the mark to be awarded.

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 13 Abbreviations AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed) CWO Correct Working Only ISW Ignore Subsequent Working SOI Seen Or Implied SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) WWW Without Wrong Working AWRT Answer Which Rounds To

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 13 Question Answer Marks Guidance 1 2 30.6 120.52  =  = x x   2 2 1 30.6 120.52 0.49643 7 8   = − =     s M1 With their 2 139 , , 280   x x . CI = 2 30.6 1.895 8 8  s M1 Use correct formula with a t value. B1 1.895 seen.   3.35, 4.30 A1 Accept with inequality signs or open brackets. . 4 Question Answer Marks Guidance 2(a) [GX(t) = 2 1 , 5 + + pt qt G’X(t) = 2 , ] 2 1.1 + + = p qt p q B1 1 4 1, 5 5 + + = + = p q p q and solve M1 1 3 , 2 10   = =     p q A1 G’’X(t) = 2q = 3 5 , Var(X) = 2 0.6 1.1 1.1 0.49 + − = A1 4

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 13 Question Answer Marks Guidance 2(b) GZ(t) = 2 2 1 1 3 2 1 1 5 2 10 3 2     + +  +         t t t t M1 Wrong p and q (or swapped) or missing t, M1A0. p and q must have their numerical values from part (a). ( ) 2 3 4 5 1 4 10 8 5 3 30 + + + + t t t t t A1 Accept answer in any equivalent form. 2 2(c) 16 8 30 15 = oe B1 FT FT their fully expanded form of PGF. 1 2(d) (Z =) 2 B1 Correct work only, must see correct fully expanded polynomial form of PGF. 1 Question Answer Marks Guidance 3(a) 40 86 105 64 20 315 2.1, 150 150 + + + + = = = x 2.1 0.42 5 = = p B1 Must see either 315 or 2.1. AG 1

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 13 Question Answer Marks Guidance 3(b) Number of seeds that germinate 0 1 2 3 4 5 Number of pots 12 40 43 35 16 4 Expected frequency 9.846 35.6475 51.627 37.3845 13.536 1.9605 B1 Calculate expected frequencies (must be seen) at least 2 correct to at least 2 decimal places. B1 At least 4 correct to at least 2 decimal places. Combine last two columns M1 20, 15.50, may be implied by answer 3.90 – 3.91. Chi-squared contributions: 0.4712 0.5314 1.4416 0.1521 1.3088 M1 At least 2 correct, may be implied by answer 3.90 – 3.91. Test statistic = 3.905 A1 accept 3.90 – 3.910 H0: Binomial B(5, 0.42) fits the data H1: Binomial B(5, 0.42) does not fit the data B1 Allow ‘Binomial’ for ‘B(5, 0.42)’ Allow ‘Number of seeds that germinate can be modelled by B(5, 0.42)’ Critical value is 6.251 B1 Must come from combined columns. Allow 7.779. ‘3.905’ < ‘6.251’ Accept H0 M1 Reject H1, not significant. Insufficient evidence to suggest that B(5, 0.42) is a not a good fit (to the data) A1 Correct work only, including hypotheses, level of uncertainty in language. 9

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 13 Question Answer Marks Guidance 4(a) ( ) 3 0 2, 1 F( ) ( 1 1 2 5, 63 1 5. x X x x x       = − −        ≤≤ M1 Integrate ( )  3 2 1 3 3 63 − + + x x x c oe. A1 Correct F(x) (with c 2 63 = − evaluated). A1 0 and 1 correct, inequalities correct. 3 4(b) G(y) = ( ) 0.75 1 1 63 − y M1 Use transformation correctly on function . May see ( ) ( ) ( ) 3 2 0.25 0.25 0.25 1 1 1 2 1 1 1 63 21 21 63 + − + + + − y y y 0.25 1 1 256, 84 g( ) 0 otherwise. y y y −   =    ≤ ≤ M1 Differentiate and change variable in domain. A1 All correct and simplified. 3 4(c) ( ) 0.75 1 1 0.5 63 − = y M1 Equate their G(y) to 0.5 and attempt to solve. y = 104 A1 AWRT 104. 2

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 13 Question Answer Marks Guidance 4(d) ( ) ( ) 256 256 0.25 0.75 1.75 1 1 1 1 1 4 84 84 84 7 −   = =        y y dy y dy y M1 Use their ( ) g , y y integrate, ignore limits. 111 A1 5461 49 2

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 11 of 13 Question Answer Marks Guidance 5   2 2 1 30.8 29.0 0.055765 34 35   = − =     xs   2 2 1 62.4 76.8 0.201763 59 60   = − =     ys B1 237 4250 1488 7375 Allow B1 for correct expression for pooled variance.   2 0.055765 0.201763 0.004956 35 60 = + = s M1 A1 Pooled variance M0A0. H0:   = x y H1:    x y B1 30.8 62.4 35 60 − = z s 0.25 2.27 0.004956 − =− M1 A1 Pooled variance M0A0. Corresponding probability is 0.9885 M1 FT their z . One tail: ( ) 100 1 '0.9885' 1.15  − = M1 Allow for 1 – ‘0.9885’. 1.15  A1 Allow ⩾. 9

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 12 of 13 Question Answer Marks Guidance 6 Signed differences from 51.0 0.3 1.0 2.4 - 1.8 - 1.7 0.1 1.2 - 3.8 2.0 - 2.5 - 1.6 - 0.7 - 0.2 0.6 - 1.9 1.3 0.8 1.4 - 3.1 - 2.1 - 0.4 0.9 M1 Allow up to 2 errors. 3 9 19 -15 -14 1 10 -22 17 -20 -13 -6 -2 5 -16 11 7 12 -21 -18 -4 8 M1 Attempt at ranking of signed differences from 51.0. Smaller sum is 102 A1 151 H0: population median is 51.0 (seconds) H1: population median is less than 51.0 (seconds) B1 Allow m. Normal approximation: mean =126.5 variance = 948.75 B1 Both. '102' 0.5 126.5 948.75 + − M1 Allow no continuity correction. 0.779 − A1 Accept 0.779. Accept p-value = 0.218. (A0 for z = – 0.795 or p = 0.213 from no continuity correction). ' ' 0.779 1.645 − − accept H0 M1 Or ‘0.218’ > 0.05 or ‘0.782’ < 0.95

9231/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 13 of 13 Question Answer Marks Guidance 6 Insufficient evidence to suggest that (population) median is less than 51.0 seconds/ Insufficient evidence to support the claim A1 Correct work only, except possibly hypotheses, in context, level of uncertainty in language. Alternative approach using critical values. 9