Cambridge A Level Mathematics - Further 9231 — 2025 May/June Paper 4 · Variant 3

This document has 12 pages. [Turn over * 0 7 2 1 9 3 2 2 0 2 * Cambridge International AS & A Level FURTHER MATHEMATICS 9231/43 Paper 4 Further Probability & Statistics May/June 2025 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 [ ]. DC (DE) 347145/2 © UCLES 2025 ,










, * 0000800000001 * ¬Wz> 4mHuOªEŠ^{6€W ¬‰UsS¢”—inƒcy0‰w‚ ¥•UUu•¥•e•¥ 5 •5U

2 9231/43/M/J/25 © UCLES 2025 1 A person’s eye colour may be categorised as “brown”, “blue” or “other”. A scientist claims that these eye colours are uniformly distributed and hence are equally likely to occur in the population. A survey of 120 people from this population found that 38 people had brown eyes, 52 people had blue eyes and 30 people had eyes which were neither brown nor blue. Use the data to carry out a goodness of fit test at the 5% significance level to test the scientist’s claim. [6] … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000002 * ,










, ĬÕú¾Ġ´íÈõÏĪÅĊßù¸þ× Ĭĉ×ôÒĦĘñÒăõÀÁÆݱïĂ ĥÅÅÕµõåõĕÕµÅąõĥÕĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

3 9231/43/M/J/25 © UCLES 2025 [Turn over 2 A farmer is investigating whether using a new fertiliser will increase the yield of tomato plants. The farmer selects 40 tomato plants at random and gives them the new fertiliser. The crop mass, x kg, of each of these 40 plants is recorded. The farmer selects a further 60 tomato plants at random and gives them a standard fertiliser. The crop mass, y kg, of each of these 60 plants is recorded. The results are summarised as follows. / x 168 = / x 720 2 = / y 228 = / y 900 2 = Find a 90% confidence interval for the difference in mean crop mass associated with each type of fertiliser. [7] … … … … … … … … … … … … … … … … … … … … … … … * 0000800000003 * ,










, Ĭ×ú¾Ġ´íÈõÏĪÅĊßû¸þ× ĬĉØóÚĤĜāçõČĉĕ®ù±ÿĂ ĥŵĕõĕÅĕąåĥÅąĕąĕąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

4 9231/43/M/J/25 © UCLES 2025 3 A continuous random variable X has probability density function f given by ( ) , , ( ) f x x x kx k x 0 0 8 1 1 8 otherwise, 1 G G G = - Z [ \ ] ] ]] where k is a constant. (a) Show that k 25 1 = . [2] … … … … … … (b) Find the median value of X. [3] … … … … … … … … … … … … … … … * 0000800000004 * ,










, ĬÕú¾Ġ´íÈõÏĪÅĊÝù¸Ā× ĬĉØòÚĪĪĈÔûăĂ·ĒÛá÷Ă ĥõĥĕµĕŵĥąĕÅÅĕåĕĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

5 9231/43/M/J/25 © UCLES 2025 [Turn over The random variable Y is defined by Y X 3 = . (c) Find the probability density function of Y. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000005 * ,










, Ĭ×ú¾Ġ´íÈõÏĪÅĊÝû¸Ā× Ĭĉ×ñÒĠĦøåýîÇģĪÿáćĂ ĥõĕÕõõåÕõõÅÅÅõÅÕąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

6 9231/43/M/J/25 © UCLES 2025 4 A researcher claims that older people take longer to react to a sudden loud noise than younger people. To investigate this, the researcher randomly selects 6 people over 50 years old and 8 people under 25 years old and records their reaction times, in milliseconds, to a sudden loud noise. The reaction times are as follows. Over 50 198 212 217 229 235 242 Under 25 178 181 183 192 203 209 223 231 Carry out a Wilcoxon rank-sum test at the 5% significance level to test the researcher’s claim. [8] … … … … … … … … … … … … … … … … … … … … … … … * 0000800000006 * ,










, ĬÙú¾Ġ´íÈõÏĪÅĊàûµþ× ĬĉØñÍĦþîÝĊÿ®ăĒĞĉ÷Ă ĥåµÕõÕåõµĕĥÅąõĥĕÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

7 9231/43/M/J/25 © UCLES 2025 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000007 * ,










, ĬÛú¾Ġ´íÈõÏĪÅĊàùµþ× Ĭĉ×òÕĤĂþÜðòû×ĪºĉćĂ ĥåÅĕµµÅĕåĥµÅąĕąÕÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

8 9231/43/M/J/25 © UCLES 2025 5 A doctor is investigating the concentration of blood glucose in patients at risk of developing type 2 diabetes, where blood glucose is measured in appropriate units. The doctor claims that a particular intervention reduces the concentration by more than k units on average. A group of 8 at risk patients is selected at random and each patient follows the intervention for six months. The blood glucose concentrations before and after the intervention are given in the following table. Patient A B C D E F G H Before 183 165 172 165 143 176 161 153 After 164 148 164 149 134 153 155 148 (a) Use a t-test at the 5% significance level to find the range of values of k for which the result of the test is to reject the null hypothesis. [7] … … … … … … … … … … … … … … … … … … … * 0000800000008 * ,










, ĬÙú¾Ġ´íÈõÏĪÅĊÞûµĀ× Ĭĉ×óÕĪôċßòùôõÆĜęïĂ ĥĕĕĕõµÅµÅÅÅÅÅĕåÕÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

9 9231/43/M/J/25 © UCLES 2025 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … (b) State an assumption necessary for the test in part (a) to be valid. [1] … … … * 0000800000009 * ,










, ĬÛú¾Ġ´íÈõÏĪÅĊÞùµĀ× ĬĉØôÍĠðûÚĈĈµá®ÀęÿĂ ĥĕĥÕµÕåÕÕµĕÅÅõÅĕÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

10 9231/43/M/J/25 © UCLES 2025 6 A bag contains 7 red balls and 3 blue balls. Kieran selects 2 balls at random, without replacement. The number of red balls selected by Kieran is denoted by X, and the number of different colours present in Kieran’s selection is denoted by Y. (a) Find the probability generating functions, ( )t GX of X and ( )t GY of Y. [4] … … … … … … … … … … … … … … … … … … … The random variable Z is the sum of the number of red balls and the number of different colours present in Kieran’s selection. Kieran claims that the probability generating function of Z is equal to ( ) ( ) t t G G X Y # . (b) Explain why Kieran is incorrect. [1] … … … * 0000800000010 * ,










, ĬÙú¾Ġ´íÈõÏĪÅĊßû·þ× ĬĉÖóÐĬüčáùòÚęÈðáćĂ ĥÅõÕõµĥµĥõÅąÅµåÕĥÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

11 9231/43/M/J/25 © UCLES 2025 (c) Find the probability generating function of Z, expressing your answer as a polynomial in t. [4] … … … … … … … … … … … … … … … … … (d) Use the probability generating function of Z to find E(Z ). [2] … … … … … … … … … * 0000800000011 * ,









, ĬÛú¾Ġ´íÈõÏĪÅĊßù·þ× ĬĉÕôØĞøĝØÿÿď½°ìá÷Ă ĥÅąĕµÕąÕõąĕąÅÕÅĕõÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

12 9231/43/M/J/25 © UCLES 2025 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. Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … * 0000800000012 * ,









, ĬÙú¾Ġ´íÈõÏĪÅĊÝû·Ā× ĬĉÕñØĨĆĬãāĈĘğĔĊ±ÿĂ ĥõÕĕõÕąõĕåĥąąÕĥĕĥÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

This document consists of 16 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/43 Paper 4 Further Probability & Statistics May/June 2025 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 May/June 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 16 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.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 16 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 or sign 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.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 16 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning More information required Accuracy mark awarded zero Accuracy mark awarded one Independent accuracy mark awarded zero Independent accuracy mark awarded one Independent accuracy mark awarded two Benefit of the doubt Blank Page Incorrect Dep Used to indicate DM0 or DM1

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 16 Annotation Meaning DM1 Dependent on the previous M1 mark(s) Follow through Indicate working that is right or wrong Highlighter Highlight a key point in the working Ignore subsequent work Judgement Judgement Method mark awarded zero Method mark awarded one Method mark awarded two Misread Omission or Other solution Off-page comment Allows comments to be entered at the bottom of the RM marking window and then displayed when the associated question item is navigated to. On-page comment Allows comments to be entered in speech bubbles on the candidate response. Judgment made by the PE Premature approximation Special case

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 16 Annotation Meaning Indicates that work/page has been seen Error in number of significant figures Correct Transcription error Correct answer from incorrect working

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 16 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.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 16 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

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 16 Question Answer Marks Guidance 1 H0: uniform distribution fits the data. H1: uniform distribution does not fit the data. B1 Must mention data and uniform distribution (or uniform distribution clearly described). Expected frequencies are 40, 40, 40. B1 ( ) ( ) ( )   2 2 2 38 40 52 40 30 40 0.1 3.6 2.5 40 40 40 − − − + + = + + OR 2 2 2 38 52 30 120 40 40 40 + + − M1 Expression of correct form. May see individual terms not summed. 6.2 A1 SC unsupported 6.2 scores M0 B1. Note: 6.208 scores M0 A0 XP. [2 degrees of freedom, so critical value is 5.991.] '6.2' 5.991  , reject H0 (significant). M1 Compare their 6.2 with 5.991 and appropriate conclusion (may be in terms of H1). Reject H0 can be implied by an attempt at an appropriate conclusion in context. Sufficient evidence to suggest that the scientist’s claim is incorrect. OR Sufficient evidence eye colour is not uniformly distributed. OR Sufficient evidence to suggest that uniform distribution is not a good fit. A1 Correct conclusion in context from correct working ignoring hypotheses. Level of uncertainty in language is used (for example, not ‘prove’). 6

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 16 Question Answer Marks Guidance 2 ( ) 2 2 168 1 24 39 40 65 720 0.36923 xs   = − = =   ( ) 2 2 228 168 1 59 60 295 900 0.56949 ys   = − = =   M1 Correct method to find one unbiased estimate, may be un-simplified. A1 Both unbiased estimates correct to 3sf or better. 2 '0.36923' '0.56949' 40 60 s = + M1 M0 for pooled estimate. 2 359 19175 0.0187 s   = =   A1 SOI ( ) 168 228 40 60 CI 1.645' 's = −  M1 Correct form with any z-value, FT their .s B1 1.645 (accept 1.64 or 1.65) used in CI formula.   0.175, 0.625 A1 Accept with inequality signs or open brackets. Condone   0.625,0.175 . Do not accept 0.4 0.225  . 7

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 16 Question Answer Marks Guidance 3(a) ( ) ( ) ( ) 1 8 1 8 2 2 1 1 2 2 0 1 0 1 1 1 2 2 d 8 d 1 8 1 64 32 8 1 kx x k x x kx kx kx k k k k k       + − = + − =         + − − − =   M1 Linear equation in terms of k formed following attempt to integrate and use correct limits. 25 1 k = , leading to 1 25 k = . A1 AG, shown convincingly. No errors seen. Alternative method for question 3(a) 1 1 2 2 1 7 7 1 k k  +  = M1 Linear equation in terms of k formed by considering areas. 25 1 k = , leading to 1 25 k = . A1 AG, shown convincingly. No errors seen. 2 3(b) ( ) 1 1 2 0 1 d 8 d m kx x k x x + − =   *M1 Use of definite integrals and attempt at integration to form equation with 0.5 and their k. Could be in terms of k. 2 16 39 0 − + = m m or DM1 Quadratic equation in m . 3 = m A1 Must reject second solution if seen. CAO Alternative method for question 3(b) 1 1 2 2 (8 ) (8 ) m k m  −  − = *M1 Use areas to form equation with 0.5 and their k. Could be in terms of k. ( ) 2 8 25 − = m DM1 Quadratic equation in m . 3 = m A1 Must reject second solution if seen. CAO 3

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 16 Question Answer Marks Guidance 3(c) 2 1 50 2 8 7 1 25 50 25 [0 1] F( ) [1 8] x x x x x x    =  − −  ≤ ≤ ≤ [ F( ) 0 x = for 0 x  and F( ) 1 x = for 8 x  ] M1 Limits not required. Sight of expressions for 0 1 x  ≤ and 1 8 x ≤ ≤ sufficient. B1 For 7 25 − . 6 1 50 3 6 8 7 1 25 50 25 [0 1] G( ) [1 2] y y y y y y    =  − −  ≤ ≤ ≤ [ G( ) 0 y = for 0 x  and G( ) 1 y = for 2 y  ] M1 Both parts with y substituted in their F( )x . Limits not required. Sight of expressions for 0 1 y  ≤ and 1 2 y ≤ ≤ sufficient. ( ) 5 3 25 2 5 3 24 25 25 0 1 g 1 2 0 otherwise. y y y y y y    = −    ≤ ≤ ≤ M1 Differentiation of their G( )y . Limits not required. Sight of expressions for 0 1 y  ≤ and 1 2 y ≤ ≤ sufficient. A1 Fully correct with correct limits. Alternative method for question 3(c) Let 3 p( ) y y = , then 1 3 q( ) p ( ) y y y − = = . M1 Find inverse function. 2 'q ( ) 3 y y = . A1 Differentiate inverse function. May see 2 3 3x . ( ) 3 2 1 25 3 2 1 25 (3 ) [0 1] g (8 )(3 ) [1 2] y y y y y y y    =  −  ≤ ≤ ≤ or ( ) 2 3 2 3 1 25 1 25 (3 ) [0 1] g (8 )(3 ) [1 8] x x x x x x x    =  −  ≤ ≤ ≤ ( g( ) 0 y = otherwise) (g( ) 0 x = otherwise) M2 M2 for using 'q ( ) f ( ) y y  or 'q ( ) f ( ) x x  twice. M1 for one expression. Limits not required. Sight of expressions for 0 1 y  ≤ and 1 2 y ≤ ≤ or 0 1 x  ≤ and 1 8 x ≤ ≤ sufficient. ( ) 5 3 25 2 5 3 24 25 25 0 1 g 1 2 0 otherwise. y y y y y y    = −    ≤ ≤ ≤ A1 Fully correct with correct limits. 5

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 16 Question Answer Marks Guidance 4 H0: population medians are equal for under 25s and over 50s (or 25 50 m m   = ). H1: population median for under 25s < population median time for over 50s (or 25 50 m m    ). B1 B0 for defining m as median, not population median. 198 5 178 1 212 8 181 2 217 9 183 3 229 11 192 4 235 13 203 6 242 14 209 7 223 10 231 12 M1 Attempt at ranks using the integers 1 to 14 (ranks may be reversed). Sum of ranks for sample of size 6: 6 60 R = A1 Note: 6 30 R = from reverse ranks. ( ) 6 6 6 8 1 ' ' [ 30] R + + − = M1 Note: ( ) 6 6 6 8 1 ' ' 60 R + + − = from reverse ranks. [ min(60,30) ] 30 W = = A1 Critical value for 6, 8 n m = = is 31. B1 '30' 31 ≤ , reject H0 (significant). M1 Compare their W with 31 and appropriate conclusion (may be in terms of H1). Reject H0 may be implied by attempt at appropriate conclusion in context. Sufficient evidence to suggest that older people take longer to react. OR Sufficient evidence to support researcher’s claim. A1 Correct conclusion in context from correct working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). 8

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 16 Question Answer Marks Guidance 5(a) 0 1 H : , H : B A B A k k     − = −  B1 Allow 0 1 H : , H : d d k k   =  if defined or consistent with working. Differences, d : 19, 17, 8, 16, 9, 23, 6, 5. M1 Allow one error. ( ) ( ) 2 2 2 2519 1 1 7 8 56 103, 1641, 12.875, 1641 103 44.982 d d d d s  =  =   = = − = =   M1 Sample mean and variance for differences. 2 12.875 8 d k t s − = M1 Correct form for t with their d and their 2 ds . M0 if k is omitted. A1 2 12.875 1.895 8 ds k − ≥ M1 Accept ≥, > or =.   0 8.38 k ≤ ≤ A1 Accept 8.38  k . 7 5(b) Population of differences are normally distributed. B1 Accept underlying distribution or population. B0 for "population mean normally distributed". B0 for "(underlying) data is normal". 1

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 15 of 16 Question Answer Marks Guidance 6(a) Probabilities: 7 7 1 15 15 15 , , for 0, 1, 2 heads. B1 ( ) 2 7 7 1 15 15 15 G X t t t = + + B1 FT FT their probabilities provided 1.  = p Probabilities: 8 7 15 15 , for 1, 2 colours. B1 SC B1 for 8 7 15 15 , for 0,1 colours or 0,2 colours. ( ) 2 8 7 15 15 GY t t t = + B1 FT FT their probabilities provided 1.  = p SC B1 for ( ) 8 7 15 15 GY t t = + from 0,1 colours or ( ) 2 8 7 15 15 GY t t = + from 0,2 colours. 4 6(b) X and Y are not independent. B1 1

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 16 of 16 Question Answer Marks Guidance 6(c) x 0 1 2 y 1 2 1 = + z x y 1 3 3 ( ) P = Z z 1 15 7 15 7 15 B1 Correct 3 cases identified correctly where probabilities are not zero from 1,2 colours. ( ) ( ) ( ) ( ) , 0,1 , 1,2 or 2,1 x y = . Award SC B1 for either of the following: ( ) ( ) ( ) ( ) , 0,0 , 1,1 or 2,0 x y = from 0,1 colours. ( ) ( ) ( ) ( ) , 0,0 , 1,2 or 2,0 x y = from 0,2 colours. M1 At least one non-zero probability correct. Allow for 7 1 14 15 15 15 , , or . A1 All probabilities correct. ( ) 3 1 14 15 15 GZ t t t = + A1 FT FT their probabilities provided 1.  = p Allow A1 FT for either of the following: ( ) 1 14 15 15 2 GZ t t = + from 0,1 colours. ( ) 2 3 7 7 1 15 15 15 GZ t t t = + + from 0, 2 colours. 6(d) ( ) 2 1 42 15 15 ' GZ t t = + ( ) 1 42 15 15 ' G 1 Z = + M1 Attempt to differentiate ( ) GZ t and find ( ) ' G 1 Z . FT their ( ) GZ t . 43 15   2.87 = A1 Allow A1 FT for either of the following: 28 15   1.87 = from 0,1 colours. 35 15   2.33 = from 0,2 colours. 2