Cambridge A Level Mathematics - Further 9231 — 2024 May/June Paper 4 · Variant 3

This document has 12 pages. [Turn over * 8 6 6 1 2 5 4 7 9 2 * Cambridge International AS & A Level FURTHER MATHEMATICS 9231/43 Paper 4 Further Probability & Statistics May/June 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 [ ]. DC (DE) 331918/1 © UCLES 2024 ,

  
, * 0019655481101 * ¬Wz> 3mEyi‰E¢l—V€W ¬8XiD¦0<_‡9p.¢s‚ ¥uU•5UE•E¥uU EEUeU

2 9231/43/M/J/24 © UCLES 2024 1 A college uses two assessments, X and Y, when interviewing applicants for research posts at the college. These assessments have been used for a large number of applicants this year. The scores for a random sample of 9 applicants who took assessment X are as follows. 21.4 24.6 25.3 22.7 20.8 21.5 22.9 21.3 22.3 The scores for a random sample of 10 applicants who took assessment Y are as follows. 20.9 23.5 24.8 21.9 23.4 24.0 23.8 24.1 25.1 25.8 The interviewer believes that the population median score from assessment X is lower than the population median score from assessment Y. Carry out a Wilcoxon rank-sum test, at the 1% significance level, to test whether the interviewer’s belief is supported by the data. [7] … … … … … … … … … … … … … … … … … … … … * 0019655481102 * ,

  
, ĬÕú¾Ġ³íÅùéĉÅĢéĕØþ× Ĭ¸ÖêÁĢ¼ÎØðñòāĆþÚûĂ ĥåÅĕõµąõõååĕąąåĕąÕ 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/24 © UCLES 2024 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655481103 * ,

  
, Ĭ×ú¾Ġ³íÅùéĉÅĢéėØþ× Ĭ¸ÕéÉĨ¸ÞáĊĀ·ÕîÚÚċĂ ĥåµÕµÕĥĕĥÕõĕąĥÅÕĕÕ 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/24 © UCLES 2024 2 A rowing club has a large number of members. A random sample of 12 of these members is taken and the pulse rate, x beats per minute (bpm), of each is measured after a 30-minute training session. A 98% confidence interval for the population mean pulse rate, n bpm, is calculated from the sample as . . 64 22 68 66 1 1 n . (a) Find the values of x / and x2 / . [6] … … … … … … … … … … … … … (b) State an assumption that is necessary for the confidence interval to be valid. [1] … … … … … … … … … … … * 0019655481104 * ,

  
, ĬÕú¾Ġ³íÅùéĉÅĢëĕØĀ× Ĭ¸ÕìÉĞÆëÖĈć°÷ÒüÊóĂ ĥĕĥÕõÕĥµąõąĕÅĥĥÕąÕ 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/24 © UCLES 2024 [Turn over 3 There are three bus companies in a city. The council is investigating whether the buses reliably arrive at their destination on time. The results from random samples of buses from each company are summarised in the following table. Bus company A B C Total Arrival Early 22 22 10 54 On time 30 52 42 124 Late 28 26 18 72 Total 80 100 70 250 Test, at the 5% significance level, whether the reliability of buses is independent of bus company. [7] … … … … … … … … … … … … … … … … … … … * 0019655481105 * ,

  
, Ĭ×ú¾Ġ³íÅùéĉÅĢëėØĀ× Ĭ¸ÖëÁĬÊÛãòúùãêàÊăĂ ĥĕĕĕµµąÕĕąÕĕÅąąĕĕÕ 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/24 © UCLES 2024 4 The random variable X has probability generating function ( )t GX given by ( )t ( ) , ct t 1 GX 5 = + where c is a constant. (a) Find the value of c. [1] … … … … … … (b) Find the value of ( ) X E . [2] … … … … … … … … … … … … … … … … … * 0019655481106 * ,

  
, ĬÙú¾Ġ³íÅùéĉÅĢêėÕþ× Ĭ¸Õë¾ĢâÑëõċĄÃÒ½ĢóĂ ĥŵĕµĕąõÕĥõĕąąåÕÅÕ 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/24 © UCLES 2024 [Turn over The random variable Y is the sum of two independent values of X. (c) Write down the probability generating function of Y and hence find Var(Y ). [4] … … … … … … … … … … … … … … (d) Find ( ) Y 5 P = . [2] … … … … … … … … … … … * 0019655481207 * ,

  
, ĬÛù¿Ġ³íÅùéĉÅĢéėÕþ× Ĭ¸ÕéÁĝÑ·Üø÷·üĒ½ÒċĂ ĥõĕÕµµåĕµĥÕĕÅÅåĕõÕ 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/24 © UCLES 2024 5 The continuous random variable X has cumulative distribution function F given by , x 2 1 ( ) ( ) ( ) , , x x x x x 12 2 2 4 1 24 8 4 8 F 2 2 1 G G G = - - - x 1 8 2 0 Z [ \ ] ] ] ] ]] . (a) Sketch the graph of the probability density function of X. [3] (b) Find ( ) X E . [3] … … … … … … … … … … * 0019655481208 * ,

  
, ĬÙù¿Ġ³íÅùéĉÅĢëĕÕĀ× Ĭ¸ÕìÁħã²ßúð°Ú®ğÂóĂ ĥÅÅÕõµåµÕÅåĕąÅąĕĥÕ 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/24 © UCLES 2024 [Turn over (c) Find the exact value of the interquartile range of X. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655481209 * ,

  
, ĬÛù¿Ġ³íÅùéĉÅĢëėÕĀ× Ĭ¸ÖëÉġßÂÚĀāùþÆ»ÂăĂ ĥŵĕµÕÅÕŵõĕąåĥÕõÕ 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/24 © UCLES 2024 6 Seva is investigating the lengths of the tails of adult wallabies in two regions of Australia, X and Y. He chooses a random sample of 50 adult wallabies from region X and records the lengths, x cm, of their tails. He also chooses a random sample of 40 adult wallabies from region Y and records the lengths, y cm, of their tails. His results are summarised as follows. x 1080 = / x 23480 2 = / y 940 = / y 22 220 2 = / It cannot be assumed that the population variances of the two distributions are the same. (a) Find a 90% confidence interval for the difference between the population mean lengths of the tails of adult wallabies in regions X and Y. [6] … … … … … … … … … … … … … … … … … … … … … … * 0019655481210 * ,

  
, ĬÙù¿Ġ³íÅùéĉÅĢêĕ×þ× Ĭ¸ØìÌĥëèáñ÷Ė¶°ċúċĂ ĥĕåĕõµąµõõåÕąĥąĕÕÕ 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/24 © UCLES 2024 The population mean lengths of the tails of adult wallabies in regions X and Y are X n cm and Y n cm respectively. (b) Test, at the 10% significance level, the null hypothesis X .1 1 Y n n - = against the alternative hypothesis X .1 1 Y 2 n n - . State your conclusion in the context of the question. [4] … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655481211 * ,

  , ĬÛù¿Ġ³íÅùéĉÅĢêė×þ× Ĭ¸×ëÄģçØØćĊÓĢÈÏúûĂ ĥĕÕÕµÕĥÕĥąõÕąąĥÕÅÕ 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/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. Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … * 0019655481212 * ,

  , ĬÙù¿Ġ³íÅùéĉÅĢìĕ×Ā× Ĭ¸×êÄĩÕÑãĉāÜÄĬíĪăĂ ĥåąÕõÕĥõąåąÕÅąÅÕÕÕ 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 15 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/43 Paper 4 Further Probability & Statistics May/June 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 May/June 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 15 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 2024 © Cambridge University Press & Assessment 2024 Page 3 of 15 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.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 15 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 2024 © Cambridge University Press & Assessment 2024 Page 5 of 15 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 2024 © Cambridge University Press & Assessment 2024 Page 6 of 15 Question Answer Marks Guidance 1 21.4 4 20.9 2 24.6 15 23.5 11 25.3 18 24.8 16 22.7 8 21.9 6 20.8 1 23.4 10 21.5 5 24.0 13 22.9 9 23.8 12 21.3 3 24.1 14 22.3 7 25.1 17 25.8 19 M1 A1 Attempt at ranking from 1 to 19. A1 for all correct. Test statistic: 70 A1 If ranks reversed, then sum of ranks = 110, and then 9(9 + 10 + 1) – 110 = 70. H0: population median of X = population median of Y H1: population median of X < population median of Y B1 Accept H0: X Y m m  H1: X Y m m  . Do not accept any other notation. Critical value for m = 9, n = 10 is 61 B1 Using normal approximation implies last 3 marks are B0M0A0, max 4 out of 7.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 15 Question Answer Marks Guidance 1 ‘70’ > ‘61’ accept H0 M1 ‘70’ must come from ranks 1 – 19, ‘61’ must be a critical value from the correct table. ft conclusion for their 70 and their 61. Condone Reject H1 . ‘Accept H0’ can be implied by a conclusion in context that is consistent with their 70 and their 61. Insufficient evidence to suggest that population medians of X and Y are not equal or Insufficient evidence to suggest that population median of X is less than median of Y or Insufficient evidence to support the interviewer’s belief A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not ‘prove’ or equivalent. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’. 7

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 15 Question Answer Marks Guidance 2(a)   1 68.66 64.22 66.44 2 x    M1 12 66.44 797.28 x     A1 May come from use of incorrect t value. Accept AWRT 797.   1 68.66 64.22 2 12 ts   M1 Must be with a t value (not z value). With 2.718 t  gives 2 8.0055 s  A1 Accept AWRT 8.01; may be recovered by final answer exactly 53059(.34). 2.8294 s  . Use   2 2 2 1 11 12 x s x             to find an expression for 2x  M1 Must see an expression for 2x  . 2 53059 x   A1 AWRT 53100 but not from wrong working. 6 2(b) Population is normally distributed B1 Underlying distribution is normal B1. Pulse rates are normal B1. Population mean is normally distributed B0. (Underlying) data is normal B0. 1

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 15 Question Answer Marks Guidance 3 Bus company A Bus company B Bus company C Arrived early 22 (17.28) 22 (21.60) 10 (15.12) Arrived on time 30 (39.68) 52 (49.60) 42 (34.72) Arrived late 28 (23.04) 26 (28.80) 18 (20.16) M1 A1 Calculate expected values, at least 4 correct to 3sf. These must be seen. A1 for all correct to at least 3sf. 1.2893 0.00741 1.7338 2.3615 0.1161 1.5265 1.0678 0.2722 0.2314 M1 At least 4 correct values (to 2sf) or expressions. May be implied by 8.600 – 8.615. Test statistic = 8.61 A1 8.600 – 8.615 Unsupported 8.600 – 8.615 scores M0A0 M1A1. H0: Reliability of buses is independent of bus company H1: Reliability of buses is not independent of/dependent on bus company B1 Condone ‘No association/relationship between…’ ‘Association/relationship between…’ ‘8.61’ < 9.488 Accept H0 M1 ‘8.61’ from an attempt at chi-squared, 9.488 must be correct. ft conclusion from their 8.61. Condone ‘Reject H1’. ‘Accept H0’ can be implied by a conclusion in context that is consistent with their 8.61.

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 15 Question Answer Marks Guidance 3 There is insufficient evidence to support/ suggest that reliability of buses is dependent/not independent of bus company A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not ‘prove’ or equivalent. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’ 7 Question Answer Marks Guidance 4(a) 1 32 c  B1 Accept 0.03125 or 0.0313. 1 4(b) GX (t) =   5 1 , ct t  G’X(t) =       4 5 5 1 1 c t t t    M1 Must use product rule OR differentiate their expansion of G Allow one error in their differentiation. Condone missing c. E(X) = G’X (1) =   80 32 3.5 c   A1 Correct work only. 2

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 11 of 15 Question Answer Marks Guidance 4(c) GY (t) =   10 2 2 1 c t t  *M1 Or   2 2 3 4 5 6 7 8 9 10 11 12 10 45 120 210 252 210 120 45 10 c t t t t t t t t t t t           G'Y(t) =       9 10 2 2 10 1 2 1 c t t t t    G''Y(t) =         8 9 10 2 2 90 1 40 1 2 1 ) c t t t t t      DM1 Differentiate their GY (t) twice, allow one error. G'Y(1) = 7, G’’Y(1) = 44.5 Var(Y) = G’’Y(1) + G’Y(1)    2 G’ 1  = 2.5 M1 A1 Use correct formula and substitute values for t = 1 (may be in terms of c). A1 for correct work only. Alternative solution for question 4(c) Var (Y) = 2Var(X) *M1 Differentiate GX (t) twice DM1 Note that G'X(t) has already been found in part (b).    2 Var( ) '' (1) ' (1) G ' 1 X X X X G G    M1 Use correct formula and substitute values for t = 1. 1.25  2 = 2.5 A1 4

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 12 of 15 Question Answer Marks Guidance 4(d) P(Y = 5) = coefficient of 5t in expansion of their   10 2 2 1 c t t  Required term is 2 120 c  Expansion =   2 2 2 3 1 10 45 120 c t t t t     M1 Identify required term. 15 128 A1 0.117, 120 1024 or any equivalent fraction. 2 Question Answer Marks Guidance 5(a) 2 6 x  8 12 x  M1 Attempt at differentiating both parts of CDF to obtain linear expressions. May be implied by first A1. Sketch A1 Correct shape (Two line segments touch the x-axis and meet at a point above the x- axis). A1 Correct shape with 2 and 8 and at least one of 4 or 1/3 correctly labelled. 3

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 13 of 15 Question Answer Marks Guidance 5(b)     4 8 1 1 6 12 2 4 2 d 8 d x x x x x x      B1 FT FT their PDF, with correct limits. 1 1 1 1 3 2 2 3 6 3 12 3 4 x x x x            M1 Attempt at integration with their PDF (not CDF) no limits needed. May be implied by correct answer. 14 3 A1 Accept 4.67 or any equivalent fraction. 3 5(c) LQ:   2 1 1 12 4 2 , 2 3 x x     B1 Allow 3.73. UQ:   2 8 3 1 , 8 6 24 4 x x      B1 Allow 5.55. IQR = UQ – LQ, with values found for both quartiles M1 6 6 3   A1 Exact answer required. 4

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 14 of 15 Question Answer Marks Guidance 6(a) 2 2 1 1080 23480 49 50 xs         (= 3.102) 2 2 1 940 22220 39 40 ys         (= 3.333) M1 152 9 10 3 Both.   2 3.102 3.333 0.1454 50 40 s    M1 2137 2 14700 0.1454 s   A1 CI: 2 1080 940 1.645 50 40 s   or 2 940 1080 1.645 40 50 s   M1 B1 Use of correct formula with a z value. B1 for 1.645 ( accept 1.64 or 1.65) used in CI formula.   1.27, 2.53 or [–2.53, –1.27] A1 (1.27, 2.53) A1. 1.27 2.53 Y X      A1 (ignore symbol used inside the inequality). Condone [2.53, 1.27] or [–1.27, –2.53] A1. Final answer 1.9 0.627  A0. Pooled estimate: max score M1 M0A0 M1B1A0. First M1 for correct form of pooled estimate. 6

9231/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 15 of 15 Question Answer Marks Guidance 6(b) 1.9 1.1 z s   M1 1.9 1.1 or 7.87 z s     implies M1. 2.098 A1 AWRT 2.10 ‘2.098’ > 1.28(2) Reject H0 or   Φ 2.098 0.9821 > 0.9 Reject H0 or 1    Φ 2.098 = 0.0179 < 0.1 Reject H0 M1 ‘2.098’ from their attempt at z, 1.28(2) must be correct. Consistent signs in comparison. FT conclusion from their 2.098. Condone ‘Accept H1’. ‘Reject H0’ can be implied by a conclusion in context that is consistent with their 2.098. Sufficient evidence to suggest that the lengths of tails of wallabies in region Y are, on average, more than 1.1 cm longer than the tails of wallabies in region X A1 Correct work only, conclusion in context with level of uncertainty in language. Not ‘prove’ or equivalent. Accept ‘enough evidence’. Condone ‘some evidence’. 4 Note that it is possible to use part (a) but it must be made clear explicitly that rejection at 5% level implies rejection at 10% level.