Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 6 · Variant 2

This document has 12 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level MATHEMATICS 9709/62 Paper 6 Probability & Statistics 2 October/November 2025 1 hour 15 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 [ ]. * 0 1 3 0 4 0 5 0 4 2 * DC (CJ) 344299/2 © UCLES 2025 ,










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2 9709/62/O/N/25 © UCLES 2025 1 The number, X, of used computers donated to a charity has a constant average rate of 2.4 computers per week. (a) State a necessary condition for X to have a Poisson distribution. [1] … … … Now assume that X has a Poisson distribution. (b) Calculate the probability that the number of computers donated during a 4-week period is more than 6 and less than 9. [3] … … … … … … … … (c) Use a suitable approximating distribution to calculate the probability that more than 50 computers are donated during a 20-week period. [4] … … … … … … … … … … … * 0000800000002 * ,










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3 9709/62/O/N/25 © UCLES 2025 [Turn over 2 The random variable X has a normal distribution with mean 10 and standard deviation 3. The independent random variable Y has a Poisson distribution with mean 4. (a) Find the standard deviation of X Y + . [3] … … … … … … … … … (b) Find the standard deviation of X Y 5 - . [3] … … … … … … … … … … … … … … … … * 0000800000003 * ,










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4 9709/62/O/N/25 © UCLES 2025 3 The times, in minutes, taken by students to complete a test have mean n and standard deviation v. The times taken by a random sample of 100 students are noted and are used to calculate a 95% confidence interval for n. (a) Given that the end points of the 95% confidence interval are 31.02 and 33.98, correct to 4 significant figures, calculate the value of v. [3] … … … … … … (b) The calculation of the confidence interval required the use of the Central Limit theorem. Explain why it is valid to use the Central Limit theorem in this case. [1] … … … (c) A researcher calculates a number, r, of 95% confidence interval for n. Find the largest value of r such that the probability that all r confidence intervals contain the true value of n is greater than 0.5. [4] … … … … … … … … … … … * 0000800000004 * ,










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5 9709/62/O/N/25 © UCLES 2025 [Turn over 4 An inspector believes that 18% of cups made at a certain factory contain flaws. The factory owner claims that the true percentage is less than 18%. The inspector examines a random sample of 40 cups and finds that 3 of them contain flaws. (a) Stating a necessary assumption, use a binomial distribution to test the factory owner’s claim at the 5% significance level. [6] … … … … … … … … … … … … … … … … … (b) Explain why it would not be appropriate to use the Poisson approximation to the binomial distribution to carry out the test in part (a). [1] … … … … … … * 0000800000005 * ,










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6 9709/62/O/N/25 © UCLES 2025 5 A random variable X has probability density function given by ( ) ( ) , f otherwise. x k x x x 2 0 2 0 2 3 G G = - ) (a) Show that k 4 3 = . [3] … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000006 * ,










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7 9709/62/O/N/25 © UCLES 2025 [Turn over The median of X is denoted by m. (b) (i) Write down the value of ( ) P X m G . [1] … … … … (ii) Hence find ( ( ) ) P X X m E G G . [5] … … … … … … … … … … … … … … … … … … … … … * 0000800000007 * ,










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8 9709/62/O/N/25 © UCLES 2025 6 The weekly profit, in dollars, made by a certain firm has a normal distribution. In the past, the weekly profit had the distribution N(736, 262). Following a change in management, the mean weekly profit for 35 randomly chosen weeks is $725. (a) Stating a necessary assumption, test at the 2% significance level whether the mean weekly profit has decreased. [6] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000008 * ,










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9 9709/62/O/N/25 © UCLES 2025 The mean weekly profit for another random sample of 35 weeks is found and a similar test is carried out at the 2% significance level. (b) State the probability of a Type I error. [1] … … … … (c) Given that the mean weekly profit is now in fact $718, find the probability of a Type II error. [5] … … … … … … … … … … … … … … … … … … … … … * 0000800000009 * ,










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10 9709/62/O/N/25 © UCLES 2025 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000010 * ,










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12 9709/62/O/N/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. BLANK PAGE * 0000800000012 * ,









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This document consists of 16 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International A Level MATHEMATICS 9709/62 Paper 6 Probability & Statistics 2 October/November 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 October/November 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 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.

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 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 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.

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 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

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 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 Indicates that work/page has been seen

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 16 Annotation Meaning Error in number of significant figures Correct Transcription error Correct answer from incorrect working

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 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.

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 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

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 16 Question Answer Marks Guidance 1(a) Computers are donated independently or singly or randomly B1 Context i.e. ‘computers’ or ‘donations’ must be mentioned. ‘Events/trials/X/It is independent’ B0. ISW after a correct statement. 1 1(b) λ = 9.6 B1 e−9.6( 7 9.6 7! + 8 9.6 8! ) = e-9.6 (1490.97 + 1789.16) = 0.10098 + 0.12118 M1 Allow one end error and incorrect λ. Accept fully correct sigma notation. = 0.222 (3 sf) A1 As final answer. SC Unjustified 0.222 scores B1M0B1. 3 1(c) N(48, 48) B1 SOI. 50.5 48 48 − [= 0.361] M1 For standardising with their N (…, …) Ignore missing or incorrect cc for M1. 1 − Φ(‘0.361’) M1 For area consistent with their working. = 0.359 (3 sf) A1 4

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 16 Question Answer Marks Guidance 2(a) Var(Y) = 4 B1 SOI. Accept sd = 2 for B1 if var = 4 not seen. Var(X + Y) = 32 + ‘4’ M1 SOI. FT their Var(Y). sd of X + Y = 13 = 3.61 A1 3 2(b) Var(5X ) = 52 × 32 B1 May be implied. Var(5X − Y) = Var(5X) + Var(Y) = “52 × 32 ” + “4” M1 ft their Var (5X), and their Var(Y) from 2(a). sd of 5X − Y = 229 or 15.1 (3 sf) A1 3 Question Answer Marks Guidance 3(a) Sample mean = 32.5 soi or Width = 2.96 B1 SOI. Could be implied. 32.5 + 1.96 × 100  = 33.98 or 2.96 = 2 × 1.96 × 100  M1 OE. Factor of 2 error can still score M1. Accept z=1.96 or 1.645 for M1. σ = 7.55 (3 sf) A1 (Final answer of 15.1 scores 2/3). 3

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 16 Question Answer Marks Guidance 3(b) Because the sample size is large. B1 Accept n is large. Accept n > 30 condone > 50 ‘Number of students is large’ B0 need ‘sample’ or 100. 1 3(c) 0.95r > 0.5 Allow ‘=’ M1 OR 0.95r evaluated for any positive integer r M1. rln 0.95 > ln 0.5 Allow ‘=’ Correctly take logs and use log rule A1 OE. 0.9513 = 0.51… A1. r < 13.5…. Allow ‘=’ A1 0.9514 = 0.48… A1. Largest value of r is 13 A1 (Condone incorrect inequality signs throughout). 4

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 16 Question Answer Marks Guidance 4(a) Assume flaws in cups are independent OR prob of containing a flaw is the same for all cups B1 OE in context (e.g. prob… is consistent) but p = 0.18 is B0. H0: P(contains flaw) = 0.18 H1: P(contains flaw) < 0.18 B1 Both. Allow ‘p’ [P(X ⩽ 3) =] 0.8240 + 40×0.8239×0.18 + 40C2×0.8238×0.182 + 40C3×0.8237×0.183 =0.0003569 + 0.0031338 + 0.013414 + 0.037298 M1 0.0542 (3 sf) A1 Unsupported 0.0542 scores M0B1. 0.0542>0.05 M1 Valid comparison – must be a tail probability Or if CR found (⩽2) then comparison with 3 M1. [Accept Ho] Insufficient evidence to accept factory owners claim. [Insufficient evidence that percentage less than 18%] A1FT In context, not definite, no contradictions no use of ‘p’ unless defined. e.g. not ‘%age is not less than 0.18’ or ‘%age is 18%’ 2TT scores max B1B0M1A1M1(with 0.025) is A0. Normal approx’n: (First B1B1* if earned) then: CV = 3.5 40 0.18 40 0.18 0.82) −    = ±1.523 SCB1 (need cc). 1.523< 1.645 or 0.0639>0.05 No evidence less than 18% SCB1. *hypotheses µ=7.2 and µ<7.2 can score if Normal approx. used. 6

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 16 Question Answer Marks Guidance 4(b) np = 7.2 which is too large, or > 5. Or n = 40 which is not greater than 50. Or p = 0.18 which is not small or > 0.1. B1 Context required (7.2 or 40 or 0.18). One correct condition required ISW. 1 Question Answer Marks Guidance 5(a) 2 2 3 0 (2 ) d k x x x −  = 1 M1 Attempt to integrate f(x) & = 1, ignore limits. k 3 4 2 3 4 2 0   −   x x = 1 A1 Correct integration and limits. k × 4 3 = 1 k = 3 4 A1 AG. Correctly obtained, at least one step shown and no errors seen. 3 5(b)(i) 0.5 B1 Could be in a correct expression. 1

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 16 Question Answer Marks Guidance 5(b)(ii) E(X) = 2 3 4 3 4 0 (2 ) d x x x −  M1 Attempt to integrate xf(x), ignore limits condone k or missing k. = 4 5 3 2 4 4 5 2 0   −   x x = 6 5 A1 P(X < E(X)) = '1.2' 2 3 3 4 0 (2 ) d x x x −  M1 Attempt to integrate f(x), limits from 0 to their E(X) (OR integrate f(x) limits from their E(X) to 2). = 3 4 3 2 4 3 4 '1.2' 0   −   x x = 594 1250 or 297/625 or 0.4752 A1 OR 0.5248 or 656/1250 or 328/625. P(E(X) ⩽ X ⩽ m) = 0.5 − ‘0.4752’ OR 0.5248 – 0.5. = 0.0248 or 31/1250 A1 CWO. SC solutions attempting to find the value of m (1.22854) 1.229 Must be correct. Can score M1A1 then B1 for integrate correctly from 1.2 to 1.229 (accept 3sf here for 1.229) and 0.0248 to 0.0252 obtained B1. (Incorrect values of m could score first M1A1 only). 5

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 16 Question Answer Marks Guidance 6(a) Assume sd is still 26 or is unchanged B1 H0: Population mean = 736 H1: Population mean < 736 B1 Both. Allow ‘µ’, but not just ‘mean’. ± 26 35 725 736 − M1 For standardising (ignore cc attempts). = −2.503 A1 (± ) Accept 3sf if nothing better seen. −2.503 < −2.054(or 5 or 6) or 0.0062 < 0.02 M1 Valid comparison. [Reject Ho] There is sufficient evidence that (mean) weekly profit has decreased A1 In context, not definite, no contradictions. No use of µ unless defined. e.g. not ‘Mean weekly profit has decreased.’. 2TT scores max B1B0M1A1M1 (comparison 2.326 or 0.01) A0. CV method: 726.973=727 M1 A1 > 725 and concl M1A1. Condone: 734.027 =734 M1 A1 < 736 and concl M1A1. 6 6(b) 0.02 B1 1

9709/62 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 16 Question Answer Marks Guidance 6(c) 26 35 736 − a = −2.054 or -2.055 M1 For standardising to find CV Must be – 2.054 OE. (2TT in part a accept –2.326 in expression for M1). a = 726.973 A1 CWO Accept 727 or better (working may be in 6(a)). 26 35 '726.973' 718 − (= 2.042) M1 For standardising with their cv and 718. = 1 – ɸ('2.042') M1 For area consistent with their working. = 0.0203 to 0.0207 A1 CWO. Note: (736–718)/(26/√35) or (725–718)/(26/√35) or similar scores M0A0M1 and possibly M1. Max 2/5. 5