Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 5 · Variant 2

This document has 12 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level MATHEMATICS 9709/52 Paper 5 Probability & Statistics 1 October/November 2024 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 [ ]. * 6 8 8 9 3 9 6 5 2 3 * DC (CE) 337396/1 © UCLES 2024 ,










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3 9709/52/O/N/24 © UCLES 2024 [Turn over 1 At a college, the students choose exactly one of tennis, hockey or netball to play. The table shows the numbers of students in Year 1 and Year 2 at the college playing each of these sports. Tennis Hockey Netball Year 1 16 22 12 Year 2 24 18 28 One student is chosen at random from the 120 students. Events X and N are defined as follows: X: the student is in Year 1 N: the student plays netball. (a) Find ( ). X N P [1] … … (b) Find ( ). N X P [1] … … (c) Determine whether or not X and N are independent events. [1] … … … … One of the students who plays netball takes 8 shots at goal. On each shot, the probability that she will succeed is 0.15, independently of all other shots. (d) Find the probability that she succeeds on fewer than 3 of these shots. [3] … … … … … … … * 0000008000003 * ,










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4 9709/52/O/N/24 © UCLES 2024 2 (a) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC. [1] … … … … … … … … (b) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC in which there are no more than two letters between the two As. [3] … … … … … … … … … … … … … … … … … … * 0000008000004 * ,










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5 9709/52/O/N/24 © UCLES 2024 [Turn over 3 A fair coin and an ordinary fair six-sided dice are thrown at the same time. The random variable X is defined as follows. • If the coin shows a tail, X is twice the score on the dice. • If the coin shows a head, X is the score on the dice if the score is even and X is 0 otherwise. (a) Draw up the probability distribution table for X. [3] … … … … … … … … … … … … (b) Find ( ) X ar V . [3] … … … … … … … … … … … * 0000008000005 * ,










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6 9709/52/O/N/24 © UCLES 2024 4 The heights, in metres, of white pine trees are normally distributed with mean 19.8 and standard deviation 2.4 . In a certain forest there are 450 white pine trees. (a) How many of these trees would you expect to have height less than 18.2 metres? [4] … … … … … … … … … … … … … The heights, in metres, of red pine trees are normally distributed with mean 23.4 and standard deviation v. It is known that 26% of red pine trees have height greater than 25.5 metres. (b) Find the value of v. [3] … … … … … … … … … * 0000008000006 * ,










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7 9709/52/O/N/24 © UCLES 2024 [Turn over 5 In a class of 21 students, there are 10 violinists, 6 guitarists and 5 pianists. A group of 7 is to be chosen from these 21 students. The group will consist of 4 violinists, 2 guitarists and 1 pianist. (a) In how many ways can the group of 7 be chosen? [2] … … … … … … … … On another occasion a group of 5 will be chosen from the 21 students. The group must contain at least 2 violinists, at least 1 guitarist and at most 1 pianist. (b) In how many ways can the group of 5 be chosen? [4] … … … … … … … … … … … … … … … * 0000008000007 * ,










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8 9709/52/O/N/24 © UCLES 2024 6 Teams of 15 runners took part in a charity run last Saturday. The times taken, in minutes, to complete the course by the runners from the Falcons and the runners from the Kites are shown in the table. Falcons 38 39 42 44 46 48 50 51 52 56 58 59 64 69 76 Kites 32 40 40 45 47 48 52 54 58 59 59 60 61 63 65 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with the Falcons on the left-hand side. [4] (b) Find the median and the interquartile range of the times for the Falcons. [3] … … … … … … … * 0000008000008 * ,










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9 9709/52/O/N/24 © UCLES 2024 [Turn over Let x and y denote the times, in minutes, of a runner from the Falcons and a runner from the Kites respectively. It is given that x 792 = / , 504 x 43 2 = / , y 783 = / , . 223 y 42 2 = / (c) Find the mean and the standard deviation of the times taken by all 30 runners from the two teams. [3] … … … … … … … … … … … … … … … … … … … … … … … * 0000008000009 * ,










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10 9709/52/O/N/24 © UCLES 2024 7 In a game, players attempt to score a goal by kicking a ball into a net. The probability that Leno scores a goal is 0.4 on any attempt, independently of all other attempts. The random variable X denotes the number of attempts that it takes Leno to score a goal. (a) Find ( ) X 5 P = . [1] … … … (b) Find ( ) X 3 7 P G G . [2] … … … … … … (c) Find the probability that Leno scores his second goal on or before his 5th attempt. [3] … … … … … … … … … … … … … … * 0000008000010 * ,










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11 9709/52/O/N/24 © UCLES 2024 Leno has 75 attempts to score a goal. (d) Use a suitable approximation to find the probability that Leno scores more than 28 goals but fewer than 35 goals. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000008000011 * ,









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12 9709/52/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. … … … … … … … … … … … … … … … … … … … … … … 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. * 0000008000012 * ,









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This document consists of 18 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level MATHEMATICS 9709/52 Paper 5 Probability & Statistics 1 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.

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 18 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.

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 18 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.

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 18 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.

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 18 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

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 18 Question Answer Marks Guidance 1(a) [P(X|N) =] 12 40 B1 0.3, 3 10 , 30% OE. 1 1(b) [P(N|X) =] 12 50 B1 0.24, 6 25 OE. 1 1(c) P(N ( ) ( ) 12 40 50 ) , , 120 120 120  = = = X P N P X   40 50 5 12 , 0.138 8 , 0.1 120 120 36 120  =  Not independent B1 ( ) ( ) ( ) ( ) , and or and P N P X P N X P N X  notation seen and equated to the values for ( ) ( ) ( ) ( ) , and or and P N P X P N X P N X  . Calculation stated and evaluated. Not independent clearly stated. 5 12 36 120  does not need to be stated. All values OE. Condone consistent use of A, B etc. If values for P(N), P(X) stated, accept P(N)×P(X) = 5 36 . 1

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 18 Question Answer Marks Guidance 1(d) Method 1 [P(0, 1, 2) =] (0.85)8 + 8C1(0.85)7(0.15) + 8C2(0.85)6(0.15)2 [= 0.27249 + 0.38469 + 0.23760] M1 One term of form 8Cx ( ) ( ) 8 1 − − x x p p . With 0 1, 0    p x or 8. A1 Correct unsimplified expression, no terms omitted leading to final answer. = 0.895 B1 0.8945 ⩽ p ⩽ 0.895. Method 2 [P(0, 1, 2) =] 1 – {8C3(0.85)5(0.15)3 + 8C4(0.85)4(0.15)4 + 8C5(0.85)3(0.15)5 + 8C6(0.85)2(0.15)6 + 8C7(0.85)(0.15)7 + (0.15)8} M1 One term of form 8Cx ( ) ( ) 8 1 − − x x p p With 0 1, 0    p x or 8. A1 Correct unsimplified expression. Condone omission of final bracket ‘}’. If other brackets omitted, allow recovery if 1 – 0.1052[…] seen. = 0.895 B1 0.8945 ⩽ p ⩽ 0.895. 3 Question Answer Marks Guidance 2(a) 9! 181440 2!   =     B1 Exact value must be seen. CAO. 1

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 18 Question Answer Marks Guidance 2(b) Method 1 Scenario number of letters between As A A [^ ^ ^ ^ ^ ^ ^] 7! × 8 or 8! [ = 40320] A ^ A [^ ^ ^ ^ ^ ^] 7! × 7 [ = 35280] A ^ ^ A [^ ^ ^ ^ ^] 7! × 6 [ = 30240] Total: 7!  (8 + 7 + 6) B1 Correct outcome/value for 1 identified scenario, accept unsimplified. M1 Add values of 3 correct scenarios, no incorrect/repeated scenarios. = 105 840 A1 If M1 not awarded, SC B1 for 105840 WWW. Method 2 Scenario number of letters between As A A [^ ^ ^ ^ ^ ^ ^] 8! [ = 40320] A ^ A [^ ^ ^ ^ ^ ^] 7P1 × 7! or 7C1 × 7! [ = 35280] A ^ ^ A [^ ^ ^ ^ ^] 7P2 × 6! or 7C2 × 2 × 6! [ = 30240] Total: 8! + 7P1 × 7! +7P2 × 6! or 8! + 7C1 × 7! +7C2 × 2 × 6! B1 Correct outcome/value for 1 identified scenario, accept unsimplified. M1 Add values of 3 correct scenarios, no incorrect/repeated scenarios. =105 840 A1 If M1 not awarded, SC B1 for 105840 WWW.

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 18 Question Answer Marks Guidance 2(b) Method 3 Scenario number of letters between As A ^ ^ ^ A [^ ^ ^ ^] 7! × 5 [ = 25200] A ^ ^ ^ ^ A [^ ^ ^] 7! × 4 [ = 20160] A ^ ^ ^ ^ ^ A [^ ^] 7! × 3 [ = 15120] A ^ ^ ^ ^ ^ ^ A [^] 7! × 2 [ = 10080] A ^ ^ ^ ^ ^ ^ ^ A 7! [× 1] [ = 5040] Total = ( ) 9! 7! 5 4 3 2 1 2!−  + + + + B1 Correct outcome/value for 1 identified scenario, accept unsimplified. M1 their 2(a), or correct, subtract values of 5 correct scenarios, no incorrect/repeated scenarios. =105 840 A1 If M1 not awarded, SC B1 for 105840 WWW. 3

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 18 Question Answer Marks Guidance 3(a) x 0 2 4 6 8 10 12 P(X = x) 3 12 , 1 4 0.25 2 12 , 1 6 0.167 2 12 2 12 1 12 0.0833 1 12 1 12 B1 Table with correct x values and at least 2 correct probabilities. Condone any additional x values if probability stated as 0. B1 Four more probabilities correctly linked to the correct x value, need not be in table, accept unsimplified. B1 7 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3 SF. SC B1 7 or more probabilities summing to 1 placed in a probability distribution table. 3

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 11 of 18 Question Answer Marks Guidance 3(b) [E(X) =]  2 2 2 1 1 1 0 2 4 6 8 10 12 12 12 12 12 12 12 +  +  +  +  +  +   4 8 12 8 10 12 54 0 4.5 12 12 12 12 12 12 12   + + + + + + = =     M1 Accept unsimplified expression. May be calculated in variance. Accept 1 2 2 5 1 1 3 3 3 6 + + + + + OE for the M mark FT their table with 7 or 8 probabilities summing to 0.999 ⩽ total ⩽ 1 (0 < p < 1). FT acceptable at the bold partially evaluated stage. [Var(X) =]  2 2 2 2 2 2 2 2 2 2 1 1 1 0 2 4 6 8 10 12 ( 4.5) 12 12 12 12 12 12 +  +  +  +  +  +  −their        0 2 4 2 16 2 36 1 64 1 100 1 144 81 12 4   +  +  +  +  +  +  −     2 35 4.5   = −   M1 Appropriate variance formula using their (E(X))2 value. FT their table with 6 or more probabilities. (0 < p < 1) which need not sum to 1 or with an expression no more evaluated than shown. FT acceptable at the bold partially evaluated stage with their probabilities. = 3 14.75, 14 4 A1 CAO Accept 59 4 . If either or both M marks not awarded, SC B1 for correct answer WWW 3

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 12 of 18 Question Answer Marks Guidance 4(a) [P(X < 18.2) = P( 18.2 19.8 ] 2.4 −  Z ) M1 Use of ± standardisation formula with 18.2, 19.8 and 2.4 substituted appropriately, no continuity correction. Condone ( ) 2 2 2.4  or ( ) 2.4  . = ( ) ( ) Φ 0.6667 1 Φ 0.6667  − = −    = 1 – 0.7477 M1 Calculating the appropriate probability areas (leading to their final answer, expect < 0.5). Note: 0.432 is z-value of 0.667 so is not an appropriate probability area (M0). = 0.252(3) A1 AWRT 0.252 SOI, accept 0.2525. If one or both M marks not awarded, SC B1 for AWRT 0.252 SOI, accept 0.2525. [Expected number = 0.2523 450 113.5,  = ] = 113 or 114 B1FT Strict FT their at least 4 figure probability × 450. (Check with calculator). One integer answer only. No indication of ‘approximation’, e.g. , , about . 4 4(b) [P(X >25.5) = 0.26 so P(Z > 25.5 23.4) 0.26  − = ] 25.5 23.4 0.643  − = B1 0.643 ⩽ z ⩽ 0.6435 or −0.6435 ⩽ z ⩽ −0.643 seen. M1 ± standardisation formula with 25.5, 23.4, σ equating to a z- value, (not 1 – their z-value…). Condone continuity correction ±0.05, not 2 , not . 3.27 = A1 3.26 ⩽ σ ⩽ 3.27. Do not award for improper fractions. 3

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 13 of 18 Question Answer Marks Guidance 5(a) 10C4  6C2 5C1 M1 10Ca  6Cb  5Cc , a + b + c = 7, a,b,c integers. No other terms present but condone × 6 or × 3!. [ = 210 15 5]   = 15750 A1 2 5(b) Scenario V G P VVVVG 4 1 0 10C4  6C1 [ 5C0] [1260 ] VVVGG 3 2 0 10C3  6C2 [ 5C0] [1800 ] VVGGG 2 3 0 10C2  6C3 [ 5C0] [900] VVVGP 3 1 1 10C3  6C1 5C1 [3600 ] VVGGP 2 2 1 10C2  6C2  5C1 [3375 ] M1 One product using 2 or 3 combinations with upper numbers correct and lower numbers summing to 5 and linked to a correct identified scenario. Condone the consistent use of permutations. B1 2 identified outcomes evaluated accurately, accept unsimplified. M1 Add values of 5 correct scenarios, no incorrect/repeated scenarios. Total = 10935 A1 If either or both Ms not awarded, SC B1 for 10935 WWW 4

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 14 of 18 Question Answer Marks Guidance 6(a) Falcons Kites 9 8 3 2 8 6 4 2 4 0 0 5 7 8 9 8 6 2 1 0 5 2 4 8 9 9 9 4 6 0 1 3 5 6 7 Key: 1|5|4 means 51 minutes for Falcons and 54 minutes for Kites B1 Correct stem, ignore extra values (not in reverse, not split). If a split stem-and-leaf plot is used (i.e. stem values are repeated) the remaining B marks are available. B1 Correct Falcons labelled on left, leaves in order from right to left and lined up vertically, no commas or other punctuation. B1 Correct Kites labelled on same diagram, leaves in order and lined up vertically, no commas or other punctuation. Penalise each error only once in question. E.g. commas in both sets of data. B1 Correct key, for their diagram, need both teams names and ‘mins’ at least once here, or in leaf headings or title. If 2 separate diagrams drawn max marks B1 if both stems correct. B1 if Falcons correct to the left of the stem. B1 if both keys correct including ‘mins’ and team name. 4 6(b) Median = 51 [minutes] B1 Accept Q2, must be identified. [IQR =] 59 – 44 M1 58 ⩽ UQ ⩽ 64 – 42 ⩽ LQ ⩽ 46. Implied if both quartile values are stated and an appropriate IQR calculated accurately. = 15 [minutes] A1 WWW 3

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 15 of 18 Question Answer Marks Guidance 6(c) 792 783 1575 1 105 mean 52.5, 52 , 30 30 2 2 +   = = =     B1 1575 30 is not acceptable for this mark. 2 ' ' 85727  = x sd2 =  ( ) 2 43504 42223 792 783 Variance 30 30   + +   = −           2 85727 1575 30 30   −    [= 101.3167] M1 Accept unsimplified variance formula. FT their mean. Ignore any square root leading to sd for this mark. ( ) 101.3167 10.1 = = A1 AWRT. Must be identified, e.g. sd, s, std d, . Condone ‘short’ square root signs. If M1 not awarded, SC B1 for, 6079 101.3167 oe 1 0.1 60 = = or . 3 Question Answer Marks Guidance 7(a) ( ) 4 [ 0.6 0.4 ] 0.  = 0518[4], 162 3125 B1 1

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 16 of 18 Question Answer Marks Guidance 7(b) Method 1 [P( 7) P( 2  −  X X ) =] (1 – 0.67) – (1 – 0.62) M1 ( ) ( ) 7 2 2 7 1 1 − − − − p p or p p seen, 0 < p <1. [= 0.36 – 0.02799 ] = 0.332[0… ], 25938 78125 A1 If M0 awarded SC B1 0.3320064 or 25938 78125 CAO. Method 2 [P(X = 3,4,5,6,7) = ] 2 3 4 5 6 0.4 0.6 0.4 0.6 0.4 0.6 0.4 0.6 0.4 0.6  +  +  +  +  M1 ( ) ( ) ( ) ( ) ( ) 2 3 4 5 6 1 1 1 1 1 −  + −  + −  + −  + −  p p p p p p p p p p seen, 0 < p <1. [= 0.144 + 0.0864 + 0.05184 + 0.031104 + 0.0186624] = 0.332[0… ], 25938 78125 A1 If M0 awarded SC B1 0.3320064 or 25938 78125 CAO. Method 3 – geometric series [P(X = 3,4,5,6,7) = ] ( ) 5 0.144 1 0.6 1 0.6 0.4 − − or M1 ( ) 5 0.144 1 1 − − p p seen 0 < p < 1. =0.332[0…], 25938 78125 A1 If M0 awarded SC B1 0.3320064 or 25938 78125 CAO. 2

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 17 of 18 Question Answer Marks Guidance 7(c) Method 1 2nd goal scored on: 2nd attempt (0.4)2 [ = 0.16] 3rd attempt (0.4)2 (0.6) × (2 or 2C1) [ = 0.192] 4th attempt (0.4)2 (0.6)2 × (3 or 3C1) [ = 0.1728] 5th attempt (0.4)2 (0.6)3 × (4 or 4C1) [ =0.13824] M1 2 correct unsimplified outcomes. Condone not identified but not incorrectly identified. M1 Add values for 4 identified correct scenarios. Condone adding values of 2nd, 3rd and 4th attempts only. No incorrect scenarios. = 0.663, 2072 3125 A1 If either or both M marks not awarded, SC B1 for 0.663, 2072 3125 WWW condone 1 index error. Method 2 5C2(0.4)2(0.6)3 +5C3(0.4)3(0.6)2 +5C4(0.4)4(0.6)1 +5C5(0.4)5 [0.3456 + 0.2304 + 0.0768 + 0.01024] or 1 – (5C0 (0.6)5 + 5C1(0.4)1(0.6)4) M1 At least 2 correct unsimplified terms. M1 Add values for 4 terms of the form 5Ca(0.4)a(0.6)5-a or 1 – sum of 2 terms of the form 5Ca(0.4)a(0.6)5-a. =0.663, 2072 3125 A1 If either or both M marks not awarded, SC B1 for 0.663 www condone 1 index error. 3

9709/52 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 18 of 18 Question Answer Marks Guidance 7(d) [Mean = 75 0.4 ] 30  = [Variance = 75 0.4 0.6 ]18   = B1 30 and 18 seen, allow unsimplified. May be seen in standardisation formula. ( 18, 3 2, 4.2426 4.243   = implies correct variance) Withold mark if variance clearly identified as standard deviation. P(28 < X < 35) = P( 28.5 30 34.5 30) 18 18 − −   Z M1 Substituting their µ and positive σ into one ±standardising formula (any number for 28.5 or 34.5), not σ2, not √σ. M1 Using continuity corrections 27.5 or 28.5 and 34.5 or 35.5 in their 2 separate standardisation formula. ( ) ( ) Φ 1.0607 Φ 0.3536 1 = + −   = 0.8556 + 0.6383 – 1 Or 0.8556 – (1 – 0.6383) Or 0.8556 – 0.3617 Or (0.8556 – 0.5) + (0.6383 – 0.5) Or 0.3556 + 0.1383 M1 Appropriate area Φ, from final process. Must be a probability. = 0.494 A1 AWRT. 5