Cambridge A Level Mathematics - Further 9231 — 2018 May/June Paper 2 · Variant 2

*3727275507* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/22 Paper 2 May/June 2018 3 hours Candidates answer on the Question Paper. Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions in the space provided. If additional space is required, you should use the lined page at the end of this booklet. The question number(s) must be clearly shown. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Where a numerical value is necessary, take the acceleration due to gravity to be 10 m s−2. The use of a calculator is expected, where appropriate. Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 22 printed pages and 2 blank pages. JC18 06_9231_22/RP © UCLES 2018 [Turn over

2 1 A bullet of mass m kg is fired horizontally into a fixed vertical block of material. It enters the block horizontally with speed 250 m s−1 and emerges horizontally with speed 70 m s−1 after 0.04 s. The block offers a constant horizontal resisting force of magnitude 450 N. Find the value of m. [3] … … … … … … … … … 2 A particle P moves on a straight line in simple harmonic motion. The centre of the motion is O. The points A and B are on the line, on opposite sides of O, with OA = 1.6 m and OB = 1.2 m. The ratio of the speed of P at A to its speed at B is 3 : 4. (i) Find the amplitude of the motion. [4] … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

3 The maximum speed of P during its motion is 1 30 m s−1. (ii) Find the period of the motion. [2] … … … … … … … … … … … (iii) Find the time taken for P to travel directly from A to B. [3] … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

4 3 Two identical uniform small spheres A and B, each of mass m, are moving towards each other in a straight line on a smooth horizontal surface. Their speeds are u and ku respectively, and they collide directly. The coefficient of restitution between the spheres is e. Sphere B is brought to rest by the collision. (i) Show that e = k −1 k + 1. [3] … … … … … … … … … … … … (ii) Given that 60% of the total initial kinetic energy is lost in the collision, find the values of k and e. [6] … … … … … … … … … © UCLES 2018 9231/22/M/J/18

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6 4 A uniform rod AB has length 2a and weight W. The end A rests on rough horizontal ground and the end B rests against a smooth vertical wall. The rod is in a vertical plane that is perpendicular to the wall. The angle between the rod and the horizontal is 1. A particle of weight 5W hangs from the rod at the point C, with AC = xa, where 0 < x < 1. (i) By taking moments about A, show that the magnitude of the normal reaction at B is W5x + 1 2 tan 1 . [3] … … … … … … … … … … … The particle of weight 5W is now moved a distance a up the rod, so that AC = x + 1a. This results in the magnitude of the normal reaction at B being double its previous value. The system remains in equilibrium with the rod at angle 1 with the horizontal. (ii) Show that x = 4 5. [3] … … … … … … … © UCLES 2018 9231/22/M/J/18

7 The coefficient of friction between the rod and the ground is 2 3. (iii) Given that the rod is about to slip when the particle of weight 5W is in its second position, find the value of tan 1. [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

8 5 Axis l A B C Three thin uniform rings A, B and C are joined together, so that each ring is in contact with each of the other two rings. Ring A has radius 2a and mass 3M; rings B and C each have radius 3a and mass 2M. The rings lie in the same plane and the centres of the rings are at the vertices of an isosceles triangle. The object consisting of the three rings is free to rotate about the horizontal axis l which is tangential to ring A, in the plane of the object and perpendicular to the line of symmetry of the object (see diagram). (i) Show that the moment of inertia of the object about the axis l is 180Ma2. [7] … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

9 … … … … … … … … … (ii) Show that small oscillations of the object about the axis l are approximately simple harmonic, and state the period. [5] … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

10 6 The continuous random variable X has distribution function given by Fx = D 1 −e−0.4x x ≥0, 0 otherwise. (i) Find PX > 2. [2] … … … … … … … … (ii) Find the interquartile range of X. [4] … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

11 7 A large number of athletes are taking part in a competition. The masses, in kg, of a random sample of 7 athletes are as follows. 98.1 105.0 92.2 89.8 99.9 95.4 101.2 Assuming that masses are normally distributed, test, at the 10% significance level, whether the mean mass of athletes in this competition is equal to 94 kg. [7] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

12 8 A manufacturer produces three types of car: hatchbacks, saloons and estates. Each type of car is available in one of three colours: silver, blue and red. The manufacturer wants to know whether the popularity of the colour of the car is related to the type of car. A random sample of 300 cars chosen by customers gives the information summarised in the following table. Colour of car Silver Blue Red Hatchback 53 36 41 Type of car Saloon 29 40 31 Estate 28 24 18 Test at the 10% significance level whether the colour of car chosen by customers is independent of the type of car. [8] … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

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14 9 At a ski resort, the probability of snow on any particular day is constant and equal to p. The skiing season begins on 1 November. The random variable X denotes the day of the skiing season on which the first snowfall occurs. (For example, if the first snowfall is on 5 November, then X = 5.) The variance of X is 4 9. (i) Show that 4p2 + 9p −9 = 0 and hence find the value of p. [4] … … … … … … … … … … … … … … … (ii) Find the probability that the first snowfall will be on 3 November. [1] … … … … … … © UCLES 2018 9231/22/M/J/18

15 (iii) Find the probability that the first snowfall will not be before 4 November. [2] … … … … … … … … (iv) Find the least integer N so that the probability of the first snowfall being on or before the Nth day of November is more than 0.999. [4] … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

16 10 The times taken to run 400 metres by students at two large colleges P and Q are being compared. There is no evidence that the population variances are equal. The time taken by a student at college P and the time taken by a student at college Q are denoted by x seconds and y seconds respectively. A random sample of 50 students from college P and a random sample of 60 students from college Q give the following summarised data. Σx = 2620 Σx2 = 138 200 Σy = 3060 Σy2 = 157 000 (i) Using a 10% significance level, test whether, on average, students from college P take longer to run 400 metres than students from college Q. [9] … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

17 … … … … … … … … … … … … … … (ii) Find a 90% confidence interval for the difference in the mean times taken to run 400 metres by students from colleges P and Q. [3] … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

18 11 Answer only one of the following two alternatives. EITHER A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle is held so that the string is taut, with OP horizontal. The particle is projected downwards with speed / 2 5ag ! and begins to move in a vertical circle. The string breaks when its tension is equal to 11 5 mg. (i) Show that the string breaks when OP makes an angle 1 with the downward vertical through O, where cos 1 = 3 5. Find the speed of P at this instant. [6] … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

19 (ii) For the subsequent motion after the string breaks, find the distance OP when the particle P is vertically below O. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

20 OR The regression line of y on x, obtained from a random sample of 6 pairs of values of x and y, has equation y = 0.25x + k, where k is a constant. The values from the sample are shown in the following table. x 4 5 7 8 10 14 y 5 8 p 7 p 9 (i) Find the value of p and the value of k. [6] … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

21 (ii) Find the product moment correlation coefficient for the data. [2] … … … … … … … … … … … … (iii) Test, at the 5% significance level, whether there is evidence of positive correlation between the variables. [4] … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18 [Turn over

22 Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/22/M/J/18

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24 BLANK PAGE 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2018 9231/22/M/J/18

IGCSE™ is a registered trademark. This document consists of 17 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 May/June 2018 MARK SCHEME Maximum Mark: 100 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 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 2 of 17 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 descriptors 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.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 3 of 17 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.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 4 of 17 Mark Scheme Notes Marks are of the following three types: 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. • 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 (or dep*) 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. • The symbol 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 and B marks are not given for fortuitously ‘correct’ answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 17 The following abbreviations may be used in a mark scheme or used on the scripts: 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 – often written by a ‘fortuitous’ answer ISW Ignore Subsequent Working SOI Seen or implied SR Special Ruling (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) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become ‘follow through’ marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR –2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 17 Question Answer Marks Guidance 1 EITHER: m (250 – 70) = 450 × 0⋅04 M1 A1 Find m from change in momentum = Ft m = 18 180 = 0⋅1 A1 (can allow M1 if 250 + 70 used; similarly below) OR: a = (250 – 70)/0⋅04 [= 4500] (M1) Find deceleration a (ignore sign) m = 450 a = 0⋅1 (M1 A1) Find m from F = ma 3 Question Answer Marks Guidance 2(i) vA 2 = ω 2 (a2 – 1⋅62) and vB 2 = ω 2 (a2 – 1⋅22) B1 Use v2 = ω 2 (a2 – x2) at A and B (may be implied) 9 16 = ( ) ( ) 2 2 2 2 1 6 1 2 a . a . − − or ( ) ( ) 2 2 2 56 1 44 a . a . − − M1 A1 Find eqn. for a from ratio of vA to vB 7a2 = 40⋅96 – 12⋅96 or 10 × 2⋅8, a2 = 4, a = 2 [m] A1 and hence a (error in 9 16 will lose all A1s, so max 4 11) 4 2(ii) ω = 1 3 a π = 6 π , T = 2 w π = 12 [s] M1 A1 Find ω and hence period T from vmax = ωa and T = 2 w π 2

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 17 Question Answer Marks Guidance 2(iii) ω –1 sin–1 1 6. a       + ω –1 sin–1 1 2. a       [= ω –1 2 π ] or ω –1 cos–1 1 6. a −       – ω –1 cos–1 1 2. a       or 1 2 T – ω –1 cos–1 1 6. a       – ω –1 cos–1 1 2. a       M1 Find time from A to B from x = a sin ωt or a cos ωt All terms must be correct (FT on a, ω) for M1 = ω –1 (0⋅9273 + 0⋅6435) or ω –1 (2⋅498 – 0⋅9273) or 6 – ω –1 (0⋅6435 + 0⋅927) (AEF throughout) A1 = 1⋅771 + 1⋅229 or 4⋅771 – 1⋅771 or 6 – 1⋅229 – 1⋅771, A1 (OR use geometry of circular motion with radius a and angular velocity ω) 3 Question Answer Marks Guidance 3(i) mvA = mu – mku (AEF) M1 Use conservation of momentum (m may be omitted) vA = – e(u + ku) M1 Use Newton’s restitution law – e (1 + k) = 1 – k , e = ( ) ( ) 1 1 k k − + AG A1 Combine to find/verify e 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 17 Question Answer Marks Guidance 3(ii) KEI – KEF = 0⋅6 KEI or 0⋅4 KEI = KEF M1 Attempt to relate initial and final KEs 0⋅4 × 1 2 mu2 (1 + k2) = 1 2 mvA 2 = 1 2 mu2 (1 – k)2 M1 Substitute for KEs ( 1 2 m may be omitted) 0⋅6k2 – 2k + 0⋅6 = 0 or 3k2 – 10k + 3 = 0 (AEF) A1 Simplify to quadratic eqn in k (3 k – 1) ( k – 3) = 0 so solutions are 3 and ⅓ A1 Find both solutions of quadratic eqn [Reject k = 1 3 ] so k = 3 A1 (Implicitly) reject one solution to find possible value of k since 0 < e < 1 or vA < 0 e = 2 4 = 1 2 (FT on k) A1√ Find single corresponding value of e 6 Question Answer Marks Guidance 4(i) RB 2a sin θ = Wa cos θ + 5Wxa cos θ M1 A1 Moments at A RB = W ( ) 5 1 2tan x + θ AG A1 Simplify to give RB 3 4(ii) RB ′ = W ( ) 5 5 1 2tan x + + θ (AEF) B1 Find new RB ′ by moments at A 5x + 6 = 2 (5x + 1) , x = 4 5 AG M1 A1 Find/verify x from RB ′ = 2 RB 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 17 Question Answer Marks Guidance 4(iii) RA = 6W, FA ′ = RB ′ or 2 RB [= 10 2tan W θ = 4W] B1, B1 Find RA and new FA ′ by resolutions (may be earned earlier) 2 3 = A A F R ′ M1 Relate RA and FA ′ using µ = F R = 5 tan 6 W W       θ so tan θ = 5 4 or 1⋅25 M1 A1 Find tan θ SC Allow M1 M1 for ground smooth and wall rough (max 2 5 ) 5

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 10 of 17 Question Answer Marks Guidance 5(i) IA = 1 2 3M (2a)2 [= 6 Ma2] *M1 Find MI of A // to axis l at A’s centre by ⊥ axis theorem IA ′ = IA + 3M (2a)2 [= 18 Ma2] (dep *M1) M1 A1 Find MI of A about axis l IB = 1 2 2M (3a)2 [= 9 Ma2] **M1 Find MI of B (or C ) // to axis l at its centre by ⊥ axis theorem IB ′ = IB + 2M (6a)2 [= 81 Ma2] (dep **M1) M1 A1 Find MI of B (or C ) about axis l I = (18 + 2 × 81) Ma2 = 180 Ma2 AG A1 Verify MI of object about axis l (A0 if inadequate explanation) SC: IA ′ = 1 2 {3M (2a)2 + 3M (2a)2 } [= 12 Ma2] (M1) SC: Invalidly applying theorems in wrong order IB ′ = 1 2 {2M (3a)2 + 2M (6a)2 } [= 45 Ma2] (M1) (max 2 7 ) 7

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 11 of 17 Question Answer Marks Guidance 5(ii) [–] I 2 2 d dt θ = 3Mg × 2a sin θ + 2 × 2Mg × 6a sin θ or 7Mg × 30 7 a sin θ [ = 30Mga sin θ ] *M1 A1 Use eqn of circular motion to find d2θ/dt2 where θ is angle of plane of object with vertical 2 2 d dt θ = 6 g a   −    θ or 0 167 . g a   −    θ (M1 dep *M1) M1 A1 Approximate sin θ by θ to show SHM (no ‘–’ is M1 A0) T = 2 6 g a π       = 6 2 a g   π     or 15⋅4 a g       or 4⋅87√a (AEF) B1√ Find period T from T = 2π/ω (FT on ω 2; requires some simplification) 5

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 12 of 17 Question Answer Marks Guidance 6(i) P(X > 2) = 1 – F(2) = exp (–0⋅8) = 0⋅449 M1 A1 Find P(X > 2). M0 for F(2) 2 6(ii) F(Q) = 1 – exp (–0⋅4x) = 1 4 or 3 4 M1 Formulate equation for either quartile value Q Q1 = (ln 4/3) / (0⋅4) [= 0⋅7192 ] (AEF) A1 Find one [lower] quartile Q1 Q3 = (ln 4) / (0⋅4) [= 3⋅466] (AEF) A1 Find other [upper] quartile Q3 Q3 – Q1 [= (ln 3) / (0⋅4)] = 2⋅75 A1√ Find interquartile range (FT on Q1, Q3; allow Q1 – Q3) 4 Question Answer Marks Guidance 7 x = 681⋅6/7 = 97⋅37 (allow 97⋅4 for this B1) B1 Find sample mean s2 = (66 536⋅1 – 681⋅62 /7) / 6 [ = 27⋅96 or 5⋅2872 ] M1 Estimate population variance (allow biased here: 23⋅96 or 4⋅8952) H0: µ = 94, H1: µ ≠ 94 (AEF) B1 State hypotheses (B0 for x …) t = ( x – 94)/(s/√7) = 1⋅69 M1 A1 Find value of t t7, 0.95 = 1⋅94[3] B1 State or use correct tabular t-value (or can compare x with 94 + 3⋅88 = 97⋅88 ) [Accept H0:] Mean mass is equal to 94 kg (AEF) B1√ Consistent conclusion (FT on both t-values) 7

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 13 of 17 Question Answer Marks Guidance 8 H0: Colour is independent of type or no association between colour and type (AEF) B1 State (at least) null hypothesis in full Ei: 47⋅67 43⋅33 39 36⋅67 33⋅33 30 25⋅67 23⋅33 21 (to 3 s.f.) M1 A1 Find expected values Ei (A0 if rounded to integers) X 2 = 0⋅597 + 1⋅241 + 0⋅103 + 1⋅603 + 1⋅333 + 0⋅033 + 0⋅212 + 0⋅019 + 0⋅429 = 5⋅57 (to 3 s.f.) M1 A1 Find value of χ 2 from Σ (Ei – Oi)2 / Ei [or Σ Oi 2 / Ei – n ] (allow 5⋅64 for this A1 if integer values of Ei used) χ4, 0.9 2 = 7⋅779 or 7⋅78 B1 State or use correct tabular χ 2 value Accept H0 if X 2 < tabular value (AEF) M1 Valid method for reaching conclusion 5⋅57 [± 0⋅01] < 7⋅78 so independent or no association (AEF) A1 Correct (abbreviated) conclusion, from approx. correct values 8 Question Answer Marks Guidance 9(i) (1 – p)/p2 = 4/9, 4p2 + 9p – 9 = 0 AG M1 A1 Find given eqn. for p using Var(X) = (1 – p)/p2 (4p – 3)(p + 3) = 0, p = ¾ M1 A1 Solve quadratic for p (A0 if p = –3 not [implicitly] rejected) 4 9(ii) P(X = 3) = (1 – p)2 p = (¼)2 ¾ = 3/64 or 0⋅0469 B1 Find P(X = 3) 1

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 14 of 17 Question Answer Marks Guidance 9(iii) P(X ⩾ 4) = (1 – p)3 = (¼)3 or 1 – (1 + ¼ + (¼)2) ¾ = 1/64 or 0⋅0156 M1 A1 Find P(X ⩾ 4) 2 9(iv) 1 – (1 – p)N > 0⋅999 M1 Formulate condition for N ( 1 – (1 – p)N – 1 is M0 ) 0⋅001 > (¼) N A1 (< or = can earn M1 M1 only, max 2/4) N > log 0⋅001 / log 0⋅25 M1 Rearrange and take logs (any base) to give bound N > 4⋅98, Nmin = 5 A1 Find Nmin 4

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 15 of 17 Question Answer Marks Guidance 10(i) H0: µ P = µ Q , H1: µ P > µ Q B1 State hypotheses (B0 for x …) sP 2 = (138 200 – 26202/50) / 49 = 18⋅61 or 912/49 and sQ 2 = (157 000 – 30602/60) / 59 = 15⋅93 or 940/59 M1 A1 Estimate both popn. variances (to 3 s.f.) (allow biased here: 18⋅24 and 15⋅67) s2 = sP 2/50 + sQ 2/60 = 0⋅6378 or 0⋅79862 M1 A1 Estimate combined variance (to 3 s.f.; may be implicit) z 0.9 = 1⋅28[2] *B1 State or use correct tabular z value z = ( y – x ) / s = (52⋅4 – 51) / s = 1⋅75 z > tabular value so [accept H1 and] M1 A1 Calculate value of z (or –z) (or can compare y – x = 1⋅02 with 1⋅4) college P students take more time (AEF) B1√ Correctly stated conclusion (FT on z, dep *B1) [SC: s2 = (912 + 940)/108 = 17⋅15 or 4⋅1412 z = 1⋅4 / s√(1/50 + 1/60) = 1⋅77 ] SC: Using pooled estimate of common variance can earn B1 M1 A1 (may be implied) M0 *B1 M0 B1√ (max 5/9) 9 10(ii) x – y ± z s (or y – x ± z s) M1 Find confidence interval for difference z0.95 = 1⋅64[5] A1 Use appropriate tabular value 1⋅4 ± 1⋅31 or [0⋅09, 2⋅71] or – 1⋅4 ± 1⋅31 or [–2⋅71, –0⋅09] [SC: e.g. x – y ± z s√(1/50 + 1/60) = 1⋅4 ± 1⋅30 ] A1 Evaluate confidence interval (either form) SC: Allow M1 A1 if common variance used (max 2/3) 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 16 of 17 Question Answer Marks Guidance 11A(i) ½mv2 = ½mu2 + mga cos θ and M1 Use conservation of energy (A0 if no m) and T = mv2/a + mg cos θ T = mu2/a + 3mg cos θ M1 A1 find tension T at P1 by using F = ma (A1 for both correct) Combine to verify cos θ (11/5)mg = 2mg/5 + 3mg cos θ , cos θ = 3/5 AG M1 A1 by substituting T = (11/5) mg and u = √(2ag/5) v2 = u2 + 2ag cos θ = (2/5 + 6/5)ag = 8ag/5 v = √(8ag/5) or √ (1⋅6ag) or 4√a (AEF) B1 Find v at P1 (can assume cos θ = 3/5) 6 11A(ii) t = (a sin θ ) / (v cos θ ) = 4a/3v [= ⅓√a] M1 A1 Find time t to P2 by considering horizontal motion h = (v sin θ ) t + ½ gt2 = 16a/15 + 5a/9 = 73a/45 or 1⋅62a M1 A1 Find ht. fallen at P2 by considering vertical motion OP2 = h + a cos θ = 20a/9 or 2⋅22a M1 A1 Find OP2 6 11B(i) Σ x = 48, Σ y = 29 + 2p, Σ xy = 242 + 17p, Σ x2 = 450, [Σ y2 = 219 + 2p2] Sxy = 242 + 17p – 48 × (29 + 2p)/6 = 10 + p Sxx = 450 – 482/6 = 66 [Syy = 219 + 2p2 – (29 + 2p)2/6 = (8p2 – 116p + 473)/6] M1 A1 Find required values 0⋅25 = Sxy / Sxx = (10 + p)/66, p = 6⋅5 M1 A1 Find p from gradient in eqn. of regression line (29 + 2p)/6 = 0⋅25 × 48/6 + k, k = 7 – 2 = 5 M1 A1 Find k from means and regression line 6

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 17 of 17 Question Answer Marks Guidance 11B(ii) [Σ y = 42, Σ y2 = 303⋅5, Syy = 303⋅5 – 422/6 = 9⋅5] r = Sxy / √(Sxx Syy) = 16⋅5 / √(66 × 9⋅5) = 0⋅659 M1 *A1 Find correlation coefficient r 2 11B(iii) H0: ρ = 0, H1: ρ > 0 B1 State both hypotheses (B0 for r …) EITHER: r6, 5% = 0⋅729 *B1 State or use correct tabular one-tail r-value Accept H0 if r < tab. r-value (AEF) M1 State or imply valid method for conclusion OR: tr = r√((n – 2) / (1 – r2)) = 1⋅75, t4,0⋅95 = 2⋅132 (*B1) (Rarely seen) Accept H0 if rt < tab. t-value (AEF) (M1) No positive correlation (AEF) A1 Correct conclusion (dep *A1, *B1) 4