Cambridge A Level Mathematics - Further 9231 — 2015 May/June Paper 2 · Variant 3

*4337893235* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2015 3 hours Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. 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. 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 5 printed pages and 3 blank pages. JC15 06_9231_23/RP © UCLES 2015 [Turn over

2 1 Two uniform small smooth spheres, A and B, of equal radii and masses 2 kg and 3 kg respectively, are at rest and not in contact on a smooth horizontal plane. Sphere A receives an impulse of magnitude 8 N s in the direction AB. The coefficient of restitution between the spheres is e. Find, in terms of e, the speeds of A and B after A collides with B. [5] Given that the spheres move in opposite directions after the collision, show that e > 2 3. [1] 2 P u ! A B A uniform sphere P of mass m is at rest on a smooth horizontal table. The sphere is projected along the table with speed u and strikes a smooth vertical barrier A at an acute angle !. It then strikes another smooth vertical barrier B which is at right angles to A (see diagram). The coefficient of restitution between P and each of the barriers is e. Show that the final direction of motion of P makes an angle 1 20 −! with the barrier B and find the total loss in kinetic energy as a result of the two impacts. [7] 3 A particle moves on a straight line AOB in simple harmonic motion, where AB = 2a m. The centre of the motion is O and the particle is instantaneously at rest at A and B. The point M is the mid-point of OB. The particle passes through M moving towards O and next achieves its maximum speed one second later. Find the period of the motion. [4] Find the distance of the particle from O when its speed is equal to one half of its maximum speed. [3] At an instant 2.5 seconds after the particle passes through M moving towards O, the distance of the particle from O is ï2 m. Find, in metres, the amplitude of the motion. [2] © UCLES 2015 9231/23/M/J/15

3 4 30Å A B C D E F 2a 2a 6a The diagram shows a central cross-section CDEF of a uniform solid cube of weight W and with edges of length 2a. The cube rests on a rough horizontal plane. A thin uniform rod AB, of weight W and length 6a, is hinged to the plane at A. The rod rests in smooth contact with the cube at C, with angle CAD equal to 30Å. The rod is in the same vertical plane as CDEF. The coefficient of friction between the plane and the cube is -. Given that the system is in equilibrium, show that - ≥3 25ï3. [6] Find the magnitude of the force acting on the rod at A. [4] 5 P A B O a ka 4a The end B of a uniform rod AB, of mass 3M and length 4a, is rigidly attached to a point on the circumference of a uniform disc. The disc has centre O, mass 2M and radius a, and ABO is a straight line. The disc and the rod are in the same vertical plane. A particle P, of mass M, is attached to the rod at a distance ka from A, where k is a positive constant (see diagram). Show that the moment of inertia of this system, about a fixed horizontal axis l through A perpendicular to the plane of the disc, is 67 + k2Ma2. [5] The system is free to rotate about l and performs small oscillations of period 40 O@a g A . Find the possible values of k. [6] © UCLES 2015 9231/23/M/J/15 [Turn over

4 6 The reliability of the broadband connection received from two suppliers, A and B, is classified as good, fair or poor by a random sample of householders. The information collected is summarised in the following table. Reliability Good Fair Poor Supplier A 65 63 33 B 51 44 44 Test, at the 5% significance level, whether reliability is independent of supplier. [8] 7 For a random sample of 10 observations of pairs of values x, y, the equation of the regression line of y on x is y = 3.25x −4.27. The sum of the ten x values is 15.6 and the product moment correlation coefficient for the sample is 0.56. Find the equation of the regression line of x on y. [7] Test, at the 5% significance level, whether there is evidence of non-zero correlation between the variables. [4] 8 A large number of long jumpers are competing in a national long jump competition. The distances, in metres, jumped by a random sample of 7 competitors are as follows. 6.25 7.01 5.74 6.89 7.24 5.64 6.52 Assuming that distances are normally distributed, test, at the 5% significance level, whether the mean distance jumped by long jumpers in this competition is greater than 6.2 metres. [7] The distances jumped by another random sample of 8 long jumpers in this competition are recorded. Using the data from this sample of 8 long jumpers, a 95% confidence interval for the population mean, - metres, is calculated as 5.89 < - < 6.75. Find the unbiased estimates for the population mean and population variance used in this calculation. [5] 9 The continuous random variable X has probability density function given by fx = D 0 x < 2, ae−x−2 x ≥2, where a is a constant. Show that a = 1. [1] Find the distribution function of X and hence find the median value of X. [5] The random variable Y is defined by Y = eX. Find (i) the probability density function of Y, [4] (ii) PY > 10. [2] © UCLES 2015 9231/23/M/J/15

5 10 Answer only one of the following two alternatives. EITHER P O A u a One end of a light inextensible string of length 3 2a is attached to a fixed point O on a horizontal surface. The other end of the string is attached to a particle P of mass m. The string passes over a small fixed smooth peg A which is at a distance a vertically above O. The system is in equilibrium with P hanging vertically below A and the string taut. The particle is projected horizontally with speed u (see diagram). When P is at the same horizontal level as A, the tension in the string is T. Show that T = 2m a u2 −ag. [3] The ratio of the tensions in the string immediately before, and immediately after, the string loses contact with the peg is 5 : 1. (i) Show that u2 = 5ag. [6] (ii) Find, in terms of m and g, the tension in the string when P is next at the same horizontal level as A. [5] OR The times taken, in hours, by cyclists from two different clubs, A and B, to complete a 50 km time trial are being compared. The times taken by a cyclist from club A and by a cyclist from club B are denoted by tA and tB respectively. A random sample of 50 cyclists from A and a random sample of 60 cyclists from B give the following summarised data. ΣtA = 102.0 Σt2 A = 215.18 ΣtB = 129.0 Σt2 B = 282.3 Using a 5% significance level, test whether, on average, cyclists from club A take less time to complete the time trial than cyclists from club B. [10] A test at the !% significance level shows that there is evidence that the population mean time for cyclists from club B exceeds the population mean time for cyclists from club A by more than 0.05 hours. Find the set of possible values of !. [4] © UCLES 2015 9231/23/M/J/15

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8 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 2015 9231/23/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 2015 series 9231 FURTHER MATHEMATICS 9231/23 Paper 2, maximum raw mark 100 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 23 © Cambridge International Examinations 2015 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 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.

Page 3 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 23 © Cambridge International Examinations 2015 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) 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 MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) 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.

Page 4 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 1 Find speed uA [m s–1] of A after initial impulse: 2uA = 8, uA = 4 B1 For A & B use conservation of momentum, e.g.: 2vA + 3vB = 2uA [= 8] M1 Use Newton’s law of restitution (consistent signs): vB – vA = euA [= 4e] M1 Combine to find vA and vB : vA = (2–3e)uA /5, vB = 2(1+e)uA /5 A1 (AEF) vA = 4(2 – 3e)/5, vB = 8(1 + e)/5 A1 Use vA < 0: 2 – 3e < 0, e > 2/3 AG B1 5 1 6 2 EITHER: Find/ state comps. of speed after 1st colln.: u cos α // to A and eu sin α ⊥ to A M1 A1 Find/ state comps. of speed after 2nd colln.: eu sin α // to B and eu cos α ⊥ to A1 Find final angle with barrier B: AG tan–1 (eu cos α / eu sin α) = ½π – α A1 OR: Relate angle β1 after 1st colln. to α: v1 sin β1 = eu sin α v1 cos β1 = u cos α tan β1 = e tan α (M1 A1) Relate angle β2 after 2nd colln. to β1: v2 sin β2 = ev1 cos β1 v2 cos β2 = v1 sin β1 tan β2 = e / tan β1 (A1) Find final angle β2 with barrier B: AG tan–1 (e / e tan α) = ½π – α (A1) EITHER: Find total loss or gain in KE, e.g.: ½ m{u2 – (eu sin α)2 – (eu cos α)2} M1 A1 Find total loss in KE (A0 if wrong sign): = ½ m (1 – e2) u2 A1 OR: Find or state final speed ufinal : ufinal = e u (B1) Find total loss in KE (A0 if wrong sign): ½ m(u2 – ufinal 2) = ½ m (1 – e2) u2 (M1 A1) 4 3 7 3 State or imply posn. where max. speed next attained: Centre O M1 EITHER: Find ω from SHM eqn. x = a sinωt:, e.g.: 1 = ω –1 sin–1 ½ , ω = π/6 M1 A1 Find period T from T = 2π/ω: [ft on ω] T = 2π/(π/6) = 12 [s] B1 OR: Find ω from SHM eqn. x = acosωt, e.g.: 1 = ω –1 cos–1 0 – ω –1 cos–1 ½ ω = π/2 – π/3 = π/6 M1 A1 Find period T from T = 2π/ω: [ft on ω] T = 2π/(π/6) = 12 [s] B1 OR: Find period T from SHM eqn, e.g.: 1 = T/4 – (T/2π) cos–1 ½ M1 A1 = T (1/4 – 1/6), T = 12 [s] A1 Find distance x from O when speed is half max: ω2 (a2 – x2) = ¼ vmax 2 M1 Use maximum speed: vmax = aω [= πa/6] M1 (note that x is independent of ω) x2 = ¾ a2 x = √3a/2 or a√(3/4) or 0.866a A1 EITHER: Find a from x = a sin ω t: 2.5 = ω –1 sin–1 (√2/a) – ω –1 sin–1 (–½) or 1⋅5 = ω –1 sin–1 (√2/a) sin–1 (√2/a) = π/4, a = 2 [m] M1 A1 OR: Find a from x = cos ω t: 2.5 = ω –1 cos–1 (–√2/a) – ω –1 cos–1 ½ or 1⋅5 = ω –1 cos–1 (–√2/a) – ω –1 cos–1 0 cos–1 (–√2/a) = 3π/4, a = 2 [m] M1 A1 4 3 2 9

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 4 Find reaction at C using moments for rod about A: RC × 2a / sin 30° = W × 3a cos 30° RC = 3√3W / 8 M1 A1 Find friction F on cube by resolving horizontally: F = RC sin 30° [= 3√3W / 16] B1 Find reaction R on cube by resolving vertically: R = W + RC cos 30° [= 25W / 16] B1 Use F Y µR (allow F < µR for M1): AG 3√3/16 Y 25µ /16, µ [ 3√3/25 M1 A1 (M0 if F = µR with no further explanation) Find horizontal force X at A (ignore sign): X = F [= 3√3W / 16] B1 Find vertical force Y at A (ignore sign): Y = 2W – R or W – RC cos 30° [= 7W/16] B1 Find magnitude of force at A: √(X2 + Y2) = (√19/8) W or 0.545W M1 A1 10 5 State or find MI of rod about A: Irod = (1 + ⅓) 3M (2a)2 or ⅓ 3M (4a)2 [= 16Ma2] B1 Find MI of disc about A: Idisc = ½ 2M a2 + 2M (5a)2 [= 51 Ma2] M1 A1 State MI of particle about A: Iparticle = M (ka)2 B1 Sum to find MI of system about A: I = (16 + 51 + k2) Ma2 = (67 + k2) Ma2 AG A1 Use eqn of circular motion to find d2θ/dt2 where [–] I d2θ/dt2 = 3Mg × 2a sin θ θ is angle of AO with vertical: + 2Mg × 5a sin θ + Mg × ka sin θ (A0 if cos θ used) = (16 + k) Mga sin θ M1 A1 [Approximate sin θ by θ and] find ω2 in SHM eqn: ω2 = (16 + k)g / (67 + k2) a M1 A1 (A0 if cos θ ≈ θ used) Equate 2π/ω to 4π√(a/g) to find possible values of k: ω2 = ¼ g/a, 16 + k = ¼(67 +k2) k2 – 4k + 3 = 0 , k = 1, 3 M1 A1 5 6 11 6 State (at least) null hypothesis (AEF): H0: Reliability is independent B1 Find expected values (to 1 d.p.): 62.253 57.423 41.323 (lose A1 if rounded to integers) 53.747 49.577 35.677 M1 A1 Calculate value of χ2: χ2 = 0.1212 + 0.5416 + 1.6764 + 0.1404 + 0.6274 + 1.9417 (allow 5.03 if 1 d.p. exp. values used) = 5⋅05 M1 A1 State or use correct tabular χ2 value (to 3 s.f.): χ2, 0.95 2 = 5.99[1] B1 Valid method for reaching conclusion: Accept H0 if χ2 < tabular value M1 Correct conclusion, from correct values: (AEF) Reliability is indep. of supplier A1 8 8 7 Find gradient b2 of line x on y from b1b2 = r2: 3.25 b2 = 0.562 , b2 = 0.0965 M1 A1 Find y from line y on x: y = 3.25 × (15.6/10) – 4.27 = 0.8 M1 A1 Find regression line of x on y: (x – x = b2 (y – y ) (M0 for y on x) (x – 1.56) = 0.0965 (y – 0.8) M1 A1 x = 0.0965y + 1.48 A1 State both hypotheses (B0 for r …): H0: ρ = 0, H1: ρ ≠ 0 B1 State or use correct tabular two–tail r–value: r10, 5% = 0.632 *B1 State or imply valid method for reaching conclusion: Accept H0 if 0.56 < r–value (AEF) M1 Correct conclusion (AEF, dep *B1): There is no non–zero correlation A1 7 4 11

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 8 Calculate sample mean: x = 45.29 / 7 = 6.47 M1 Estimate population variance: s1 2 = (295.36 – 45.292 /7) / 6 (allow biased here: 0.3334 or 0.57742) = 0.389 or 0.62362 M1 State hypotheses: (AEF; B0 for x ) H0: µ = 6⋅2, H1: µ > 6.2 B1 Calculate value of t (or –t; to 3 s.f.): t = ( x – 6.2)/(s1/√7) = 1.15 M1 A1 State or use correct tabular t–value (to 3 s.f.): t6, 0.95 = 1.94[3] B1 (or can compare x with 6.2 + 0.458 = 6.66) Consistent conclusion (AEF, ft on both t–values): [Accept H0] Popn. mean not greater than 6.2 B1 Find estimate of population mean: ½ (5.89 + 6.75) = 6.32 M1 A1 Find estimate of population variance s2 2: t √(s2 2/8) = ½ (6.75 – 5.89) [= 0.43] M1 SR Allow M1 for t √(s2 2/7), max 4/5 Use of correct tabular value: t7, 0.975 = 2.36[5] A1 (Use of 2.36 may lose final A1) [s2 = 0.5143 or 4√2/11] (to 3 d.p.) s2 2 = 0.2645 or 32/121 A1 7 5 12 9 Find a by equating 2 f ( )d x x ∞∫ to 1: ( 2) 2 e x a ∞ − − −     = a = 1 AG B1 Find or state distribution function F(x) for x [ 2: F(x) = ∫ f(x) dx = – ae–(x –2) + c F(2) = 0 so F(x) = 1 – e–(x –2) M1 A1 State F(x) for x < 2: F(x) = 0 for x < 2 A1 Find median value m from F(m) = ½: 1 – e–(m –2) = ½ , em –2 = 2 m = 2 + ln 2 or 2 – ln ½ or 2.69 M1 A1 1 5 (i) Find or state G(y) from Y = eX for x [ 2: G(y) = P(Y < y) = P(eX < y) = P(X < ln y) = F(ln y) = 1 – e–(ln y –2) = 1 – e2/y M1 A1 Find g(y) by differentiation (allow e2 = 7⋅39): g(y) = e2/y2 A1 State corresponding range of y: for y [ e2 [g(y) = 0 for y < e2] A1 4 (ii) Find P(Y > 10): P(Y > 10) = 1 – G(10) M0 for G(10) or 1 – g(10) = e2/10 or 0.739 M1 A1 2 12 10A Use conservation of energy when P is level with A: ½mv1 2 = ½mu2 – ½mga B1 Use F = ma radially at this point: T1 = mv1 2/½ a B1 Eliminate v1 2 to find T1: T1 = (2m/a) (u2 – ag) AG B1 3 (i) State or imply that string loses contact when vertical M1 Use conservation of energy at this point: ½mv2 2 = ½mu2 – mga B1 Use F = ma radially just before this point: T2 = mv2 2/½ a – mg B1 Use F = ma radially just after this point: T3 = mv2 2/(3a/2) – mg B1 EITHER: Find v2 2 and hence u2 from T2/T3 = 5: mv2 2/½ a – mg = 5{mv2 2/(3a/2) – mg}, v2 2 = 3ag u2 = 3ag + 2ag = 5ag AG M1 A1 OR: Eliminate v2 2 in T2/T3 = 5 and find u2: (m/a) (2u2 – 5ag) = 5(m/a) (2u2/3 – 7ag/3) 4u2/3 = 20ag/3, u2 = 5ag AG (M1 A1) 6

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total (ii) Find cos θ, where θ = ∠AOP, when P is level with A: cos θ = a/(3a/2) = 2/3 B1 Use conservation of energy at this point: ½mv4 2 = ½mu2 – ½mga or ½mu2 – mg{(3a/2) cos θ – ½ a} or ½mv2 2 + ½mga or ½mv2 2 + mg (3a/2)(1 – cos θ) B1 [v4 2 = 4ag] Use F = ma radially at this point: T4 = mv4 2/(3a/2) – mg cos θ B1 Eliminate v4 2 to find T4: T4 = 4mag × 2/3a – 2mg/3 = 8mg/3 – 2mg/3 = 2mg M1 A1 5 14 10B State hypotheses (B0 for At …), e.g.: H0: µA = µB , H1: µA < µB B1 Estimate both popn. variances (to 3 d.p..): sA 2 = (215.18 – 1022/50) / 49 and sB 2 = (282.3 – 1292/60) / 59 M1 A1 (allow biased here: 0.1420 or 0.37682 sA 2 = 0.1449 or 71/490 or 0.38072 and 0.0825 or 0.28722) and sB 2 = 0.0839 or 99/1180 or 0.28972 A1 EITHER: Estimate combined variance: s2 = sA 2/50 + sB 2/60 M1 (to 3 s.f.) = 0.004296 or 0.065552 A1 Calculate value of z (or –z): z = (tB – tA) / s (to 3 s.f.) = (2.15 – 2.04) / s = 1.68 M1 A1 OR: Pooled estimate of common variance: s2 = (49 sA 2 + 59 sB 2) / 108 (note sA 2 and sB 2 not needed explicitly) or (215.18 – 1022/50 + 282.3 – 1292/60) / 108 (to 3 s.f.) = 0.1116 or 0.3342 (M1 A1) Calculate value of z (or –z): z = (tB – tA) / s √(1/50 + 1/60) (to 3 s.f.) = 0.11 / 0.06396 = 1⋅72 (M1 A1) State or use correct tabular z value: (to 3 s.f.) z 0.95 = 1.645(allow 1.658) *B1 (or can compare Bt – At = 0.11 with 0.105) Correct conclusion (AEF, ft on z, dep *B1): z > tabular value so club A takes less time B1 Calculate value of z (either sign): z = (0.11 – 0.05) /s = 0.06 /s (to 3 s.f.) = 0.9154[or 0.9381] M1 A1 Find Φ(z) and set of possible values of α : Φ(z) = 0.820[or 0.826] (M0 for α ~ 82 or 0.82) (to 1 d.p.) α [ (or >) 18.0[or 17.4] M1 A1 SR Allow B1 (max 3/4) for α < 18.0 [or 17.4] 10 4 14