Cambridge A Level Mathematics - Further 9231 — 2018 Oct/Nov Paper 2 · Variant 1

*2427535828* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/21 Paper 2 October/November 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 11_9231_21/RP © UCLES 2018 [Turn over

2 1 A particle P oscillates in simple harmonic motion between the points A and B, where AB = 6 m. The period of the motion is 1 20 s. Find the speed of P when it is 2 m from B. [3] … … … … … … … … … … … … 2 Two uniform small smooth spheres A and B have equal radii and masses 5m and 2m respectively. Sphere A is moving with speed u on a smooth horizontal surface when it collides directly with sphere B which is moving towards it with speed 2u. The coefficient of restitution between the spheres is e. (i) Show that the speed of B after the collision is 1 7u1 + 15e and find an expression for the speed of A. [4] … … … … … … … … … © UCLES 2018 9231/21/O/N/18

3 In the collision, the speed of A is halved and its direction of motion is reversed. (ii) Find the value of e. [2] … … … … … … … … … (iii) For this collision, find the ratio of the loss of kinetic energy of A to the loss of kinetic energy of B. [3] … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

4 3 A B a x 6a A uniform disc, of radius a and mass 2M, is attached to a thin uniform rod AB of length 6a and mass M. The rod lies along a diameter of the disc, so that the centre of the disc is a distance x from A (see diagram). (i) Find the moment of inertia of the object, consisting of disc and rod, about a fixed horizontal axis l through A and perpendicular to the plane of the disc. [4] … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18

5 The object is free to rotate about the axis l. The object is held with AB horizontal and is released from rest. When AB makes an angle 1 with the vertical, where cos 1 = 3 5, the angular speed of the object is O@2g 5a A . (ii) Find the possible values of x. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

6 4 A uniform rod AB of length 4a and weight W is smoothly hinged to a vertical wall at the end A. The rod is held at an angle 1 above the horizontal by a light elastic string. One end of the string is attached to the point C on the rod, where AC = 3a. The other end of the string is attached to a point D on the wall, with D vertically above A and such that angle ACD = 21. A particle of weight 1 2W is attached to the rod at B. It is given that tan 1 = 8 15. (i) Show that the tension in the string is 17 12W. [4] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18

7 (ii) Find the magnitude and direction of the reaction at the hinge. [5] … … … … … … … … … … … … … … … … (iii) Given that the natural length of the string is 2a, find its modulus of elasticity. [2] … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

8 5 The fixed points A and B are on a smooth horizontal surface with AB = 2.6 m. One end of a light elastic spring, of natural length 1.25 m and modulus of elasticity , N, is attached to A. The other end is attached to a particle P of mass 0.4 kg. One end of a second light elastic spring, of natural length 1.0 m and modulus of elasticity 0.6, N, is attached to B; its other end is attached to P. The system is in equilibrium with P on the surface at the point E. (i) Show that AE = 1.4 m. [4] … … … … … … … … … … … The particle P is now displaced slightly from E, along the line AB. (ii) Show that, in the subsequent motion, P performs simple harmonic motion. [5] … … … … … … … … … © UCLES 2018 9231/21/O/N/18

9 … … … … … … … … … … (iii) Given that the period of the motion is 1 70 s, find the value of ,. [3] … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

10 6 The continuous random variable X has probability density function f given by fx = t 1 80 @ 3x −8 x A 4 ≤x ≤16, 0 otherwise. (i) Find the distribution function of X. [3] … … … … … … … … … … The random variable Y is defined by Y = ïX. (ii) Find the probability density function of Y. [3] … … … … … … … … … … © UCLES 2018 9231/21/O/N/18

11 7 The random variable T is the lifetime, in hours, of a particular type of battery. It is given that T has a negative exponential distribution with mean 500 hours. (i) Write down the probability density function of T. [1] … … … … (ii) Find the probability that a randomly chosen battery of this type has a lifetime of more than 750 hours. [3] … … … … … … … (iii) Find the median value of T. [3] … … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

12 8 The weekly salaries of employees at two large electronics companies, A and B, are being compared. The weekly salaries of an employee from company A and an employee from company B are denoted by $x and $y respectively. A random sample of 50 employees from company A and a random sample of 40 employees from company B give the following summarised data. Σx = 5120 Σx2 = 531 000 Σy = 3760 Σy2 = 375 135 (i) The population mean salaries of employees from companies A and B are denoted by $-A and $-B respectively. Using a 5% significance level, test the null hypothesis -A = -B against the alternative hypothesis -A ≠-B. [8] … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18

13 … … … … … … … … … … … … … … … … … … (ii) State, with a reason, whether any assumptions about the distributions of employees’ salaries are needed for the test in part (i). [1] … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

14 9 There are a large number of students at a particular college. The heights, in metres, of a random sample of 8 students are as follows. 1.75 1.72 1.62 1.70 1.82 1.75 1.68 1.84 You may assume that heights of students are normally distributed. (i) Test, at the 5% significance level, whether the population mean height of students at this college is greater than 1.70 metres. [7] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18

15 … … … … … … … … … … … (ii) Find a 95% confidence interval for the population mean height of students at this college. [3] … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

16 10 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 = 1.1664 + 0.4604x. It is given that Σx2 = 1419.98 and Σy2 = 439.68. The mean value of y is 6.24. (i) Find the equation of the regression line of x on y. [6] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18

17 (ii) Find the product moment correlation coefficient. [2] … … … … … … … … … (iii) Test at the 5% significance level whether there is evidence of positive correlation between the two variables. [4] … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

18 11 Answer only one of the following two alternatives. EITHER A particle P of mass m is free to move on the smooth inner surface of a fixed hollow sphere of radius a. The centre of the sphere is O and the point C is on the inner surface of the sphere, vertically below O. The points A and B on the inner surface of the sphere are the ends of a diameter of the sphere. The diameter AOB makes an acute angle ! with the vertical, where cos ! = 4 5, with A below the horizontal level of B. The particle is projected from A with speed u, and moves along the inner surface of the sphere towards C. The normal reaction forces on the particle at A and C are in the ratio 8 : 9. (i) Show that u2 = 4ag. [6] … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18

19 (ii) Determine whether P reaches B without losing contact with the inner surface of the sphere. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/18 [Turn over

20 OR A machine is used to produce metal rods. When the machine is working efficiently, the lengths, x cm, of the rods have a normal distribution with mean 150 cm and standard deviation 1.2 cm. The machine is checked regularly by taking random samples of 200 rods. The latest results are shown in the following table. Interval 146 ≤x < 147 147 ≤x < 148 148 ≤x < 149 149 ≤x < 150 Observed frequency 1 2 23 52 150 ≤x < 151 151 ≤x < 152 152 ≤x < 153 153 ≤x < 154 69 36 15 2 As a first check, the sample is used to calculate an estimate for the mean. (i) Show that an estimate for the mean from this sample is close to 150 cm. [2] … … … … As a second check, the results are tested for goodness of fit of the normal distribution with mean 150 cm and standard deviation 1.2 cm. The relevant expected frequencies, found using the normal distribution function given in the List of Formulae (MF10), are shown in the following table. Interval x < 147 147 ≤x < 148 148 ≤x < 149 149 ≤x < 150 Observed frequency 1 2 23 52 Expected frequency 1.24 8.32 30.94 59.50 150 ≤x < 151 151 ≤x < 152 152 ≤x < 153 153 ≤x 69 36 15 2 59.50 30.94 8.32 1.24 (ii) Show how the expected frequency for 151 ≤x < 152 is obtained. [3] … … … … … © UCLES 2018 9231/21/O/N/18

21 (iii) Test, at the 5% significance level, the goodness of fit of the normal distribution to the results. [7] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/21/O/N/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 9231/21/O/N/18

23 BLANK PAGE © UCLES 2018 9231/21/O/N/18

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 9231/21/O/N/18

This document consists of 14 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/21 Paper 2 October/November 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 October/November 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 2 of 14 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.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 3 of 14 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.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 4 of 14 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 – 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) Question Answer Marks Guidance 1 ω = 2π/T = 4 B1 Find ω from period T (may be implied) v = ω√(a2 – x2) = 4√(32 – 12) = 8√2 or 11.3 [m s–1] M1A1 Find speed v when BP = 2 3 Question Answer Marks Guidance 2(i) 5mvA + 2mvB = 5mu – 4mu = mu (AEF) M1 Use momentum (allow m omitted) vB – vA = e(u + 2u) = 3eu M1 Use Newton’s law (M0 if LHS signs inconsistent) vA = (u/7) (1 – 6e) A1 Combine to find/verify speeds of A and B after colln. vB = (u/7) (1 + 15e) AG A1 (ignore signs) 4

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 14 Question Answer Marks Guidance 2(ii) (u/7) (1 – 6e) = – ½ u, e = ¾ or 0.75 M1A1 Combine to find e from vA = – ½ u (M0 if dirn. of motion not reversed) 2 2(iii) KEA = ½ × 5m {u2 – (½ u)2} and [= (15/8)mu2] KEB = ½ × 2m {(2u)2 – (7u/4)2} [= (15/16)mu2] M1A1 Find loss of KE for A and B KEA / KEB = (15/8) / (15/16) = 2:1 or 2/1 or 2 A1 Combine to find ratio 3 Question Answer Marks Guidance 3(i) IAB = ⅓ M (3a)2 + M (3a)2 or (4/3) M (3a)2 [= 12 Ma2] B1 Find or state MI of rod AB about axis l Idisc = ½ 2 M a2 + 2M x2 [= Ma2 + 2M x2] M1A1 Find MI of disc about axis l I = 13 M a2 + 2M x2 A1 Find MI of object about axis l 4 3(ii) ½ I ω2 = Mg × 3a cos θ + 2Mg x cos θ M1A1 Find ω2 or angular speed ω at angle θ by energy, with cos θ = 3/5 ω2 = 6(3a + 2x)g / 5(13 a2 + 2x2) A1 6a(3a + 2x) = 2(13 a2 + 2x2) x2 – 3ax + 2a2 = 0, x = a or 2a M1A1 Equate ω2 to 2g/5a and solve quadratic for x 5

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 6 of 14 Question Answer Marks Guidance 4(i) T × 3a sin 2θ = W × 2a cos θ + ½ W × 4a cos θ M1A1 Take moments for rod about A 6a T sin θ cos θ = 4a W × a cos θ M1 Verify tension T T = 2W / 3 sin θ = 2W / 3(8/17) = 17W / 12 AG A1 using sin θ = 8/17, cos θ = 15/17 4 4(ii) EITHER: X = T cos θ = (5/4) W or 1.25W B1 Find horizontal component X of force at A Y = W + ½ W – T sin θ = (5/6) W or 0.833W B1 Find vertical component Y of force at A F = √(X2 + Y2) = (5√13 / 12) W or 1.50W B1 Find magnitude of F Upward force at angle to horizontal of tan–1 Y/X = tan–1 2/3 = 33.7° or 0.588 radians M1A1 Find direction of F (AEF; A0 if direction unclear) OR: RP = 1.5 W sin θ + T cos 2θ = (305/12 × 17) W or 1.45W (B1 Find component RP parallel to AB of force at A RN = 1.5 W cos θ – T cos 2θ = (5/34) W or 0.147W B1 Find component RN normal to AB of force at A F = √( RP 2 + RN 2) = (5√13 / 12) W or 1.50W B1 Find magnitude of F Upward force at angle to AB of tan–1 RN /RP = tan–1 6/61 = 5.6° or 0.098 radians M1A1) Find direction of F (AEF; A0 if direction unclear) 5

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 14 Question Answer Marks Guidance 4(iii) T = λ (CD – 2a)/2a M1 Find modulus λ using Hooke’s Law CD = 3a so λ = 2T = 17W / 6 or 2.83W A1 2 Question Answer Marks Guidance 5(i) λ (AE – 1.25) / 1.25 = M1A1 Verify AE by equating equilibrium tensions 0.6λ (2.6 – AE – 1.0) / 1.0 (AEF) A1 SC: vertical motion can earn M1 only AE – 1.25 = 1.2 – 0.75 AE , AE = 1.4 AG A1 4 5(ii) [±] m d2x/dt2 = – λ (0.15 + x) / 1.25 + 0.6λ (0.2 – x) / 1.0 or + λ (0.15 – x) / 1.25 – 0.6λ (0.2 + x) / 1.0 M1A2 Apply Newton’s law at 1.4 + x or 1.4 – x from A (lose A1 for one incorrect tension term) d2x/dt2 = – (1.4 λ / 0.4) x = – 3.5 λx M1A1 Simplify to give SHM eqn. in standard form (A0 if no minus sign, or dirn. of acceln. is undefined) SC: B2 if result stated without derivation (max 2/5) 5 5(iii) ω = 2π / (π/7) [= 14] = √(3.5 λ) M1A1 Find eqn. for λ using T = 2π/ω with FT on ω from SHM eqn. λ = 142 / 3.5 = 56 A1 Solve for λ 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 8 of 14 Question Answer Marks Guidance 6(i) F(x) = ∫ f(x) dx = (1/80) (2x3/2 – 16x1/2) [+ c] M1 Find or state distribution function F(x) for 4 ⩽ x ⩽ 16 using F(4) = 0 or F(16) = 1 to find c if necessary = (1/80) (2x3/2 – 16x1/2 + 16) or (1/40) (x3/2 – 8x1/2 + 8) or x3/2/40 – x1/2/5 + 1/5 (AEF) A1 State F(x) for other values of x F(x) = 0 (x < 4), F(x) = 1 (x > 16) A1 3 6(ii) EITHER: G(y) [= P(Y < y) = P(√X < y) = P(X < y 2) ] = F(y 2) = (1/40) (y 3 – 8y + 8) (AEF) M1A1 Find or state G(y) for 2 ⩽ y ⩽ 4 from Y = √X (allow < or ⩽ throughout; FT on constant term) OR: Use x = y2 to find f(x) = (1/80) (3y – 8/y) and dx/dy = 2y (M1A1) Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy g(y) [= G′(y)] = (1/40) (3y 2 – 8) or (3/40) y 2 – 1/5 [ for 2 ⩽ y ⩽ 4, g(y) = 0 otherwise ] A1 Find g(y) in simplified form for 2 ⩽ y ⩽ 4 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 9 of 14 Question Answer Marks Guidance 7(i) f(t) = (1/500) exp (– t/500) [0 otherwise] (AEF) B1 State pdf of T for t ⩾ 0 1 7(ii) F(t) = 1 – exp (– t/500) M1 Find or imply F(t) P(T > 750) = 1 – F(750) = exp (– 750/500) = 0.223 M1A1 Find P(T > 750) (M0 for F(750)) 3 7(iii) 1 – exp (– m/500) = ½ or exp (– m/500) = ½ M1 Find median value m of T from F(m) = ½ m = 500 ln 2 = 347 [hours] M1A1 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 10 of 14 Question Answer Marks Guidance 8(i) sA 2 = (531 000 – 51202/50) / 49 and [= 6712/49] sB 2 = (375 135 – 37602/40) / 39 [= 21695/39] M1 Estimate both population variances sA 2 = 136.98 and sB 2 = 556.28 (to 3 s.f. throughout) A1 (allow biased here: 134.24 and 542.4) EITHER: s2 = sA 2/50 + sB 2/40 = 16.65 or 4.082 M1A1 EITHER: Estimate combined variance z 0.975 = 1.96 *B1 State or use correct tabular z (or t) value z = ( y – x ) / s = (102.4 – 94) / s = 2.06 z > tabular value [accept H1] M1A1 Calculate value of z (or –z) (or can compare y – x = 8.4 with 8.0) so µA ≠ µB or salaries do differ (AEF) B1 Correct conclusion (FT on z, dep *B1) OR: Assume equal [population] variances s2 = (49 sA 2 + 39 sB 2) / 88 or (531 000 – 51202/50 + 375 135 – 37602/40) / 88 (B1 OR: State assumption about variances (may be in part (ii)) Find pooled estimate of common variance (M1 A1 for sA 2 and sB 2 may be implied here) = 28407/88 or 322.8 or 17.972 B1 z0.975 = 1.96 *B1 State or use correct tabular z (or t) value z = 8.4 / s√(1/50 + 1/40) = 2.20 M1A1 Calculate value of z (or –z) (or can compare y – x = 8.4 with 7.47) z > tabular value [accept H1] so µA ≠ µB or salaries do differ (AEF) B1) Correct conclusion (FT on z, dep *B1) 8

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 11 of 14 Question Answer Marks Guidance 8(ii) No assumption [of normality] needed since large samples or due to central limit theorem B1 Valid comment on required assumptions (AEF) (assumption of equal variances may earn B1 above) 1 Question Answer Marks Guidance 9(i) x = 13.88/8 = 1.735 (allow 1.73 or 1.74 for this B1) B1 Find sample mean s2 = (24.1182 – 13.882 /8) / 7 [ = 13/2500 or 0.0052 or 0.072112 ] M1 Estimate population variance (allow biased here: 0.00455 or 0.067452) H0: µ = 1.7, H1: µ > 1.7 (AEF) B1 State hypotheses (B0 for x …) t = ( x – 1.7) / (s/√8) = 1.37 M1A1 Find value of t t7, 0.95 = 1.89[5] B1 State or use correct tabular t-value (or can compare x with 1.7 + 0.048 = 1.75 ) [Accept H0:] Mean ht. not greater than 1.7 m (AEF) B1 Consistent conclusion (FT on both t-values) 7 9(ii) 13.88/8 ± t √(s2 /8) M1 Find confidence interval t7, 0.975 = 2.36[5] A1 State or use correct tabular value of t 1.73[5] ± 0.06 or [1.67, 1.80] A1 Evaluate confidence interval (either form) 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 12 of 14 Question Answer Marks Guidance 10(i) x = ( y – 1.1664) / 0.4604 = 11.02 or Σ x = 110.2 M1 Find x or Σ x using eqn. of regression line of y on x Sxx = 1419.98 – 110.22/10 = 205.576 Syy = 439.68 – 62.42/10 = 50.304 (AEF) M1 Find Sxx and Syy (may be earned in part (ii)) EITHER: b = Sxy / Syy = (Sxy / Sxx) × (Sxx / Syy) = 0.4604 × 205.576 / 50.304 = 1.88[15] M1A1 Find gradient b to 3 s.f. in x – x = b (y – y ) OR: Sxy = 0.4604 × Sxx = 94.647 b = Sxy / Syy = 94.647 / 50.304 = 1.88[15] (M1A1) (x – 11.02) = b (y – 6.24), x = 1.88y – 0.721 M1A1 and hence eqn. of regression line of x on y 6 10(ii) r = Sxy / √(Sxx Syy) = 94.647 / √(205.576 × 50.304) or √(0.4604 × b) = √(0.4604 × 1.8815) (AEF) M1 Find correlation coefficient r = 0.931 *A1 2 10(iii) H0: ρ = 0, H1: ρ > 0 B1 State both hypotheses (B0 for r …) EITHER: r10, 5% = 0.549 *B1 State or use correct tabular one-tail r-value Reject H0 if r > tab. r-value (AEF) M1 State or imply valid method for conclusion OR: tr = r√((n–2) / (1 – r2)) = 7.21, t8,0.95 = 1.86 (*B1 (Rarely seen) Reject H0 if | tr | > tab. t-value (AEF) M1) Evidence of positive correlation (AEF) A1 Correct conclusion (dep *A1, *B1) 4

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 13 of 14 Question Answer Marks Guidance 11A(i) ½mvC 2 = ½mu2 + mga (1 – cos α) [vC 2 = u2 + (2/5) ag] M1A1 Find speed vC at C by conservation of energy (A0 if no m) RA = mg cos α + mu2/a B1 Find reaction RA at A by using F = ma RC = mg + mvC 2/a B1 Find reaction RC at C by using F = ma 8 {ag + u2 + (2/5) ag} = 9 {(4/5) ag + u2) M1 Verify u2 using RC / RA = 9/8 u2 = (8 + 16/5 – 36/5) ag = 4ag AG A1 6 11A(ii) ½mvB 2 = ½mu2 – 2mga cos α or ½mvC 2 – mga (1 + cos α) [vC 2 = (22/5) ag] M1 Find speed vB at B by conservation of energy (A0 if no m) vB 2 = (4 – 16/5) ag or (22/5 – 18/5) ag = (4/5) ag *A1 ⩾ 0 so P reaches height of B (AEF; dep *A1) A1 (or valid argument based on RA) RA = m vB 2/a – mg cos α = 4mg/5 – 4mg/5 = 0 M1**A1 Find reaction RB at B by using F = ma ⩾ 0 so P still [just] in contact at B (AEF; dep **A1) A1 6 11B(i) x = (1/200) Σ x f(x) = 30 064 / 200 = 150.3[2] M1A1 Find sample mean to 4 s.f. using mid-interval values (B1 for 29 964 / 200 = 149.8[2] or 30 164 / 200 = 150.8[2]) 2

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 14 of 14 Question Answer Marks Guidance 11B(ii) P(151 ⩽ X < 152) = P((151 – 150)/1.2 ⩽ Z < (152 – 150)/1.2) = P(0.833 ⩽ Z < 1.667) = 0.9522 – 0.7975 [= 0.1547] (to 4 s.f.) M1A1 Find P(151 ⩽ X < 152) E151 = 190.44 – 159.5 or 200 × 0.1547 = 30.94 AG A1 and hence expected frequency for 151 ⩽ x < 152 3 11B(iii) H0: [Normal] distribution fits data (AEF) B1 State (at least) null hypothesis in full Oi: 3 23 52 69 36 17 Ei: 9.56 30.94 59.50 59.50 30.94 9.56 M1 Combine values consistent with all exp. values ⩾ 5 X2 = 4.501 + 2.038 + 0.945 + 1.517 + 0.828 + 5.790 M1 Find value of X2 from Σ (Ei – Oi)2 / Ei [or Σ Oi 2/Ei – n ] = 15.6 (to 3 s.f.) A1 No. n of cells: 8 7 6 5 χn-1, 0.95 2: 14.07 12.59 11.07 9.488 B1 State or use consistent tabular value χn–1, 0..95 2 (to 3 s.f.) [FT on number, n, of cells used to find X2] Accept H1 if X2 > tabular value (AEF) M1 State or imply valid method for conclusion 15.6 [± 0.1] > 11.1 so distn. doesn’t fit [data] (AEF) A1 Conclusion (requires both values correct) 7