Cambridge A Level Mathematics 9709 — 2016 May/June Paper 7 · Variant 1

*4719058569* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/71 Paper 7 Probability & Statistics 2 (S2) May/June 2016 1 hour 15 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF9) 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. The use of an electronic calculator is expected, where appropriate. 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. The total number of marks for this paper is 50. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. This document consists of 3 printed pages and 1 blank page. JC16 06_9709_71/2R © UCLES 2016 [Turn over

2 1 A six-sided die shows a six on 25 throws out of 200 throws. Test at the 10% significance level the null hypothesis: P(throwing a six) = 1 6, against the alternative hypothesis: P(throwing a six) < 1 6. [5] 2 A researcher is investigating the lengths, in kilometres, of the journeys to work of the employees at a certain firm. She takes a random sample of 10 employees. (i) State what is meant by ‘random’ in this context. [1] The results of her sample are as follows. 1.5 2.0 3.6 5.9 4.8 8.7 3.5 2.9 4.1 3.0 (ii) Find unbiased estimates of the population mean and variance. [3] (iii) State what is meant by ‘population’ in this context. [1] 3 Based on a random sample of 700 people living in a certain area, a confidence interval for the proportion, p, of all people living in that area who had travelled abroad was found to be 0.5672 < p < 0.6528. (i) Find the proportion of people in the sample who had travelled abroad. [1] (ii) Find the confidence level of this confidence interval. Give your answer correct to the nearest integer. [4] 4 In the past, the time spent by customers in a certain shop had mean 12.5 minutes and standard deviation 4.2 minutes. Following a change of layout in the shop, the mean time spent in the shop by a random sample of 50 customers is found to be 13.5 minutes. (i) Assuming that the standard deviation remains at 4.2 minutes, test at the 5% significance level whether the mean time spent by customers in the shop has changed. [5] (ii) Another random sample of 50 customers is chosen and a similar test at the 5% significance level is carried out. State the probability of a Type I error. [1] 5 The thickness of books in a large library is normally distributed with mean 2.4 cm and standard deviation 0.3 cm. (i) Find the probability that the total thickness of 6 randomly chosen books is more than 16 cm. [4] (ii) Find the probability that the thickness of a book chosen at random is less than 1.1 times the thickness of a second book chosen at random. [5] © UCLES 2016 9709/71/M/J/16

3 6 In each turn of a game, a coin is pushed and slides across a table. The distance, X metres, travelled by the coin has probability density function given by fx = T kx22 −x 0 ≤x ≤2, 0 otherwise, where k is a constant. (i) State the greatest possible distance travelled by the coin in one turn. [1] (ii) Show that k = 3 4. [3] (iii) Find the mean distance travelled by the coin in one turn. [3] (iv) Out of 400 turns, find the expected number of turns in which the distance travelled by the coin is less than 1 metre. [3] 7 (a) A large number of spoons and forks made in a factory are inspected. It is found that 1% of the spoons and 1.5% of the forks are defective. A random sample of 140 items, consisting of 80 spoons and 60 forks, is chosen. Use the Poisson approximation to the binomial distribution to find the probability that the sample contains (i) at least 1 defective spoon and at least 1 defective fork, [3] (ii) fewer than 3 defective items. [3] (b) The random variable X has the distribution Po,. It is given that PX = 1 = p and PX = 2 = 1.5p, where p is a non-zero constant. Find the value of , and hence find the value of p. [4] © UCLES 2016 9709/71/M/J/16

4 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 2016 9709/71/M/J/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 6 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/71 Paper 7 May/June 2016 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2016 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 2016 9709 71 © Cambridge International Examinations 2016 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 2016 9709 71 © Cambridge International Examinations 2016 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 2016 9709 71 © Cambridge International Examinations 2016 Note: ‘(3 sfs)’ means ‘answer which rounds to … to 3 sfs’. If correct ans seen to > 3sfs, ISW for later rounding. Penalise < 3 sfs only once in paper. 1 B(200, 1 6 ) → N( 100 3 , 250 9 ) 100 3 250 9 25.5− = –1.486 comp ‘1.486’ with 1.282 Evidence to reject H0 There is some evidence that p < 1 6 or, e.g. It is likely that p < 1 6 oe B1 M1 A1 M1 A1 ft [5] seen or implied allow with wrong or no cc (Accept alternative correct methods) or comp (‘1.486’) with 0.1 No contradictions 2 (i) Each employee has an equal chance of being chosen B1 [1] oe (ii) Est (µ) = 4 Est (σ 2) = 2 10 199.22 9 10 ( '4' ) − = 4.36 (3 sf) B1 M1 A1 [3] sub in correct formula attempted working may not be seen (iii) Distances travelled by all employees at the firm B1 [1] oe 3 (i) ((0.5672 + 0.6528) ÷ 2) = 0.61 B1 [1] (ii) ‘0.61’ + z '0.61' (1 '0.61') 350 × − = 0.6528 z = 0.0428 × 700 '0.61' (1 '0.61') × − oe = 2.321 98% confidence M1 M1 A1 A1 ft [4] oe correct rearrangement of correct equn, ft ‘0.61’ ft their z (dep on both Ms)

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9709 71 © Cambridge International Examinations 2016 4 (i) H0: µ = 12.5 H1: µ ≠ 12.5 13.5 12.5 4.2 50 − ÷ = 1.68(4) ‘1.684’ < 1.96 No evidence that mean time has changed B1 M1 A1 M1 A1 ft [5] allow 4.2 ÷ 50 comp 1.96 allow comp 1.645 if H1: µ > 12.5 or comp 1 – (‘1.684’) with 0.025 No contradictions ft their 1.684, but not comp 1.645 (ii) 0.05 B1 [1] 5 (i) T ~ N(6 × 2.4, 6 × 0.32) (= N(14.4, 0.54) 16 '14.4' '0.54' − (= 2.177) 1 – (‘2.177’) = 0.0147 (3 sf) M1 M1 M1 A1 [4] seen or implied ft their E and Var; allow without √ (Accept alternative method N(2.4,(0.3²) / 6)) correct area consistent with their working (ii) D = X1 – 1.1X2 E(D) = –0.24 Var(D) = 0.32 + 1.12 × 0.32 (= 0.1989) 0 ( 0.24) '0.1989' −− (= 0.538) (‘0.538’) = 0.705 (3 sf) B1 M1 M1 M1 A1 [5] ft their E and Var; allow without √ correct area consistent with their working 6 (i) 2 m B1 [1] allow without units (ii) k 2 2 0 (2 )d x x x − ∫ = 1 k 3 4 2 3 4 2 0 x x   −   k × 16 3 4   −   = 1 or k × 4 3 = 1 oe k = 3 4 AG M1 A1 A1 [3] attempt integ f(x) and ‘= 1’. Ignore limits correct integration and limits No errors seen (iii) 3 4 2 3 0 (2 )d x x x − ∫ = 3 4 × 4 5 2 4 5 2 0 x x   −   1.2 m oe M1 A1 A1 [3] attempt integ xf(x), condone missing k correct integration and limits, condone missing k allow without units

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9709 71 © Cambridge International Examinations 2016 (iv) 3 4 1 2 0 (2 )d x x x − ∫ (= 3 4 × ( 2 3 −1 4 )) = 5 16 or 0.3125 oe 400 × 5 16 = 125 M1 A1 A1 ft [3] attempt integ f(x), 0 to 1, condone missing k ft their 5 16 7 (a) (i) 0.01 × 80 and 0.015 × 60 (1 – e–0.8) × (1 – e–0.9) = 0.327 (3 sf) M1 M1 A1 [3] (1 – e–λ) × (1 – e–µ) any λ, µ (λ ≠ µ) allow one end error (ii) λ = 0.02 × 40 + 0.015 × 60 e–1.7 × (1 + 1.7 + 2 1.7 2 ) = 0.757 (3 sf) M1 M1 A1 [3] or their 0.8 + 0.9 (b) e–λ × λ = p and e–λ × 2 2 λ = 1.5p λ = 3 p = e–3 × 3 = 0.149 (3 sf) M1 A1 M1 A1 [4] or e–λ × 2 2 λ = 1.5 × e–λ × λ seen or implied their λ [Total for paper 50]