Cambridge A Level Mathematics 9709 — 2016 May/June Paper 7 · Variant 2

*1157967783* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/72 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_72/3R © UCLES 2016 [Turn over

2 1 The length of time, in minutes, taken by people to complete a task has mean 53.0 and standard deviation 6.2. Find the probability that the mean time taken to complete the task by a random sample of 50 people is more than 51 minutes. [4] 2 Jacques is a chef. He claims that 90% of his customers are satisfied with his cooking. Marie suspects that the true percentage is lower than 90%. She asks a random sample of 15 of Jacques’ customers whether they are satisfied. She then performs a hypothesis test of the null hypothesis p = 0.9 against the alternative hypothesis p < 0.9, where p is the population proportion of customers who are satisfied. She decides to reject the null hypothesis if fewer than 12 customers are satisfied. (i) In the context of the question, explain what is meant by a Type I error. [1] (ii) Find the probability of a Type I error in Marie’s test. [3] 3 (i) Give a reason for using a sample rather than the whole population in carrying out a statistical investigation. [1] (ii) Tennis balls of a certain brand are known to have a mean height of bounce of 64.7 cm, when dropped from a height of 100 cm. A change is made in the manufacturing process and it is required to test whether this change has affected the mean height of bounce. 100 new tennis balls are tested and it is found that their mean height of bounce when dropped from a height of 100 cm is 65.7 cm and the unbiased estimate of the population variance is 15 cm2. (a) Calculate a 95% confidence interval for the population mean. [3] (b) Use your answer to part (ii)(a) to explain what conclusion can be drawn about whether the change has affected the mean height of bounce. [1] 4 At a certain company, computer faults occur randomly and at a constant mean rate. In the past this mean rate has been 2.1 per week. Following an update, the management wish to determine whether the mean rate has changed. During 20 randomly chosen weeks it is found that 54 computer faults occur. Use a suitable approximation to test at the 5% significance level whether the mean rate has changed. [6] 5 Each box of Fruity Flakes contains X grams of flakes and Y grams of fruit, where X and Y are independent random variables, having distributions N400, 50 and N100, 20 respectively. The weight of each box, when empty, is exactly 20 grams. A full box of Fruity Flakes is chosen at random. (i) Find the probability that the total weight of the box and its contents is less than 530 grams. [5] (ii) Find the probability that the weight of flakes in the box is more than 4.1 times the weight of fruit in the box. [5] © UCLES 2016 9709/72/M/J/16

3 6 At a certain shop the demand for hair dryers has a Poisson distribution with mean 3.4 per week. (i) Find the probability that, in a randomly chosen two-week period, the demand is for exactly 5 hair dryers. [3] (ii) At the beginning of a week the shop has a certain number of hair dryers for sale. Find the probability that the shop has enough hair dryers to satisfy the demand for the week if (a) they have 4 hair dryers in the shop, [2] (b) they have 5 hair dryers in the shop. [2] (iii) Find the smallest number of hair dryers that the shop needs to have at the beginning of a week so that the probability of being able to satisfy the demand that week is at least 0.9. [3] 7 (a) x y −4 −3 −2 −1 0 1 2 3 4 5 6 The diagram shows the graph of the probability density function of a variable X. Given that the graph is symmetrical about the line x = 1 and that P0 < X < 2 = 0.6, find PX > 0. [2] (b) A flower seller wishes to model the length of time that tulips last when placed in a jug of water. She proposes a model using the random variable X (in hundreds of hours) with probability density function given by fx = T k2.25 −x2 0 ≤x ≤1.5, 0 otherwise, where k is a constant. (i) Show that k = 4 9. [3] (ii) Use this model to find the mean number of hours that a tulip lasts in a jug of water. [4] The flower seller wishes to create a similar model for daffodils. She places a large number of daffodils in jugs of water and the longest time that any daffodil lasts is found to be 290 hours. (iii) Give a reason why fx would not be a suitable model for daffodils. [1] (iv) The flower seller considers a model for daffodils of the form gx = T ca2 −x2 0 ≤x ≤a, 0 otherwise, where a and c are constants. State a suitable value for a. (There is no need to evaluate c.) [1] © UCLES 2016 9709/72/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 9709/72/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 9709/72 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 72 © 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 72 © 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 72 © Cambridge International Examinations 2016 1 6.2 50 or 2 6.2 50 51 53 6.2 50 − ÷ (= –2.281) P(z > ‘–2.281’) = φ(‘2.281’) = 0.989 (3 sf) B1 M1 M1 A1 [4] seen or implied allow without ÷√50 for finding correct area consistent with working as final answer 2 (i) Conclude less than 90% satisfied when this is not true oe B11 In context (ii) 1 – (0.915 + 15 × 0.914 × 0.1 + 15C2 × 0.913 × 0.12 + 15C3 × 0.912 × 0.13 ) = 0.0556 (3 sf) or 0.0555 M1 M1 A1 [3] Attempt (1–)P(X = 15,14,13,12) allow 1 end error Attempt fully correct expression 3 (i) Pop too big or takes too long oe or testing destroys articles oe B1 [1] or too expensive oe or pop inaccessible oe (ii) (a) z = 1.96 65.7 ± z × 15 10 = 64.9 to 66.5 (3 sf) B1 M1 A1 [3] seen Expression of correct form (must be ‘z’ must be 65.7) Must be an interval (b) CI does not include 64.7 Probably has affected (or increased) mean bounce ht. B1 [1] allow 64.7 not within CI both needed. ft their CI ft 65.7 / 64.7 mix 4 H0: λ (or µ) = 42 H1: λ (or µ) ≠ 42 Po(42) ~ N(42, 42) stated or implied 53.5 42 42 − = 1.77(4) (or 0.038 for area comparison) comp 1.96 No evidence that mean has changed B1 B1 M1 A1 M1 A1 [6] Or pop weekly mean = 2.1 etc. allow ‘population mean’ not just ‘mean’ ft their ‘42’ (Accept alt method N(2.1,2.1 / 20) allow with wrong or no cc. Accept alt method using N(2.1,2.1 / 20) with or without cc Valid comp z or 1 – (‘1.774’) with 0.025 seen allow comp 1.645 if H1: λ (or µ) > 42 No contradictions. No ft for H1: λ (or µ) > 42 Note – accept other valid methods(e.g. cv method)

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9709 72 © Cambridge International Examinations 2016 5 (i) T ~ N(520, 70) 530 520 '70' − (= 1.195) (‘1.195’) = 0.884 (3 sf) B1 B1 M1 M1 A1 [5] for N(520,..) or N(500,..) if standardising with 510 for Var = 70 seen or implied ft their E and Var; allow without √ finding correct area consistent with working CWO (ii) E(T) = –10 Var(T) = 50 + 4.12 × 20 (= 386.2) 0 ( 10) '386.2' −− (= 0.509) 1 – (‘0.509’) = 0.305 (3 sf) B1 B1 M1 M1 A1 [5] or +10 for T < 0 Seen or implied ft their E and Var; allow without √ finding correct area consistent with working CWO 6 (i) λ = 6.8 e–6.8 × 5 6.8 5! = 0.135 (3 sf) B1 M1 A1 [3] any λ (ii) (a) e-3.4(1 + 3.4 + 2 3 4 3.4 3.4 3.4 2 3! 4! + + ) = 0.744 (3 sf) M1 A1 [2] any λ, allow one end-error (b) ‘0.744’ + e–3.4 × 5 3.4 5! = 0.87(0) (3 sf) or 0.871 M1 A1 [2] or complete method, any λ, allow one end-error (iii) P(X ⩽ 6) = ‘0.870’ + e–3.4 × 6 3.4 6! = 0.94 Need 6 hair driers M1 A1 A1 [3] or complete method, any λ fully correct un-simplified expression or better dep M1A1 with numerical justification (0.94 or better) 7 (a) 0.3 or 1 – 0.6 or 0.4 or 0.2 seen 0.8 M1 A1 [2] (b) (i) k 1.5 2 0 (2.25 )d − ∫ x x = 1 k 3 3 1.5 2.25 0   −   x x = 1 k × [ ] 3.375 1.125 − = 1 or k × 9 4 = 1 oe k = 4 9 AG M1 A1 A1 [3] attempt integ f(x) and ‘= 1’. Ignore limits correct integration and limits No errors seen

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9709 72 © Cambridge International Examinations 2016 (ii) 4 9 1.5 3 0 (2.25 )d − ∫ x x x = 4 9 2 4 2 4 1.5 2.25 0   −   x x = 0.5625 or 0.563 Mean no. of hours = 56.25 or 56.3 56 hrs 15 mins M1 A1 A1 A1 [4] attempt integ xf(x), ignore limits, condone missing k correct integration and limits, condone missing k ft their 0.5625 (iii) Max x is 1.5, less than 2.9 or 150 < 290 B1 [1] Needs numerical justification (iv) any a such that 2.9 ⩽ a ⩽ 5 B1 [1] Total for paper 50