Cambridge A Level Mathematics (US) 9280 — 2015 May/June Paper 7 · Variant 1

*4783481205* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS (US) 9280/71 Paper 7 Probability & Statistics 2 (S2) May/June 2015 1 hour 15 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulas (MF9) (US) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Center number, candidate number, and name on 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. JC15 06_9280_71/FP © UCLES 2015 [Turn over

2 1 x fx a a O The random variable X has probability density function, f, as shown in the diagram, where a is a constant. Find the value of a and hence show that EX = 0.943 correct to 3 significant figures. [5] 2 Sami claims that he can read minds. He asks each of 50 people to choose one of the 5 letters A, B, C, D, or E. He then tells each person which letter he believes they have chosen. He gets 13 correct. Sami says “This shows that I can read minds, because 13 is more than I would have got right if I were just guessing.” (i) State null and alternative hypotheses for a test of Sami’s claim. [1] (ii) Test at the 10% significance level whether Sami’s claim is justified. [5] 3 The daily times, in minutes, that Yu Ming takes showering, getting dressed and having breakfast are independent and have the distributions N9, 2.22, N8, 1.32 and N17, 2.62 respectively. The total daily time that Yu Ming takes for all three activities is denoted by T minutes. (i) Find the mean and variance of T. [2] (ii) Yu Ming notes the value of T on each day in a random sample of 70 days and calculates the sample mean. Find the probability that the sample mean is between 33 and 35. [4] 4 In the past, the time taken by vehicles to drive along a particular stretch of road has had mean 12.4 minutes and standard deviation 2.1 minutes. Some new signs are installed and it is expected that the mean time will increase. In order to test whether this is the case, the mean time for a random sample of 50 vehicles is found. You may assume that the standard deviation is unchanged. (i) The mean time for the sample of 50 vehicles is found to be 12.9 minutes. Test at the 2.5% significance level whether the population mean time has increased. [4] (ii) State what is meant by a Type II error in this context. [2] (iii) State what extra piece of information would be needed in order to find the probability of a Type II error. [1] © UCLES 2015 9280/71/M/J/15

3 5 The masses, m grams, of a random sample of 80 strawberries of a certain type were measured and summarized as follows. n = 80 Σm = 4200 Σm2 = 229 000 (i) Find unbiased estimates of the population mean and variance. [3] (ii) Calculate a 98% confidence interval for the population mean. [3] 50 random samples of size 80 were taken and a 98% confidence interval for the population mean, -, was found from each sample. (iii) Find the number of these 50 confidence intervals that would be expected to include the true value of -. [1] 6 A publishing firm has found that errors in the first draft of a new book occur at random and that, on average, there is 1 error in every 3 pages of a first draft. Find the probability that in a particular first draft there are (i) exactly 2 errors in 10 pages, [2] (ii) at least 3 errors in 6 pages, [3] (iii) fewer than 50 errors in 200 pages. [4] 7 The independent variables X and Y are such that X ∼B10, 0.8 and Y ∼Po3. Find (i) E7X + 5Y −2, [2] (ii) Var4X −3Y + 3, [4] (iii) P2X −Y = 18. [4] © UCLES 2015 9280/71/M/J/15

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 2015 9280/71/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 9280 MATHEMATICS (US) 9280/71 Paper 7, maximum raw mark 50 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 9280 71 © 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 9280 71 © 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 9280 71 © Cambridge International Examinations 2015 Note: “(3 sfs)” means “answer which rounds to ... to 3 sfs”. If correct ans seen to [ 3sfs, ISW for later rounding. Penalise I 3 sfs only once in paper. 1 2 1 a2 = 1 a = √2 ∫ 2 0 2dx x = 3 2 3 0 x         3 2 3     = = or 3 2 5.1 or 3 83 .2 or 0.9428 (= 0.943 AG) M1 A1 M1 A1f A1 [5] or ∫ a x x 0 d = 1 Allow 1.41 or better ignore limits correct integral and limits, but ft their a must see this numerical expression, or equiv SR Equating ∫x f(x) to 0.943 scores M1 Solving to find a = 1.41 scores A1 [Total 5] 2 (i) H0: p = 0.2 or µ = 10 H1: p > 0.2 or µ > 10 B1 [1] (ii) N(10, 8) seen or implied 50 8.0 2.0 50 125 02 or 8 10 125 × − − = 0.884 comp 1.282 Claim not justified or No evidence to support claim B1 M1 A1 M1f A1f [5] or N       × 50 8.0 2.0 ,2.0 For standardising allow with no or wrong cc Allow area comparison with 0.188 or comp 1.645 if H1 p ≠ 0.2 Allow accept H0 provided correctly defined. Follow through their test statistic ;dep 1–tail test No Contradictions SR; Use of B(50,0.2) scores B1 provided at least two probabilities calculated. M1 For finding P(X [ 13) allow one end error. A1 for 0.186 [Total: 6]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9280 71 © Cambridge International Examinations 2015 3 (i) 34 2.22 + 1.32 + 2.62 (=13.29) B1 B1 [2] Accept 13.3 or 3.652 Allow at early stage (ii) 70 ' 29 . 13 ' ' 34 ' 33− (= –2.295) 70 ' 29 . 13 ' ' 34 ' 35− (= 2.295) Φ(‘2.295’) – Φ(‘–2.295’) = Φ(‘2.295’) – (1 – Φ(‘2.295’)) oe = 0.978 (3 sf) M1 M1 M1 A1 [4] correct standardisation method for either For attempt to use tables to find the probability between two z values ,may be implied by next line For a correct method to find the area between their two z values [Total: 6] 4 (i) H0: pop mean (or µ) = 12.4 H1: pop mean (or µ) > 12.4 50 1.2 4. 12 9. 12 + − 1.684 comp cv z = 1.96 No evidence that pop mean time has increased B1 M1 A1 B1f [4] not just “mean” Allow with 50 instead of √50 or P(z K 1.684) = 0.0461 K 0.025 Allow accept H0 if correctly defined. Ft their test statistic. No contradictions (ii) Not reject (or accept) that mean time is unchanged (or is 12.4) oe although mean time has increased (or is more than 12.4) oe B1 B1 [2] (iii) True (or new) mean B1 [1] [Total: 7]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9280 71 © Cambridge International Examinations 2015 5 (i) 4200/80 (=52.5) 2 80 229 000 '52.5' 79 80   = −     (= 107.595) = 108 (3 sf) B1 M1 A1 [3] (ii) 80 ' 595 . 107 ' '5. 52 ' z ± z = 2.326 49.8 to 55.2 M1 B1 A1f [3] Correct form – must be z-value – allow one side only Seen ft their 52.5 and 107.595. Must be an interval (iii) 49 B1 [1] [Total: 7] 6 (i) ( ) 10 3 2 10 3 e 2 −× = 0.198 (3 sf) M1 A1 [2] P(2), allow any λ (ii)       + + − − 2 2 2 1 e 1 2 2 = 0.323 (3 sf) M1 M1 A1 [3] M1 allow any λ and/or 1end error Correct expression , correct λ (iii)       3 200 , 3 200 N 3 200 3 200 5. 49 − (= –2.102) Φ('–2.102') = 1 – Φ('2.102') = 0.0178 (3 sf) M1 M1 M1 A1 [4] seen or implied For standardising allow either wrong or no cc No sd/var mix For finding area consistent with their working [Total: 9]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9280 71 © Cambridge International Examinations 2015 7 (i) 7E(X) + 5E(Y) – 2 (= 7 × 8 + 5 × 3) – 2 = 69 M1 A1 [2] allow incorrect means (ii) Var(X) = 1.6, Var(Y) = 3 16Var(X) + 9Var(Y) (= 16×1.6 + 9 × 3) = 52.6 B1 M1 M1 A1 [4] both M1 for mult by 16 and 9; allow with ‘+ 3’ M1 for add without ‘+ 3’; allow incorrect multipliers (iii) X = 10, Y = 2 and X = 9, Y = 0 0.810 × e–3 × 2 3 2 or 10×0.89 × 0.2 × e–3 0.810 × e–3 × 2 3 2 + 10×0.89 × 0.2 × e–3 = 0.0374/5 B1 M1 M1 A1 [4] both pairs seen or implied or 0.0241 or 0.0134 (3sf) one correct product all correct [Total: 10] [Total for paper 50]