Cambridge A Level Mathematics 9709 — 2014 May/June Paper 7 · Variant 2

*4606052898* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/72 Paper 7 Probability & Statistics 2 (S2) May/June 2014 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. JC14 06_9709_72/RP © UCLES 2014 [Turn over

2 1 The weights, in grams, of a random sample of 8 packets of cereal are as follows. 250 248 255 244 259 250 242 258 Calculate unbiased estimates of the population mean and variance. [3] 2 Each day Samuel travels from A to B and from B to C. He then returns directly from C to A. The times, in minutes, for these three journeys have the independent distributions N20, 22, N18, 1.52 and N30, 1.82, respectively. Find the probability that, on a randomly chosen day, the total time for his two journeys from A to B and B to C is less than the time for his return journey from C to A. [5] 3 The number of calls per day to an enquiry desk has a Poisson distribution. In the past the mean has been 5. In order to test whether the mean has changed, the number of calls on a random sample of 10 days was recorded. The total number of calls was found to be 61. Use an approximate distribution to test at the 10% significance level whether the mean has changed. [5] 4 (i) The random variable W has the distribution Po1.5. Find the probability that the sum of 3 independent values of W is greater than 2. [3] (ii) The random variable X has the distribution Po,. Given that PX = 0 = 0.523, find the value of , correct to 3 significant figures. [2] (iii) The random variable Y has the distribution Po-, where - ≠0. Given that PY = 3 = 24 × PY = 1, find -. [3] 5 Mahmoud throws a coin 400 times and finds that it shows heads 184 times. The probability that the coin shows heads on any throw is denoted by p. (i) Calculate an approximate 95% confidence interval for p. [4] (ii) Mahmoud claims that the coin is not fair. Use your answer to part (i) to comment on this claim. [1] (iii) Mahmoud’s result of 184 heads in 400 throws gives an !% confidence interval for p with width 0.1. Calculate the value of !. [4] 6 The time, T hours, spent by people on a visit to a museum has probability density function ft = T kt16 −t2 0 ≤t ≤4, 0 otherwise, where k is a constant. (i) Show that k = 1 64. [3] (ii) Calculate the probability that two randomly chosen people each spend less than 1 hour on a visit to the museum. [4] (iii) Find the mean time spent on a visit to the museum. [3] © UCLES 2014 9709/72/M/J/14

3 7 A researcher is investigating the actual lengths of time that patients spend with the doctor at their appointments. He plans to choose a sample of 12 appointments on a particular day. (i) Which of the following methods is preferable, and why? ³ Choose the first 12 appointments of the day. ³ Choose 12 appointments evenly spaced throughout the day. [2] Appointments are scheduled to last 10 minutes. The actual lengths of time, in minutes, that patients spend with the doctor may be assumed to have a normal distribution with mean - and standard deviation 3.4. The researcher suspects that the actual time spent is more than 10 minutes on average. To test this suspicion, he recorded the actual times spent for a random sample of 12 appointments and carried out a hypothesis test at the 1% significance level. (ii) State the probability of making a Type I error and explain what is meant by a Type I error in this context. [2] (iii) Given that the total length of time spent for the 12 appointments was 147 minutes, carry out the test. [5] (iv) Give a reason why the Central Limit theorem was not needed in part (iii). [1] © UCLES 2014 9709/72/M/J/14

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. 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 2014 9709/72/M/J/14

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2014 series 9709 MATHEMATICS 9709/72 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 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 72 © Cambridge International Examinations 2014 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 GCE A LEVEL – May/June 2014 9709 72 © Cambridge International Examinations 2014 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 GCE A LEVEL – May/June 2014 9709 72 © Cambridge International Examinations 2014 1 8 x Σ = 8 2006 = 250.75 or 251 (3 s.f.) (Σx2 = 503274)         − 2 " 75 . 250 " 8 " 503274 " 7 8 = 38.5 o.e. (accept 6.204²) B1 M1 A1 [3] Any equivalent form For use of formula of correct form cao (as final answer) 2 (X + Y – Z) ~ N(8, …) µ=8 (or –8) Var (X + Y – Z) = 22 + 1.52 + 1.82 (= 9.49) ' 49 .9' 8 0− (= –2.597) Φ(‘–2.597’) = 1 – Φ(‘2.597’) = 0.0047 B1 B1 M1 M1 A1 [5] seen or implied – award at early stage For standardising (accept sd/var mixes, but variance must be a combination of at least 2 of X, Y, Z) For area consistent with their working 3 H0: Pop mean (or µ or λ) = 50 (or 5) H1: Pop mean (or µ or λ) ≠ 50 (or 5) 50 50 5. 60 − (±) = (±)1.485 OR 0.0687 OR C.V 1.485 < 1.645 or 0.0687 > 0.05 No evidence that mean changed B1 M1 A1 M1 A1 [5] Not just “mean” For standardising with N(50,50) or N(5,5/√10) Allow M1 with wrong or no continuity correction OR no √ (accept c.v method M1, A1 for 61.63 or 48.868) For valid comparison (zs or areas or cv) (S.R For cv comparison 61.63 only award final A1 if cc used) or if H1: λ > 50, 1.485 < 1.96 M1 No evid mean changed A0 (i.e. if one-tail test, max B0 M1 A1 M1 A0)

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 72 © Cambridge International Examinations 2014 4 (i) λ = 4.5 1 – e–4.5       + + 2 5.4 5.4 1 2 = 0.826 (3 s.f.) B1 M1 A1 [3] seen any λ. Allow one end error (ii) e–λ = 0.523 (–λ = ln 0.523) λ = 0.648 (3 s.f.) B1 B1 [2] (iii) e–µ × !3 3 µ = 24 × e–µ × µ 6 2 µ = 24 µ = 12 B1 M1 A1 [3] For a simplified expression in µ² with e–µ and µ cancelled and no factorials. 5 (i) p = 400 184 or 0.46 z = 1.96 " 46 .0 " ± z × 400 )" 46 .0 " 1(" 46 .0 " − = 0.411 to 0.509 B1 B1 M1 A1 [4] Used Seen Using expression of correct form Must be an interval (ii) 0.5 within CI Claim not supported or not justified B1 [1] Both needed. No contradictions. ft their (i) (iii) z × 400 )" 46 .0 " 1(" 46 .0 " − = 0.05 z = 2.006 Φ(‘2.006’) = 0.9775 α = ‘0.9775’ – (1 – ‘0.9775’) = 95.5% M1 A1 M1 A1 [4] Allow M1 for z × 400 )" 46 .0 " 1(" 46 .0 " − = 0.1 or 1 – 2(1 – ‘0.9775’)

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 72 © Cambridge International Examinations 2014 6 (i) t t t k )d 4 0 (16 3 − ∫ = 1 k         −4 8 4 2 t t 0 4 = 1 k (128 – 64) = 1 o.e. k × 64 = 1        = 64 1 k AG M1 A1 A1 [3] Int f(t) = 1 ignore limits correct integration with correct limits must be convinced (AG) (ii) t t t d ) 3 1 0 16 ( 64 1 − ∫ = 64 1         −4 8 4 2 t t 0 1 = 64 1      −4 1 8 = 256 31 or 0.121094       256 31 2 = 0.0147 (3 s.f.) o.e. M1 A1 A1 B1 [4] Int f(t) between 0 and 1 (accept 0 and a value < 1 , 1 and 4) correct integration and correct limits (ignore “k”) ft their “ 256 31 ” iii t t t d ) 4 0 16 ( 64 1 4 2 − ∫ = 64 1         −5 3 16 5 3 t t 0 4 = 64 1       − 5 1024 3 1024 = 15 32 or 2.13 (3 s.f.) o.e. M1 A1 A1 [3] Int tf(t) ignore limits correct integration and correct limits (ignore “k”)

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 72 © Cambridge International Examinations 2014 7 (i) 2nd More representative of all appointments or Lengths may vary during the day or 1st does not include later appts so not representative B1 B1 [2] Any implication that times or conditions vary throughout day, e.g. doctors get tired (ii) 0.01 o.e. Concluding that times spent are too long when they are not. B1 B1 [2] Concluding that the mean time spent is more than 10 mins when it is not. Must be in context. (iii) H0: Pop mean appt time (or µ) = 10 H1: Pop mean appt time (or µ) > 10 12 4.3 10 12 147− (±) = (±)2.292 or (0.0109 if area comparison done) “2.292” < 2.326 o.e. (No evidence to reject H0.) No reason to believe appts are too long B1 M1 A1 M1 A1 [5] Both correct. Allow µ, but not just “mean” Allow incorrect 12 147 Must have √12 (accept totals method) For valid comparison Comp “2.292” with 2.326 Or 0.0109 with 0.01 Or 147/12 with 12.28 Dep 2.326, ft their “2.292” No contradictions 10 + 2.326 × 12 4.3 M1 = 12.28 A1 12 147 < 12.28 M1 (iv) Normal population B1 [1] Must have “population” or equiv