Cambridge A Level Mathematics 9709 — 2016 May/June Paper 6 · Variant 2

*9125200877* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/62 Paper 6 Probability & Statistics 1 (S1) May/June 2016 1 hour 15 minutes Additional Materials: List of Formulae (MF9) READ THESE INSTRUCTIONS FIRST An answer booklet and a graph paper booklet are provided inside this question paper. You should follow the instructions on the front cover of both booklets. If you need additional answer paper or graph paper ask the invigilator for a continuation booklet or graph paper booklet. 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. This document consists of 3 printed pages, 1 blank page and 2 inserts. JC16 06_9709_62/2R © UCLES 2016 [Turn over

2 1 Ayman’s breakfast drink is tea, coffee or hot chocolate with probabilities 0.65, 0.28, 0.07 respectively. When he drinks tea, the probability that he has milk in it is 0.8. When he drinks coffee, the probability that he has milk in it is 0.5. When he drinks hot chocolate he always has milk in it. (i) Draw a fully labelled tree diagram to represent this information. [2] (ii) Find the probability that Ayman’s breakfast drink is coffee, given that his drink has milk in it. [3] 2 When visiting the dentist the probability of waiting less than 5 minutes is 0.16, and the probability of waiting less than 10 minutes is 0.88. (i) Find the probability of waiting between 5 and 10 minutes. [1] A random sample of 180 people who visit the dentist is chosen. (ii) Use a suitable approximation to find the probability that more than 115 of these people wait between 5 and 10 minutes. [5] 3 A particular type of bird lays 1, 2, 3 or 4 eggs in a nest each year. The probability of x eggs is equal to kx, where k is a constant. (i) Draw up a probability distribution table, in terms of k, for the number of eggs laid in a year and find the value of k. [3] (ii) Find the mean and variance of the number of eggs laid in a year by this type of bird. [3] 4 When people visit a certain large shop, on average 34% of them do not buy anything, 53% spend less than $50 and 13% spend at least $50. (i) 15 people visiting the shop are chosen at random. Calculate the probability that at least 14 of them buy something. [3] (ii) n people visiting the shop are chosen at random. The probability that none of them spends at least $50 is less than 0.04. Find the smallest possible value of n. [3] 5 The following are the maximum daily wind speeds in kilometres per hour for the first two weeks in April for two towns, Bronlea and Rogate. Bronlea 21 45 6 33 27 3 32 14 28 24 13 17 25 22 Rogate 7 5 4 15 23 7 11 13 26 18 23 16 10 34 (i) Draw a back-to-back stem-and-leaf diagram to represent this information. [5] (ii) Write down the median of the maximum wind speeds for Bronlea and find the interquartile range for Rogate. [3] (iii) Use your diagram to make one comparison between the maximum wind speeds in the two towns. [1] © UCLES 2016 9709/62/M/J/16

3 6 The time in minutes taken by Peter to walk to the shop and buy a newspaper is normally distributed with mean 9.5 and standard deviation 1.3. (i) Find the probability that on a randomly chosen day Peter takes longer than 10.2 minutes. [3] (ii) On 90% of days he takes longer than t minutes. Find the value of t. [3] (iii) Calculate an estimate of the number of days in a year (365 days) on which Peter takes less than 8.8 minutes to walk to the shop and buy a newspaper. [3] 7 (a) Find the number of different arrangements which can be made of all 10 letters of the word WALLFLOWER if (i) there are no restrictions, [1] (ii) there are exactly six letters between the two Ws. [4] (b) A team of 6 people is to be chosen from 5 swimmers, 7 athletes and 4 cyclists. There must be at least 1 from each activity and there must be more athletes than cyclists. Find the number of different ways in which the team can be chosen. [4] © UCLES 2016 9709/62/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 9709/62/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 Subsidiary and Advanced Level MATHEMATICS 9709/62 Paper 6 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 AS/A Level – May/June 2016 9709 62 © 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 AS/A Level – May/June 2016 9709 62 © 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 AS/A Level – May/June 2016 9709 62 © Cambridge International Examinations 2016 Qu Answer Marks Notes 1 (i) M1 A1 [2] Correct shape with either one branch after HC or 2 branches with 0 prob seen correct Labelled and clear annotation All probs correct (ii) P(C milk) = ( ) ( ) P coffee milk P milk ∩ = 0.28 0.5 0.65 0.8 0.28 0.5 0.07( 1) × × + × + × = 0.14 0.73 =0.192 M1 M1 A1 [3] Attempt at P(coffee∩ milk)as a two-factor prod only seen as num or denom of a fraction Summing appropriate three 2-factor products seen anywhere (can omit the 1) Correct answer oe 2 (i) 0.72 B1 [1] (ii) np = 180×0.72, npq = 180×0.72×0.28 X~N(129.6, 36.288) P(x >115) = 115.5 129.6 36.288 P z −   >     = P(z > −2.341) = 0.990 B1 M1 M1 M1 A1 [5] 180 × 0.72, 180 × 0.72 × 0.28 seen, their values or correct Standardising ( ±) must have sq rt cc either 115.5 or 114.5 seen Correct area, Φ from final answer attempt fully correct method 3 (i) x 1 2 3 4 P(x) k 2k 3k 4k 10k = 1 k = 1/10 B1 M1 A1 [3] Probability Distribution Table, either k or correct numerical values Summing probs involving k to = 1, 3 or 4 terms (ii) E(X) = 1/10 + 4/10 + 9/10 + 16/10 = 3 Var(X) = 1/10 + 8/10 + 27/10 + 64/10 – 32 = 1 B1 M1 A1 [3] Correct mean Correct method seen for var, their k and µ 0.07 0.28 0.65 T C HC 0.8 0.2 0.5 0.5 1 0 NM NM NM M M M

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9709 62 © Cambridge International Examinations 2016 4 (i) p = 0.66X~ B(15, 0.66) P(at least 14) = P(14, 15) = 15C14 (0.66)14(0.34) + (0.66)15 = 0.0171 M1 M1 A1 [3] Bin term 15Cxpx(1 – p)15-x seen any p Unsimplified correct expression for P(14,15) (ii) (0.87)n < 0.04 n = 24 M1 M1 A1 [3] Eqn involving 0.87, power of n, 0.04 only Solving by logs or trial and error(can be implied). Must be exponential equation 5 (i) Bronlea Rogate 6 3 0 4 5 7 7 7 4 3 1 0 1 3 5 6 8 8 7 5 4 2 1 2 3 3 6 3 2 3 4 5 4 Key 3| 1 | 5 represents 13 kph for Bronlea and 15 kph for Rogate B1 B1 B1 B1 B1 [5] Correct single stem Correct ordered leaves Bronlea Correct ordered leaves Rogate Correct overall shape Single key must have both towns and units consistent with their values (ii) median Bronlea = 23 km per hour IQ range Rogate = 23 – 7 = 16 B1 M1 A1 [3] Units not necessary Subt their LQ <14 from their UQ>14 from Rogate leaf (iii) Rogate is less windy than Bronlea B1 [1] Not a comparison of a statistic but interpretation of information 6 (i) P(x > 10.2) = P 10.2 9.5 1.3 z −   >     = P(z > 0.53846) = 1 − 0.7046 = 0.295 M1 M1 A1 [3] Standardising allow cc, sq rt, sq 1 – Φ final solution attempt (ii) z = −1.282 9.5 1.282 1.3 t − − = t = 7.83 B1 M1 A1 [3] ± rounding to 1.28 seen Standardising correctly can be ± z value here Correct answer from z = −1.282 only (iii) P(x < 8.8) = 0.2954 by symmetry Days = 365×0.2954 = 107 or 108 B1 M1 A1 [3] oe method, FT their 0.2954 from (i) Mult a probability <1 by 365 Correct answer (no decimals) 7 (a) (i) 10! 2!3! = 302400 B1 [1] Exact value only, isw rounding

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9709 62 © Cambridge International Examinations 2016 (ii) e.g. *W******W*, **W******W, W******W** 8! 3! × 3(for the Ws) = 20160 M1 M1 M1 A1 [4] 8! Seen mult or alone. Cannot be embedded (arrangements of other 8 letters). Dividing by 3! (removing repeated L’s) Mult by 3 (different W positions) may be sum of 3 terms (b) S(5) A(7) C(4) 1 3 2 : 5×7C3×4C2 = 1050 1 4 1 : 5×7C4× 4 = 700 2 3 1 : 5C2×7C3×4 = 1400 3 2 1 : 5C3×7C2×4 = 840 (Outcomes : Options) Total = 3990 M1 A1 M1 A1 [4] Mult 3 combinations, 5Cx, 7Cy, 4Cz (not 5 x 7 x 4) 2 correct options unsimplified Summing only 3 or 4 correct outcomes involving combs or perms