Cambridge A Level Mathematics 9709 — 2015 May/June Paper 6 · Variant 2

*3926170419* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/62 Paper 6 Probability & Statistics 1 (S1) May/June 2015 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. JC15 06_9709_62/2R © UCLES 2015 [Turn over

2 1 A fair die is thrown 10 times. Find the probability that the number of sixes obtained is between 3 and 5 inclusive. [3] 2 120 people were asked to read an article in a newspaper. The times taken, to the nearest second, by the people to read the article are summarised in the following table. Time (seconds) 1 −25 26 −35 36 −45 46 −55 56 −90 Number of people 4 24 38 34 20 Calculate estimates of the mean and standard deviation of the reading times. [5] 3 0 1 2 3 4 5 6 7 8 9 10 0 10 20 30 40 50 60 70 80 90 100 Time in seconds Cumulative frequency In an open-plan office there are 88 computers. The times taken by these 88 computers to access a particular web page are represented in the cumulative frequency diagram. (i) On graph paper draw a box-and-whisker plot to summarise this information. [4] An ‘outlier’ is defined as any data value which is more than 1.5 times the interquartile range above the upper quartile, or more than 1.5 times the interquartile range below the lower quartile. (ii) Show that there are no outliers. [2] © UCLES 2015 9709/62/M/J/15

3 4 0.3 Nikita buys a scarf Nikita buys a handbag 0.72 x Mother likes her present Mother does not like her present Mother likes her present Mother does not like her present Nikita goes shopping to buy a birthday present for her mother. She buys either a scarf, with probability 0.3, or a handbag. The probability that her mother will like the choice of scarf is 0.72. The probability that her mother will like the choice of handbag is x. This information is shown on the tree diagram. The probability that Nikita’s mother likes the present that Nikita buys is 0.783. (i) Find x. [3] (ii) Given that Nikita’s mother does not like her present, find the probability that the present is a scarf. [4] 5 A box contains 5 discs, numbered 1, 2, 4, 6, 7. William takes 3 discs at random, without replacement, and notes the numbers on the discs. (i) Find the probability that the numbers on the 3 discs are two even numbers and one odd number. [3] The smallest of the numbers on the 3 discs taken is denoted by the random variable S. (ii) By listing all possible selections (126, 246 and so on) draw up the probability distribution table for S. [5] 6 (a) Find the number of different ways the 7 letters of the word BANANAS can be arranged (i) if the first letter is N and the last letter is B, [3] (ii) if all the letters A are next to each other. [3] (b) Find the number of ways of selecting a group of 9 people from 14 if two particular people cannot both be in the group together. [3] 7 (a) Once a week Zak goes for a run. The time he takes, in minutes, has a normal distribution with mean 35.2 and standard deviation 4.7. (i) Find the expected number of days during a year (52 weeks) for which Zak takes less than 30 minutes for his run. [4] (ii) The probability that Zak’s time is between 35.2 minutes and t minutes, where t > 35.2, is 0.148. Find the value of t. [3] (b) The random variable X has the distribution N-, 32. It is given that PX < 7 = 0.2119 and PX < 10 = 0.6700. Find the values of - and 3. [5] © UCLES 2015 9709/62/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 9709/62/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the May/June 2015 series 9709 MATHEMATICS 9709/62 Paper 6 (paper 6), 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 AS/A Level – May/June 2015 9709 62 © 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 AS/A Level – May/June 2015 9709 62 © 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 AS/A Level – May/June 2015 9709 62 © Cambridge International Examinations 2015 1 P(3, 4, 5) = 10C3 3 1 6       7 5 6      + 10C4 4 1 6       6 5 6      + 10C5 5 1 6       5 5 6       = 0.222 M1 A1 A1 3 Bin expression of form 10Cx (p)x(1–p)10–x any x any p Correct unsimplified answer accept (0.17, 0.83), (0.16, 0.84), (0.16, 0.83), (0.17, 0.84) or more accurate Correct answer 2 mid points 13, 30.5, 40.5, 50.5, 73 Mean = 4 13 24 30.5 38 40.5 34 50.5 20 20 7 1 3 × + × + × + × + × = 5500 120 = 45.8 var = 2 2 2 2 2 4 13 24 30.5 38 40.5 34 50. 12 20 0 5 73 × + × + × + × + × – (45.8…)2 = 278620 120 – 45.8…2 = 2321.8333 – 45.8…2 sd = 14.9 M1 M1 A1 M1 A1 5 Attempt at midpoints at least 3 correct Using their midpoints i.e. cw, ucb, 1/2 cw and freqs into correct formula must be divided by 120 Correct answer from correct working Evaluating 2 2 120 their fx x ∑ − must see their 45.82 subtracted allow cw etc Correct answer 3 (i) 0 1 2 3 4 5 6 7 8 9 10 time in sec B1 B1 B1 B1 4 LQ = 2.6 med = 3.8– 3.85, UQ = 6.4– 6.6 Correct quartiles and median on graph ft linear from 2–10 End whiskers correct not through box Label need seconds and linear 2–10 axis or can have 5 values on boxplot no line provided correct (ii) 1.5 × IQR = 1.5 × 3.8 = 5.7 LQ – 5.7 = –ve, UQ + 5.7 = 12.1 i.e. > 10 So no outliers AG M1 A1 2 Attempt to find 1.5 × IQR and add to UQ or subt from LQ OR compare 1.5 × IQR with gap 3.6 between UQ and max 10 Correct conclusion from correct working need both 4 (i) 0.3×0.72 + 0.7×x = 0.783 x = 0.81 M1 A1 A1 3 Eqn with sum of two 2-factor probs =0.783 Correct equation Correct answer

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9709 62 © Cambridge International Examinations 2015 (ii) P(S given not like) = ) ( ) ( NL P NL S P ∩ = 783 .0 1 19 .0 7.0 28 .0 3.0 28 .0 3.0 − × + × × or = 0.387 (12/31) B1 M1 A1 A1 4 0.3×0.28 seen on its own as num or denom of a fraction Attempt at P(NL) either (0.3 × p1) + (0.7 × p2) or 1 – 0.783 seen anywhere Correct unsimplified P(NL) as num or denom of a fraction Correct answer 5 (i) P(2Es 1O) = 3 5 × 2 4 × 2 3 ×3C2 = 3 5 (0.6) OR P(2Es 1O) = 3 5 1 2 2 3C C C × = 10 6 = 0.6 OR 241, 247, 261, 267, 461, 467 = 6 options 124 126 127 146 147 167 246 247 267 467 Prob = 6/10 M1 M1 A1 3 M1 M1 A1 M1 M1 A1 5×4×3 seen in denom Mult a prob by 3C2 oe Correct answer 3Cx or yC2 or 2C1 oe seen mult by k [ 1 in num 5C3 seen in denom Correct answer List at least 3 of 241, 247, 261, 267, 461, 467 5C3 or list to get all 10 options in denom see below Correct answer (ii) 124 126 127 146 147 167 246 247 267 467 s 1 2 4 P(S =s) 6/10 3/10 1/10 M1 A1 B1 B1 B1 5 Attempt at listing with at least 7 correct All correct and no others or all 60 1, 2, 4 only seen in top row Any two correct All correct 6 (a) (i) N*****B Number of ways = 5! 3! = 20 B1 B1 B1 3 5! Seen in num oe or alone mult by k [ 1 3! Seen in denom can be mult by k [ 1 Correct final answer (ii) B(AAA)NNS Number of ways = !2 !5 or 5P3 = 60 M1 M1 A1 3 5! seen as a num can be mult by k [ 1 Dividing by 2! Correct final answer (b) 14C9 total options = 2002 T and M both in 12C7 = 792 Ans 2002 – 792 = 1210 OR Neither in12C9 = 220 One in 12C8 = 495 Other in 12C8 = 495 M1 B1 A1 3 M1 B1 14C9 or 14P9 in subtraction attempt 12C7 (792) seen Correct final answer Summing 2 or 3 options at least 1 correct condone 12P9 + 12P8 + 12P8 here only Second correct option seen accept another 495 or if M1 not awarded, any correct option

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9709 62 © Cambridge International Examinations 2015 total = 1210 A1 Correct final answer 7 (a) (i) prob =       − < 7.4 2. 35 30 z p = P( z < –1.106) = 1 – 0.8655 = 0.1345 0.1345×52 = 6.99 M1 M1 A1 A1 4 Standardising no sq rt no cc no sq Φ − 1 Correct ans rounding to 0.13 Correct final answer accept 6 or 7 if 6.99 not seen but previous prob 0,1345 correct (ii) Φ (t) = 0.648 z = 0.380 0.380 = 7.4 2. 35 − t t = 37.0 B1 M1 A1 3 0.648 seen standardising allow cc, sq rt,sq, need use of tables not 0.148, 0.648, 0.352, 0.852 correct answer rounding to 37.0 (b) σ σ µ 8.0 7 − = − so σ µ 8.0 7 − = − 44 .0 10 = − σ µ so σ µ 44 .0 10 = − 94 .8 = µ 42 .2 = σ B1 B1 M1 M1 A1 5 ± 0.8 seen ± 0.44 seen An eqn with z-value, µ and σ no sq rt no cc no sq Sensible attempt to eliminate µ or σ by subst or subtraction, need at least one value Correct answers