Cambridge A Level Mathematics 9709 — 2012 May/June Paper 6 · Variant 1

*1203393825* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/61 Paper 6 Probability & Statistics 1 (S1) May/June 2012 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 a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. 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. JC12 06_9709_61/FP © UCLES 2012 [Turn over

2 1 It is given that X ∼N(28.3, 4.5). Find the probability that a randomly chosen value of X lies between 25 and 30. [3] 2 Maria has 3 pre-set stations on her radio. When she switches her radio on, there is a probability of 0.3 that it will be set to station 1, a probability of 0.45 that it will be set to station 2 and a probability of 0.25 that it will be set to station 3. On station 1 the probability that the presenter is male is 0.1, on station 2 the probability that the presenter is male is 0.85 and on station 3 the probability that the presenter is male is p. When Maria switches on the radio, the probability that it is set to station 3 and the presenter is male is 0.075. (i) Show that the value of p is 0.3. [1] (ii) Given that Maria switches on and hears a male presenter, find the probability that the radio was set to station 2. [4] 3 A spinner has 5 sides, numbered 1, 2, 3, 4 and 5. When the spinner is spun, the score is the number of the side on which it lands. The score is denoted by the random variable X, which has the probability distribution shown in the table. x 1 2 3 4 5 P(X = x) 0.3 0.15 3p 2p 0.05 (i) Find the value of p. [1] A second spinner has 3 sides, numbered 1, 2 and 3. The score when this spinner is spun is denoted by the random variable Y. It is given that P(Y = 1) = 0.3, P(Y = 2) = 0.5 and P(Y = 3) = 0.2. (ii) Find the probability that, when both spinners are spun together, (a) the sum of the scores is 4, [3] (b) the product of the scores is less than 8. [3] 4 In a certain mountainous region in winter, the probability of more than 20 cm of snow falling on any particular day is 0.21. (i) Find the probability that, in any 7-day period in winter, fewer than 5 days have more than 20 cm of snow falling. [3] (ii) For 4 randomly chosen 7-day periods in winter, find the probability that exactly 3 of these periods will have at least 1 day with more than 20 cm of snow falling. [4] © UCLES 2012 9709/61/M/J/12

3 5 The lengths of the diagonals in metres of the 9 most popular flat screen TVs and the 9 most popular conventional TVs are shown below. Flat screen : 0.85 0.94 0.91 0.96 1.04 0.89 1.07 0.92 0.76 Conventional : 0.69 0.65 0.85 0.77 0.74 0.67 0.71 0.86 0.75 (i) Represent this information on a back-to-back stem-and-leaf diagram. [4] (ii) Find the median and the interquartile range of the lengths of the diagonals of the 9 conventional TVs. [3] (iii) Find the mean and standard deviation of the lengths of the diagonals of the 9 flat screen TVs. [2] 6 The lengths of body feathers of a particular species of bird are modelled by a normal distribution. A researcher measures the lengths of a random sample of 600 body feathers from birds of this species and finds that 63 are less than 6 cm long and 155 are more than 12 cm long. (i) Find estimates of the mean and standard deviation of the lengths of body feathers of birds of this species. [5] (ii) In a random sample of 1000 body feathers from birds of this species, how many would the researcher expect to find with lengths more than 1 standard deviation from the mean? [4] 7 (a) Seven friends together with their respective partners all meet up for a meal. To commemorate the occasion they arrange for a photograph to be taken of all 14 of them standing in a line. (i) How many different arrangements are there if each friend is standing next to his or her partner? [3] (ii) How many different arrangements are there if the 7 friends all stand together and the 7 partners all stand together? [2] (b) A group of 9 people consists of 2 boys, 3 girls and 4 adults. In how many ways can a team of 4 be chosen if (i) both boys are in the team, [1] (ii) the adults are either all in the team or all not in the team, [2] (iii) at least 2 girls are in the team? [2] © UCLES 2012 9709/61/M/J/12

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. University of 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. 9709/61/M/J/12

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 9709 MATHEMATICS 9709/61 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 must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 61 © University of Cambridge International Examinations 2012 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: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 61 © University of Cambridge International Examinations 2012 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: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 61 © University of Cambridge International Examinations 2012 1 5.4 3. 28 30 1 − = z = 0.8014 5.4 3. 28 25 2 − = z = –1.5556 Φ1 – ( 1 – Φ 2) = 0.7884 + 0.9401 – 1 = 0.729 M1 M1 A1 [3] Standardising at least one value, sq rt.ess; no cc Φ 1 + Φ 2 – 1 oe Correct answer 2 (i) 0.25p = 0.075 p = 0.075/0.25 = 0.3 (ii) P ( ) ) ( ) 2 ( 2 M P M and P M = = 3.0 25 .0 85 .0 45 .0 1.0 3.0 85 .0 45 .0 × + × + × × = 4875 .0 3825 .0 = 0.785 B1 [1] M1 B1 A1 A1 [4] Answer given, must show some working attempt at cond prob with single prod in num and Σ three 2-factor o.e prods in denom correct numerator of a fraction correct unsimplified denom correct answer 3 (i) p = 0.1 (ii) (a) P(X = 1, Y = 3) = 0.3× 0.2 = 0.06 P(X = 2, Y = 2) = 0.15× 0.5 = 0.075 P(X = 3, Y = 1) = 0.3× 0.3 = 0.09 P( sum is 4) = 0.225 (b) P(X = 1, Y = anything) = 0.3 P(X = 2, Y = anything) = 0.15 P(X = 3, Y = 1, 2) = 0.3× 0.8 = 0.24 P(X = 4, Y = 1) = 0.2× 0.3 = 0.06 P(X = 5, Y = 1) = 0.05× 0.3 = 0.015 P(product < 8) = 0.765 OR P(Y = 1, X = anything) = 0.3 P(Y = 2, X =1, 2, 3) = 0.5× 0.75 = 0.375 P(Y = 3, X =1, 2) = 0.2× 0.45 = 0.09 P(product < 8) = 0.765 B1 [1] M1 B1 A1 [3] M1 B1 A1 [3] M1 B1 A1 Summing 2 or 3 options One option correct unsimplified correct final answer Σ 3 or more two-factor options Two correct options Correct answer 4 (i) P(X < 5) = 1 – P(5, 6, 7) = 1 – (0.21)5(0.79)2 7C5 –(0.21)6(0.79)1 7C6 – (0.21)7 = 0.994 (ii) P(at least 1) = 1 – P(0) = 1 – (0.79)7 = 0.808 P(exactly 3 weeks) = (0.808)3(0.192)4C3 = 0.405 M1 A1 A1 [3] M1 A1 M1 A1 [4] Binomial expression with powers Σ 7 and probs Σ = 1, and 7Cr Correct unsimplified expression Correct answer Attempt to find P(at least 1) or 1– P(0 and 1) Rounding to correct answer Bin expression with powers Σ 4 and their 0.808 etc. and 4C3 Correct answer

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 61 © University of Cambridge International Examinations 2012 5 (i) Flat screen conventional 6 5 7 9 6 7 1 4 5 7 9 5 8 5 6 6 4 2 1 9 7 4 10 key 5 8 4 means 0.85 m for flat screen and 0.84 m for conventional (ii) Conventional median = 0.74 conv IQ range = 0.81 – 0.68 = 0.13 (iii) mean = 0.927 sd = 0.0882 B1 B1 B1 B1 [4] B1 M1 A1 [3] B1 B1 [2] Correct stem must be integers Correct flat screen leaves Correct conventional screen leaves Key must have units and TV type Correct median Their UQ – their LQ Correct answer Need 3 s.f. (Accept 0.0878 to 0.0889) 6 (i) σ µ − = − 6 253 .1 0.648 = σ µ − 12 µ = 9.9 σ = 3.15 or 3.16 (ii) need P(z < –1 or z > 1) = 1 – Ф1) + Ф (–1) = 2 – 2 × 0.8413 = 0.3174 number = 317 B1 B1 M1 M1 [Indpt] A1 [5] B1 M1 M1 A1 [4] Z = ± 1.253 Z = ± 0.648 Any equation with µ and σ and a reasonable z value not a prob. Allow cc or –, not √ơ or ơ2 Att. to solve by substitution or elimination z = 1 or – 1 seen Correct area i.e. 2 – 2Φ Mult their prob if sensible, by 1000 Accept 317, 317.4, 318 7 (a) (i) 7 couples in 7! ways each couple in 2 ways so 7! 7 2 × = 645120 OR 14 × 12 × 10 × 8 × 6 × 4 × 2 = 645120 (ii) 7! 2 !7 × × = 50,803,200 (50,800,000) OR 14× 6!× 7! B1 M1 A1 [3] B2 A1 B1 B1 [2] B1 B1 7! seen multiplied mult by 27 correct final answer correct unsimplified answer correct answer 7! × 7! seen Correct answer 14× 7! seen Correct answer

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 61 © University of Cambridge International Examinations 2012 (b) (i) 7C2 = 21 (ii) all in: 1 all not in: 5C4 = 5 total 6 (iii) 2 girls in: 6C2 × 3C2 = 45 3 girls in: 6C1= 6 Total 51 B1 [1] M1 A1 [2] M1 A1 [2] Considering both cases Correct answer Attempt at summing 2 and 3 girls in the team need not see 3C2 Correct answer