Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 6 · Variant 3

*9467039441* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/63 Paper 6 Probability & Statistics 1 (S1) October/November 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 11_9709_63/FP © UCLES 2012 [Turn over

2 1 In a normal distribution with mean 9.3, the probability of a randomly chosen value being greater than 5.6 is 0.85. Find the standard deviation. [3] 2 The discrete random variable X has the following probability distribution. x −3 0 2 4 P(X = x) p q r 0.4 Given that E(X) = 2.3 and Var(X) = 3.01, find the values of p, q and r. [6] 3 Ronnie obtained data about the gross domestic product (GDP) and the birth rate for 170 countries. He classified each GDP and each birth rate as either ‘low’, ‘medium’ or ‘high’. The table shows the number of countries in each category. Birth rate Low Medium High Low 3 5 45 GDP Medium 20 42 12 High 35 8 0 One of these countries is chosen at random. (i) Find the probability that the country chosen has a medium GDP. [1] (ii) Find the probability that the country chosen has a low birth rate, given that it does not have a medium GDP. [2] (iii) State with a reason whether or not the events ‘the country chosen has a high GDP’ and ‘the country chosen has a high birth rate’ are exclusive. [2] One country is chosen at random from those countries which have a medium GDP and then a different country is chosen at random from those which have a medium birth rate. (iv) Find the probability that both countries chosen have a medium GDP and a medium birth rate. [3] 4 In a survey, the percentage of meat in a certain type of take-away meal was found. The results, to the nearest integer, for 193 take-away meals are summarised in the table. Percentage of meat 1 −5 6 −10 11 −20 21 −30 31 −50 Frequency 59 67 38 18 11 (i) Calculate estimates of the mean and standard deviation of the percentage of meat in these take-away meals. [4] (ii) Draw, on graph paper, a histogram to illustrate the information in the table. [5] © UCLES 2012 9709/63/O/N/12

3 5 The random variable X is such that X ∼N(82, 126). (i) A value of X is chosen at random and rounded to the nearest whole number. Find the probability that this whole number is 84. [3] (ii) Five independent observations of X are taken. Find the probability that at most one of them is greater than 87. [4] (iii) Find the value of k such that P(87 < X < k) = 0.3. [5] 6 (a) A chess team of 2 girls and 2 boys is to be chosen from the 7 girls and 6 boys in the chess club. Find the number of ways this can be done if 2 of the girls are twins and are either both in the team or both not in the team. [3] (b) (i) The digits of the number 1 244 687 can be rearranged to give many different 7-digit numbers. How many of these 7-digit numbers are even? [4] (ii) How many different numbers between 20 000 and 30 000 can be formed using 5 different digits from the digits 1, 2, 4, 6, 7, 8? [2] (c) Helen has some black tiles, some white tiles and some grey tiles. She places a single row of 8 tiles above her washbasin. Each tile she places is equally likely to be black, white or grey. Find the probability that there are no tiles of the same colour next to each other. [3] © UCLES 2012 9709/63/O/N/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. 9709/63/O/N/12

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2012 series 9709 MATHEMATICS 9709/63 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 October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 63 © 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 Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 63 © 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 Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 63 © Cambridge International Examinations 2012 1 z = −1.036 = ë 93 6 5 − . ë = 3.57 B1 M1 A1 3 ± (1.036 to 1.037) seen Equation with 5.6 or 13.0, 9.3, ë and a z value, no cc Correct final answer 2 −3p + 2r + 4 × 0.4 = 2.3 (−3)2p + 22r + 42 × 0.4 – 2.32 = 3.01 p + q + r + 0.4 = 1 −3p + 2r = 0.7 9p + 4r = 1.9 so − 9p + 6r = 2.1 or − 6p + 4r = 1.4 4r + 6r = 1.9 + 2.1 or 9p + 6p = 1.9−1.4 r = 5 2 (0.4), p = 30 1 (0.0333) q = 0.6 – 0.4 – 0.0333 = 6 1 (0.167) B1 B1 B1 M1 A1 A1 6 Correct unsimplified equation, oe Correct unsimplified equation, oe Correct equation, oe Obtain an equation in 1 unknown One correct answer Remaining two answers correct 3 (i) 170 74       85 37 (0.435) B1 1 Correct answer (ii) 96 38       49 19 (0.396) B1 B1 2 Correct unsimplified numerator or denominator Correct answer (iii) P(high GDP and high birth rate) = 0 So they are exclusive B1* B1dep* 2 Correct reason Correct answer, CWO (iv) 54 41 × 74 42       = 666 287 3996 1722 (0.431) M1 B1 A1 3 Multiplying 2 probabilities with different numerators and denominators, only One correct probability seen Correct answer 4 (i) (3 × 59 + 8 × 67 + 15.5 × 38 + 25.5 × 18 + 40.5 × 11) / 193 = 11.4 ë 2 = (32 × 59 + 82 × 67 +….) / 193 – (11.43..)2 ë = 9.78 or 9.79 M1 A1 M1 A1 4 Attempt to calculate the mean using midpoints not ends, with frequencies, can be implied Correct mean Using px2f with mean2 subtracted numerically, can be implied Correct answer, method marks can be implied

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 63 © Cambridge International Examinations 2012 (ii) fd = 11.8, 13.4, 3.8, 1.8, 0.55 0 10 20 30 40 50 % of meat M1 A1 B1 B1 B1 5 Attempt at frequency density or scaling Correct heights seen on graph Bar lines correctly located at 5.5, 10.5, 20.5 and 30.5, no gaps, their scale which may be non-linear correct widths of bars, independent of bar lines Both axes uniform, from at least 0 to 14 if fd and 0.5 to 50.5, and labelled (fd or freq per 5% and % meat or % or meat) 5 (i) c       − 126 82 5 84. =c     − 126 82 5 83. = c(0.2227) – c (0.1336) = 0.5883 – 0.5533 = 0.0350 M1 M1 A1 3 Standardising using 83.5 or 84.5, must have square root Subtracting two probabilities, both > 0.5 or both < 0.5 Correct answer (ii) P(x > 87) = 1 - c ( ) 126 82 87 − = 1 - c (0.445) = 1 - 0.6718 = 0.3282 P(0, 1) = (0.6718)5 + 5C1 (0.3282) (0.6718)4 = 0.471 M1 A1 M1 A1 4 Standardising, no cc, must have square root Correct probability Any binomial term of form 5Cxpx(1−p)5-x,x≠0 Correct answer (iii) P(x < 87) = 0.6718  P(x < k) = 0.9718 z = 1.908 or 1.909 1.909 = ± 126 82 − k k = 103 M1 M1 A1 M1 A1 5 Finding P(x < 87), value > 0.5 Adding 0.3 to their 0.6718 or equivalent Correct z Equation with k, 82 or 81.5 or 82.5, 126 , and a z-value Correct answer rounding to 103 6 (a) twins in: 6C2 twins out: 5C2 × 6C2 Total = 15 + 150 = 165 OR all: 7C2 × 6C2 one twin: 2 × 5C1 × 6C2 Total = 315 − 150 = 165 B1 M1 A1 3 B1 M1 A1 6C2 alone or 5C2 multiplied seen or implied Summing two cases Correct final answer 7C2×6C2alone or 5C1 multiplied seen or implied 2×5C1 × 6C2 seen, subtracted Correct final answer

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 63 © Cambridge International Examinations 2012 (b) (i) ends in 2, 6 or 8: 6!/2! (= 360) ways ends in 4: 6! (= 720) ways Total = 3 × 360 + 720 = 1800 ways OR1 all: 7!/2! (= 2520) ways ends in 1 or 7: 6!/2! (= 360) ways Total = 2520 − 2 × 360 = 1800 OR2 (4A, 4B) final digit: 5 ways other digits: 6! ways and ÷ by 2! Total = 5 × 360 = 1800 B1 B1 M1 A1 4 B1 B1 M1 A1 B1 B1 M1 A1 Correct option for ending with 2 or 6 or 8.6!/2! seen anywhere, not multiplied Correct option for ending in 4 Summing 3 or 4 even options Correct final answer 7!/2! seen anywhere, not multiplied 6!/2! seen, subtracted Subtract 2 odd options from total options Correct final answer 5 seen, multiplied 6! seen and divide by 2! at some stage Multiplying their two numbers Correct final answer (ii) 5×4×3×2 or 5P4 or 5C4×4! or 5! or 5P5 or 6P5÷6 = 120 ways M1 A1 2 One of these oe Correct final answer (c) 7 3 2       2187 128 = (0.0585) M1 M1 A1 3 2/3 seen multiplied 7 probabilities multiplied together Correct final answer