Cambridge A Level Mathematics 9709 — 2010 May/June Paper 6 · Variant 2

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/62 Paper 6 Probability & Statistics 1 (S1) May/June 2010 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. © UCLES 2010 [Turn over *8471957091*

2 1 The times in minutes for seven students to become proficient at a new computer game were measured. The results are shown below. 15 10 48 10 19 14 16 (i) Find the mean and standard deviation of these times. [2] (ii) State which of the mean, median or mode you consider would be most appropriate to use as a measure of central tendency to represent the data in this case. [1] (iii) For each of the two measures of average you did not choose in part (ii), give a reason why you consider it inappropriate. [2] 2 The lengths of new pencils are normally distributed with mean 11 cm and standard deviation 0.095 cm. (i) Find the probability that a pencil chosen at random has a length greater than 10.9 cm. [2] (ii) Find the probability that, in a random sample of 6 pencils, at least two have lengths less than 10.9 cm. [3] 3 0 100 200 300 400 500 600 700 800 900 1000 0 1 2 3 4 5 6 Cumulative frequency Weight (kg) Country A Country B The birth weights of random samples of 900 babies born in country A and 900 babies born in country B are illustrated in the cumulative frequency graphs. Use suitable data from these graphs to compare the central tendency and spread of the birth weights of the two sets of babies. [6] © UCLES 2010 9709/62/M/J/10

3 4 The random variable X is normally distributed with mean µ and standard deviation σ. (i) Given that 5σ = 3µ, find P(X < 2µ). [3] (ii) With a different relationship between µ and σ, it is given that P(X < 1 3µ) = 0.8524. Express µ in terms of σ. [3] 5 Two fair twelve-sided dice with sides marked 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 are thrown, and the numbers on the sides which land face down are noted. Events Q and R are defined as follows. Q : the product of the two numbers is 24. R : both of the numbers are greater than 8. (i) Find P(Q). [2] (ii) Find P(R). [2] (iii) Are events Q and R exclusive? Justify your answer. [2] (iv) Are events Q and R independent? Justify your answer. [2] 6 A small farm has 5 ducks and 2 geese. Four of these birds are to be chosen at random. The random variable X represents the number of geese chosen. (i) Draw up the probability distribution of X. [3] (ii) Show that E(X) = 8 7 and calculate Var(X). [3] (iii) When the farmer’s dog is let loose, it chases either the ducks with probability 3 5 or the geese with probability 2 5. If the dog chases the ducks there is a probability of 1 10 that they will attack the dog. If the dog chases the geese there is a probability of 3 4 that they will attack the dog. Given that the dog is not attacked, find the probability that it was chasing the geese. [4] 7 Nine cards, each of a different colour, are to be arranged in a line. (i) How many different arrangements of the 9 cards are possible? [1] The 9 cards include a pink card and a green card. (ii) How many different arrangements do not have the pink card next to the green card? [3] Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line. (iii) How many different arrangements in total of 3 cards are possible? [2] (iv) How many of the arrangements of 3 cards in part (iii) contain the pink card? [2] (v) How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card? [2] © UCLES 2010 9709/62/M/J/10

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. 9709/62/M/J/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 9709 MATHEMATICS 9709/62 Paper 62, 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2010 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 2010 9709 62 © UCLES 2010 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 2010 9709 62 © UCLES 2010 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 2010 9709 62 © UCLES 2010 1 (i) 9. 18 = x (132/7) sd = 12.3 B1 B1 [2] (ii) median B1 [1] (iii) mode inappropriate because it is 10 and this is the lowest value. mean inappropriate because it is affected by the outlier (of 48). B1 B1 [2] Sensible reason allow if seen in (ii) Sensible reason allow if seen in (ii) not ‘outliers’ in plural 2 (i) P(x > 10.9) = ) 095 .0 11 9. 10 ( P − > z = P(z > –1.0526) = 0.8538 (0.854) M1 A1 [2] Standardising, no cc, no sq rt Rounding to correct answer (ii) P(at least 2 < 10.9) = 1 – P(0, 1) = 1 – (0.8538)6 – 6C1(0.1462)(0.8538)5 = 0.215 M1 A1ft A1 [3] Bin expression with ∑ powers = 6, 6Cx, p + q = 1. Reasonably correct unsimplified expression ft their (i) Rounding to correct answer 3 median A = 2.0 – 2.1 or A x = 2.0 – 2.1 median B = 3.8 – 3.9 or B x = 3.4 – 3.5 Country B has heavier babies on average IQ range A = 2.4 – 1.5 = 0.9 or sd = 0.5 – 0.7 IQ range B = 4.5 – 2.2 = 2.3 or sd = 1.2 – 1.4 Country B has greater spread of weights M1 A1 B1 M1 A1 A1 [6] For finding medians or using mid-pts and freqs to find means, or seen on 2 box- plots Correct medians or means for A and B Correct statement allow ’...higher median…’ etc. Finding spreads by IQ range or range or sd or 2 box-plots Correct IQ range or sd for A and B (±0.1 kg) or correct IQR on box-plots Correct statement 4 (i) P (X < 2 µ) = P       − < σ µ µ 2 z = P(z < µ/σ) = P(z < 5/3) = 0.952 M1 A1 A1 [3] Standardising, and attempt to get 1 variable, no cc, no , no sq ±5/3 seen oe Rounding to correct answer (ii)       − < =       < σ µ µ 3 2 P 3 P z X 047 .1 3 2 = − σ µ µ = –1.57σ M1 B1 A1 [3] standardising attempt resulting in z Y – some µ/σ allow       − ± σ µ µ 3 / ±1.047 seen correct single number, answer must have a minus sign and µ = …..σ 5 (i) (2,12), (3, 8), (4, 6), (6, 4), (8,3), (12,2) P(Q) = 6/144 (1/24) (0.0417) M1 A1 [2] Listing or picking out at least 3 different options from a 12 by 12 (mult) table or seeing 3, 4, 5 or 6/144 Correct answer

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9709 62 © UCLES 2010 (ii) P(both > 8) = 1/3 × 1/3 = 1/9 = P(R) (16/144) M1 A1 [2] Squaring a sensible prob or picking out 12 – 25 options Correct answer (iii) since P(R and Q) = 0 Yes, R and Q are exclusive B1* B1dep [2] o.e. in words (iv) P(R and Q) = 0 ≠ P(R) × P(Q) or P(R|Q) = 0, ≠ P(R) No, not independent B1* B1dep [2] o.e. in words 6 (i) x 0 1 2 P(X = x) 1/7 4/7 2/7 B1 B1 B1 [3] 0, 1, 2 only in table or listed with some prob 3, 4… if in table must have blank or 0 for prob One correct probability All correct (ii) E(X) = 8/7 (1.14) AG Var(X) = 12/7 – (8/7)2 = 20/49 (0.408) B1 M1 A1 [3] Legitimate correct given answer rounding to 1.14 Correct method with mean2 subt numerically no dividing by anything Correct final answer (iii) P(G | NA) = ) ( ) ( NA P NA G P ∩ 10 / 9 5 / 3 4 / 1 5 / 2 4 / 1 5 / 2 × + × × = 32 5 = (0.156) M1 M1 A1 A1 [4] Attempt at P(G ∩ NA) or P(G ∩ A) as numerator of a fraction Attempt at P(NA) or P(A) in form of summing two 2-factor products, seen anywhere Correct unsimplified denominator of a fraction Correct answer 7 (i) 362880 (363000) B1 [1] (ii) PG or GP in 8! × 2 = 80640 or 7/9 of (i) 362880 – 80640 = 282240 M1 B1 A1ft [3] Considering together and also subtracting from their (i) or using probabilities 8! × 2 or 80640 seen oe correct answer ft 40320 only (iii) 9P3 or 9C3 × 3! or 9!/6! = 504 M1 A1 [2] 9P3 or 9C3 oe seen allow extra multiplication correct final answer (iv) 8C2 × 3! or 504 – 8C3 × 3! or 8P2 × 3 = 168 M1 A1 [2] 8Cx or 8Px seen allow extra mult, or (iii)/9 or (iii)/3 correct final answer (v) PG and x in 7 × 2 × 2 ways = 28 Answer 504 – 28 = 476 M1 A1 [2] x × 2 × 2 seen or their (iii) – 7 or 7C1 or 7C2 correct answer