Cambridge A Level Mathematics 9709 — 2011 May/June Paper 6 · Variant 1

*9986337980* 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 2011 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. JC11 06_9709_61/RP © UCLES 2011 [Turn over

2 1 Biscuits are sold in packets of 18. There is a constant probability that any biscuit is broken, independently of other biscuits. The mean number of broken biscuits in a packet has been found to be 2.7. Find the probability that a packet contains between 2 and 4 (inclusive) broken biscuits. [4] 2 When Ted is looking for his pen, the probability that it is in his pencil case is 0.7. If his pen is in his pencil case he always finds it. If his pen is somewhere else, the probability that he finds it is 0.2. Given that Ted finds his pen when he is looking for it, find the probability that it was in his pencil case. [4] 3 The possible values of the random variable X are the 8 integers in the set {−2, −1, 0, 1, 2, 3, 4, 5}. The probability of X being 0 is 1 10. The probabilities for all the other values of X are equal. Calculate (i) P(X < 2), [2] (ii) the variance of X, [3] (iii) the value of a for which P(−a ≤X ≤2a) = 17 35. [1] 4 A cricket team of 11 players is to be chosen from 21 players consisting of 10 batsmen, 9 bowlers and 2 wicketkeepers. The team must include at least 5 batsmen, at least 4 bowlers and at least 1 wicketkeeper. (i) Find the number of different ways in which the team can be chosen. [4] Each player in the team is given a present. The presents consist of 5 identical pens, 4 identical diaries and 2 identical notebooks. (ii) Find the number of different arrangements of the presents if they are all displayed in a row. [1] (iii) 10 of these 11 presents are chosen and arranged in a row. Find the number of different arrangements that are possible. [3] 5 (a) The random variable X is normally distributed with mean µ and standard deviation σ. It is given that 3µ = 7σ2 and that P(X > 2µ) = 0.1016. Find µ and σ. [4] (b) It is given that Y ∼N(33, 21). Find the value of a given that P(33 −a < Y < 33 + a) = 0.5. [4] © UCLES 2011 9709/61/M/J/11

3 6 There are 5000 schools in a certain country. The cumulative frequency table shows the number of pupils in a school and the corresponding number of schools. Number of pupils in a school ≤100 ≤150 ≤200 ≤250 ≤350 ≤450 ≤600 Cumulative frequency 200 800 1600 2100 4100 4700 5000 (i) Draw a cumulative frequency graph with a scale of 2 cm to 100 pupils on the horizontal axis and a scale of 2 cm to 1000 schools on the vertical axis. Use your graph to estimate the median number of pupils in a school. [3] (ii) 80% of the schools have more than n pupils. Estimate the value of n correct to the nearest ten. [2] (iii) Find how many schools have between 201 and 250 (inclusive) pupils. [1] (iv) Calculate an estimate of the mean number of pupils per school. [4] 7 (a) (i) Find the probability of getting at least one 3 when 9 fair dice are thrown. [2] (ii) When n fair dice are thrown, the probability of getting at least one 3 is greater than 0.9. Find the smallest possible value of n. [4] (b) A bag contains 5 green balls and 3 yellow balls. Ronnie and Julie play a game in which they take turns to draw a ball from the bag at random without replacement. The winner of the game is the first person to draw a yellow ball. Julie draws the first ball. Find the probability that Ronnie wins the game. [4] © UCLES 2011 9709/61/M/J/11

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/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2011 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 2011 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 2011 9709 61 © University of Cambridge International Examinations 2011 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 2011 9709 61 © University of Cambridge International Examinations 2011 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 2011 9709 61 © University of Cambridge International Examinations 2011 1 18p = 2.7 p = 0.15 P(2, 3, 4) = 18C2 × (0.15)2(0.85)16 + 18C3(0.15)3(0.85)15 + 18C4(0.15)4(0.85)14 = 0.655 B1 M1 A1 A1 [4] Correct value for p Summing 3 binomial probs o.e Correct unsimplified answer Correct answer 2 P(pencil case | find) = M1 Attempt to use cond prob formula, must be quotient ) find ( ) find and pencilcase ( P P = 2.0 3.0 7.0 1 7.0 × + × A1 A1 Correct num of a fraction Correct denominator = 0.921 A1 [4] Correct answer 3 (i) P(any other number) = 9/70 P(X < 2) = 27/70 + 1/10 = 34/70 (17/35) (0.486) B1 B1ft [2] 9/70 Seen Ft their probs if < 1 (ii) E(X) = 108/70 (54/35) (1.543) Var(X) = ((–2)2 + … + 52) × 9 / 70 – (54 / 35)2 = 5.33 M1 M1 A1 [3] Valid attempt at E(X) (needn’t be accurate) Using a variance formula correctly with mean2 subtracted numerically, no extra division Correct final answer (iii) a = 1 B1 [1] 4 (i) Options 5 bat 5 bl 1 Wk in 10C5 × 9C5 × 2C1 = 63504 ways or 5 bat 4 bl 2 Wk in 10C5 × 9C4 × 2C2 = 31752 ways or 6 bat 4 bl 1 Wk in 10C6 × 9C4 × 2C1 = 52920 ways Total = 148176 (148000) M1 M1 A1 A1 [4] Multiplying three combinations together Summing more than one sensible option Two options correct unsimplified Correct final answer (ii) !2!4!5 ! 11 = 6930 B1 [1] Correct answer evaluated (iii) Omit a pen !2!4!4 ! 10 = 3150 M1 Summing three options Omit a diary !2!3!5 ! 10 = 2520 Omit a notebook !4!5 ! 10 = 1260 B1 One option correct Total = 6930 A1 [3] Correct final answer

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 61 © University of Cambridge International Examinations 2011 5 (a) σ µ µ − > 2 z = σ µ = σ σ 3 7 2 M1 Standardising attempt resulting in z > some µ/σ 3 7σ = 1.272 M1 B1 Substituting to eliminate µ or σ 1.272 seen σ = 0.545 µ = 0.693 A1 [4] Both answers correct (b) P(X < a + 33) = 0.75 z = 0.674 M1 A1 Using 0.75 oe 674 .0 ± seen 674 .0 21 33 33 = − + a M1 Standardising, no cc, must have sq rt a = 3.09 A1 [4] Correct answer 6 (i) pupils Median 270 M1 M1 (Indpt) A1 [3] Sensible attempt at graph using u.c.b. 2500 seen in median attempt on a CF graph Can be implied Correct answer + or – 5 (ii) 20% less than 160 M1 A1 [2] Using 20% Correct answer + or – 5 (iii) 2100 – 1600 = 500 B1 [1] (iv) (50.5 × 200 + 125.5 × 600 + 175.5 × 800 + 225.5 × 500 + 300.5 × 2000 + 400.5 × 600 + 525.5 × 300) / 5000 = 268 M1 M1 A1 A1 [4] Using an attempt at mid-points Using an attempt at frequencies Correct mid-points or frequencies Correct answer only 7 (a) (i) P(at least one 3) = 1 – P(no 3s) = 1 – (5/6)9 = 0.806 M1 A1 [2] Using 1 – none Correct answer (ii) P(at least 1 three) = 1 – (5/6)n 1 – (5/6)n > 0.9 n > 12.6 n = 13 B1 M1 M1 A1 [4] Equation or inequality involving n and 0.9 Solving attempt of sensible equation, can be trial Correct answer (b) P(R wins his 1st ball) = P(GY) = 15/56 (0.268) P(R wins 2nd ball) = P(GGGY) = 3/28 P(R wins 3rd ball) = P(GGGGGY) 3 3 4 1 5 2 6 3 7 4 8 5 × × × × × = 1/56 P(R wins) = 11/28 (0.393) M1 M1 M1 A1 [4] Using P(GY) Attempt to find P(GGGY) or P(GGGGGY) Adding three options Correct answer