Cambridge A Level Mathematics 9709 — 2012 May/June Paper 6 · Variant 3

*7694666010* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/63 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_63/RP © UCLES 2012 [Turn over

2 1 Ashfaq and Kuljit have done a school statistics project on the prices of a particular model of headphones for MP3 players. Ashfaq collected prices from 21 shops. Kuljit used the internet to collect prices from 163 websites. (i) Name a suitable statistical diagram for Ashfaq to represent his data, together with a reason for choosing this particular diagram. [2] (ii) Name a suitable statistical diagram for Kuljit to represent her data, together with a reason for choosing this particular diagram. [2] 2 The heights, x cm, of a group of young children are summarised by Σ(x −100) = 72, Σ(x −100)2 = 499.2. The mean height is 104.8 cm. (i) Find the number of children in the group. [2] (ii) Find Σ(x −104.8)2. [3] 3 (i) In how many ways can all 9 letters of the word TELEPHONE be arranged in a line if the letters P and L must be at the ends? [2] How many different selections of 4 letters can be made from the 9 letters of the word TELEPHONE if (ii) there are no Es, [1] (iii) there is exactly 1 E, [2] (iv) there are no restrictions? [4] 4 The six faces of a fair die are numbered 1, 1, 1, 2, 3, 3. The score for a throw of the die, denoted by the random variable W, is the number on the top face after the die has landed. (i) Find the mean and standard deviation of W. [3] (ii) The die is thrown twice and the random variable X is the sum of the two scores. Draw up a probability distribution table for X. [4] (iii) The die is thrown n times. The random variable Y is the number of times that the score is 3. Given that E(Y) = 8, find Var(Y). [3] © UCLES 2012 9709/63/M/J/12

3 5 Suzanne has 20 pairs of shoes, some of which have designer labels. She has 6 pairs of high-heeled shoes, of which 2 pairs have designer labels. She has 4 pairs of low-heeled shoes, of which 1 pair has designer labels. The rest of her shoes are pairs of sports shoes. Suzanne has 8 pairs of shoes with designer labels in total. (i) Copy and complete the table below to show the number of pairs in each category. [2] Designer labels No designer labels Total High-heeled shoes Low-heeled shoes Sports shoes Total 20 Suzanne chooses 1 pair of shoes at random to wear. (ii) Find the probability that she wears the pair of low-heeled shoes with designer labels. [1] (iii) Find the probability that she wears a pair of sports shoes. [1] (iv) Find the probability that she wears a pair of high-heeled shoes, given that she wears a pair of shoes with designer labels. [1] (v) State with a reason whether the events ‘Suzanne wears a pair of shoes with designer labels’ and ‘Suzanne wears a pair of sports shoes’ are independent. [2] Suzanne chooses 1 pair of shoes at random each day. (vi) Find the probability that Suzanne wears a pair of shoes with designer labels on at most 4 days out of the next 7 days. [3] 6 The lengths, in cm, of trout in a fish farm are normally distributed. 96% of the lengths are less than 34.1 cm and 70% of the lengths are more than 26.7 cm. (i) Find the mean and the standard deviation of the lengths of the trout. [5] In another fish farm, the lengths of salmon, X cm, are normally distributed with mean 32.9 cm and standard deviation 2.4 cm. (ii) Find the probability that a randomly chosen salmon is 34 cm long, correct to the nearest centimetre. [3] (iii) Find the value of t such that P(31.8 < X < t) = 0.5. [4] © UCLES 2012 9709/63/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. 9709/63/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 9709/63 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 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 63 © 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 63 © 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 63 © University of Cambridge International Examinations 2012 1 (i) stem and leaf shows shape / spread / range / all prices [OR box-and-whisker shows shape/spread/range SR histogram B1only (ii) histogram shows shape/spread/modal class [OR1 cumulative frequency graph shows shape or large number of prices [OR2 box-and-whisker shows shape/spread/range B1* B1dep* [2] B1* B1dep* [2] correct answer valid reason, e.g. small number of prices, easy to calculate median correct answer valid reason, e.g. can read median] correct answer valid reason, e.g. large number of prices, too many prices for stem-and-leaf correct answer valid reason, e.g. easy to calculate median] correct answer valid reason, e.g. can read median] 2 (i) 72/n + 100 = 104.8 or 72 + 100n = 104.8n n = 15 (ii) sd2 = 499.2/15 – (72/15)2 (= 10.24) sd2 = 2 ( 104.8) x − ∑ /15− ( ( 104.8) x − ∑ /15)2 2 ( 104.8) x − ∑ = 153.6 (154) [OR1 2 ( 100) x − ∑ − 2 × 4.8 × ( 100) x − ∑ + 15 × 4.82 = 153.6 (154) [OR2 2x ∑ = 2 ( 100) x − ∑ + 200 × x ∑− 150000 2 ( 104.8) x − ∑ = 2x ∑ −209.6 x ∑ + 15 × 104.82 = 153.6 (154) M1 A1 [2] M1 M1 A1 [3] 72/n or 100n and 104.8n seen or implied correct answer numerical use of a correct sd/variance formula, their n numerical use of different correct sd/var formula, their n correct final answer numerical 1st and 2nd terms numerical 3rd term correct final answer] numerical use of a correct expansion to find 2x ∑ numerical use of a correct expansion for 2 ( 104.8) x − ∑ correct final answer]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 63 © University of Cambridge International Examinations 2012 3 (i) 2 !3 !7 × = 1680 (ii) 6C4 = 15 (iii) 1E in 6C3 ways = 20 (iv) need 2Es in 6C2 ways = 15 ways need 3Es in 6C1 = 6 ways total = 15 + 20 + 15 + 6 = 56 ways B1 B1 [2] B1 [1] M1 A1 [2] M1 A1 M1 A1ft [4] 7! 3! or 840 seen or implied correct answer correct answer k × 6Ca or k × bC3 (k a constant) or 6Pd or eP3 seen correct final answer attempt to find ways with 2Es or 3Es 6C2 oe and 6C1 oe seen summing ways for no Es, 1E, 2Es and 3Es correct final answer, ft on their four answers 4 (i) mean = 11/6 (1 6 5 , 1.83) sd = 2 ) 6 / 11 ( 6 /) 9 9 4 1 1 1( − + + + + + = 29 / 6 (0.898) (ii) x 2 3 4 5 6 Pr 9/36 6/36 13/36 4/36 4/36 (iii) p = 1/3 np = 8 n = 24 Var = 24× 1/3× 2/3 = 16/3 (5.33) B1 M1 A1 [3] B1 B1 M1 A1 [4] B1 M1 A1ft [3] correct answer numerical use of a correct sd/variance formula correct answer all correct x values P(2) and P(6) correct considering more than 1 case for a sum of 3 or 4 or 5 P(3), P(4) and P(5) correct correct p using np = 8 to find n or 8(1 − p) to find var, 0<p<1 correct answer, ft their p

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 63 © University of Cambridge International Examinations 2012 5 (i) Designer Not designer Total H-h shoes 2 4 6 L-h shoes 1 3 4 Sports 5 5 10 Total 8 12 20 (ii) 1/20 (0.05) (iii) 10/20 (1/2, 0.5) (iv) 2/8 (1/4, 0.25) (v) P(D) = 8/20 (0.4) P(S) =10/20 (0.5) P(D∩S) = 5/20 (0.25) Not independent as P(D) × P(S) ≠ P(D∩S) [OR1 P(D|S) = ( ) ( ) P D S P S ∩ = 5 10 P(D) = 8 20 Not independent as P(D|S) ≠ P(D) [OR2 P(S|D) = ( ) ( ) P S D P D ∩ = 5 8 P(S) = 10 20 Not independent as P(S|D) ≠ P(S) (vi) P(at most 4) = 1 − 7C5(0.4)5(0.6)2 − 7C6(0.4)6(0.6)1 − (0.4)7 = 0.904 B1 B1 [2] B1ft [1] B1ft [1] B1ft [1] M1 A1ft [2] M1 M1 A1 [3] one row or column correct all correct correct answer, ft their table correct answer, ft their table correct final answer, ft their table finding P(D∩S) and comparing with their P(D) × P(S) correct conclusion, ft their table finding P(D|S) and comparing with their P(D) correct conclusion, ft their table] finding P(D|S) and comparing with their P(D) correct conclusion, ft their table] bin probability of form 7Cr pr(1 – p)7 – r, r ≠ 0 or 7 bin expression for 1 − P(5, 6, 7) or P(0, 1, 2, 3, 4), any p correct answer

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 63 © University of Cambridge International Examinations 2012 6 (i) 751 .1 1. 34 = − σ µ 524 .0 7. 26 − = − σ µ  µ = 28.4, σ = 3.25 (ii)       − Φ −       − Φ 4.2 9. 32 5. 33 4.2 9. 32 5. 34 = Φ(0.667) – Φ (0.25) = 0.7477 – 0.5987 = 0.149 (iii)      − Φ 4.2 9. 32 t −       − Φ 4.2 9. 32 8. 31 = 0.5      − Φ 4.2 9. 32 t − (1 – 0.6765) = 0.5      − Φ 4.2 9. 32 t = 0.8235 929 .0 4.2 9. 32 = − t t = 35.1 B1 B1 M1 M1 A1 [5] M1 M1 A1 [3] M1 M1 M1 A1 [4] ±1.751 seen ±0.524 seen a standardising equation with a z-value, µ and σ valid attempt to eliminate µ or σ correct answers one numerical standardising expression, no cc, no square root, can have 34 subtracting two areas correct answer using 2 standardising expressions to give an equation involving subtraction and 0.5, oe adding their tail to 0.5 oe solving a standardised equation, must be a z- value from their 0.8235 correct final answer