Cambridge A Level Mathematics 9709 — 2010 Oct/Nov 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) October/November 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 *3377769412*

2 1 The discrete random variable X takes the values 1, 4, 5, 7 and 9 only. The probability distribution of X is shown in the table. x 1 4 5 7 9 P(X = x) 4p 5p2 1.5p 2.5p 1.5p Find p. [3] 2 Esme noted the test marks, x, of 16 people in a class. She found that Σ x = 824 and that the standard deviation of x was 6.5. (i) Calculate Σ(x −50) and Σ(x −50)2. [3] (ii) One person did the test later and her mark was 72. Calculate the new mean and standard deviation of the marks of all 17 people. [3] 3 A fair five-sided spinner has sides numbered 1, 2, 3, 4, 5. Raj spins the spinner and throws two fair dice. He calculates his score as follows. • If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on the dice to get his score. • If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to get his score. Given that Raj’s score is 12, find the probability that the spinner landed on an even-numbered side. [6] 4 The weights in kilograms of 11 bags of sugar and 7 bags of flour are as follows. Sugar: 1.961 1.983 2.008 2.014 1.968 1.994 2.011 2.017 1.977 1.984 1.989 Flour: 1.945 1.962 1.949 1.977 1.964 1.941 1.953 (i) Represent this information on a back-to-back stem-and-leaf diagram with sugar on the left-hand side. [4] (ii) Find the median and interquartile range of the weights of the bags of sugar. [3] 5 The distance the Zotoc car can travel on 20 litres of fuel is normally distributed with mean 320 km and standard deviation 21.6 km. The distance the Ganmor car can travel on 20 litres of fuel is normally distributed with mean 350 km and standard deviation 7.5 km. Both cars are filled with 20 litres of fuel and are driven towards a place 367 km away. (i) For each car, find the probability that it runs out of fuel before it has travelled 367 km. [3] (ii) The probability that a Zotoc car can travel at least (320 + d) km on 20 litres of fuel is 0.409. Find the value of d. [4] © UCLES 2010 9709/62/O/N/10

3 6 (i) State three conditions that must be satisfied for a situation to be modelled by a binomial distribution. [2] On any day, there is a probability of 0.3 that Julie’s train is late. (ii) Nine days are chosen at random. Find the probability that Julie’s train is late on more than 7 days or fewer than 2 days. [3] (iii) 90 days are chosen at random. Find the probability that Julie’s train is late on more than 35 days or fewer than 27 days. [5] 7 A committee of 6 people, which must contain at least 4 men and at least 1 woman, is to be chosen from 10 men and 9 women. (i) Find the number of possible committees that can be chosen. [3] (ii) Find the probability that one particular man, Albert, and one particular woman, Tracey, are both on the committee. [2] (iii) Find the number of possible committees that include either Albert or Tracey but not both. [3] (iv) The committee that is chosen consists of 4 men and 2 women. They queue up randomly in a line for refreshments. Find the probability that the women are not next to each other in the queue. [3] © UCLES 2010 9709/62/O/N/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/O/N/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 9709 MATHEMATICS 9709/62 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 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 A LEVEL – October/November 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 A LEVEL – October/November 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 A LEVEL – October/November 2010 9709 62 © UCLES 2010 1 4p + 5p2 + 1.5p + 2.5p + 1.5p = 1 10p2 + 19p – 2 = 0 p = 0.1 or –2 p = 0.1 M1 A1 A1 [3] Summing 5 probs to = 1 can be implied For 0.1 seen with or without –2 Choosing 0.1 must be by rejecting –2 2 (i) Σ(x – 50) = 824 – 16 × 50 = 24 B1 Correct answer 2 2 2 5.6 16 ) 50 ( 16 ) 50 ( =       − Σ − − Σ x x M1 Consistent substituting in the correct coded variance formula OR valid method for Σx2 then expanding Σ(x – 50)2, 3 terms at least 2 correct Σ(x – 50)2 = 712 A1 [3] Correct answer (ii) new mean = 896/17 (= 52.7) B1 Correct answer new var = 2 2 17 ) 50 72 ( 24 17 22 712       − + − + M1 Using the correct coded variance formula with n = 17 and new coded mean2 OR their (Σx2 + 722)/17 – their new mean2 new sd = 7.94 A1 [3] Rounding to correct answer, accept 7.95 or 7.98 or 7.91

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9709 62 © UCLES 2010 3 P(E and 12) = 36 4 5 2 × = 180 8 (2/45) M1 2/5 or 3/5 mult by dice-related probability seen anywhere A1 36 4 5 2 × seen oe P(12) = 180 8 36 1 5 3 + × = 180 11 (0.0611) M1 A1ft Summing two 2-factor probs involving 2/5 and 3/5 3/5 × 1/36 + their P(E and 12), ft their P(E 12) ) 12 ( P ) 12 ( P ) 12 ( P and E E = M1dep Subst in condit prob formula, must have a fraction = 11 8 (0.727) A1 [6] Correct answer OR list Even: 2 and (4,3) or (3,4) or (2,6) or (6,2) 4 and ditto Gives 8 options Odd: 1 and (6,6) or 3 and (6,6) or 5 and (6,6) Gives 3 options Prob(E│12) = 8/11 M1 A1 M1 A1 M1 A1 List attempt evens 8 options List attempt odds 3 options (Their even)/(their total) Correct answer 4 (i) sugar flour 194 1 5 9 195 3 8 1 196 2 4 7 197 7 9 4 3 198 4 199 8 200 7 4 1 201 key 1 196 2 means 1.961 kg for sugar and 1.962 kg for flour B1 B1 B1 B1ft [4] Correct stem must be integers. (stem and leaves can be in reverse order) Correct leaves flour must be single and ordered Correct leaves sugar must be single and ordered Correct key needs all this, ft if single leaves and 1.96 etc in stem (ii) med = 1.989 kg IQ range = 2.011 – 1.977 = 0.034 kg B1 M1 A1 [3] correct median subt their LQ from their UQ, UQ > med, LQ < med Correct answer

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9709 62 © UCLES 2010 5 (i) Zotoc: 176 .2 6. 21 320 367 = − = z Ganmor: 267 .2 5.7 350 367 = − = z P(Zotoc) = 0.985 P(Ganmor) = 0.988 M1 A1 A1 [3] Standardising either car’s fuel, no cc, no sq, no √ Correct answer Correct answer (ii) z = 0.23 6. 21 320 23 .0 − = x x = 324.968 d = 4.97 B1 M1 M1ind A1 [4] ± 0.23 seen Standardising either car, no cc, no sq rt, no sq 320 + d – 320 i.e. just d on num Correct answer, –4.97 gets A0 6 (i) constant/given prob, independent trials, fixed/given no. of trials, only two outcomes B1 B1 [2] One option correct Three options correct (ii) P(8, 9, 0, 1) = 9C8(0.3)8(0.7) + (0.3)9 + (0.7)9 + 9C1(0.3)(0.7)8 = 0.196 M1 A1 A1 [3] One term seen involving (0.3)x(0.7)9 – x(9Cx) Correct unsimplified expression Correct answer (iii) mean = 90 × 0.3 = 27 var = 18.9 P(X > 35) = 1 – Φ       − 9. 18 27 5. 35 = 1 – Φ(1.955) = 0.0253 P(X < 27) =       − Φ 9. 18 27 5. 26 = 1 – Φ(0.115) = 0.4542 Total prob = 0.480 accept 0.48 B1 M1 M1 M1 A1 [5] Expressions for 27 and 18.9 (4.347) seen Standardising one expression, must have sq rt in denom, cc not necessary Continuity correction applied at least once (1 – Φ1) + (1 – Φ2) accept (0.0329 + 0.5) if no cc Rounding to correct answer

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9709 62 © UCLES 2010 7 (i) 4M 2W or 5M 1W chosen in 10C4 × 9C2 + 10C5 × 9C1 = 9828 M1 A1 A1 [3] At least 1 of 10C4 × 9C2 and 10C5 × 9C1 seen Correct unsimplified expression Correct answer (ii) 9C3 × 8C1 + 9C4 = 798 Prob = 798/9828 = 0.0812 M1 A1 [2] One of 9C3 × 8C1 and 9C4 × (8C0) seen Correct answer (iii) Albert + not T... 9C3 × 8C2 + 9C4 × 8C1 = 3360 Tracey + not A... 9C4 × 8C1 + 9C5 = 1134 Number of ways = 4494 M1 A1 A1 [3] One of 9C3 × 8C2 or 9C4 × 8C1 or 9C5 × (8C0) seen Unsimplified 3360 or 1134 seen Correct final answer (iv) 6! – 4! × 5 × 2 or 6! – 5! × 2 (= 480) OR 4! × 5 × 4 or 4! × 5P2 (= 480) prob = 480/6! = 2/3 (0.667) OR using probabilities…as above OR Women together 5!/4! (= 5) Women not together = 15 – 5 = 10 total ways MMMMWW = 6!/4!2! = 15 prob = 2/3 B1 M1 A1 [3] B1 M1 A1 6! – 4! × 5 × 2 or 6! – 5! × 2 or 4! × 5 × 4 or 4! × 5P2 dividing by 6! correct answer 5 or 10 seen Dividing by 15 Correct answer