Cambridge A Level Mathematics 9709 — 2004 Oct/Nov Paper 6 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level Advanced International Certificate of Education MATHEMATICS STATISTICS 9709/06 0390/06 Paper 6 Probability & Statistics 1 (S1) October/November 2004 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 on both sides of the paper. 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. 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. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. This document consists of 3 printed pages and 1 blank page. © UCLES 2004 [Turn over

2 1 The word ARGENTINA includes the four consonants R, G, N, T and the three vowels A, E, I. (i) Find the number of different arrangements using all nine letters. [2] (ii) How many of these arrangements have a consonant at the beginning, then a vowel, then another consonant, and so on alternately? [3] 2 The lengths of cars travelling on a car ferry are noted. The data are summarised in the following table. Length of car (x metres) Frequency Frequency density 2.80 ≤x < 3.00 17 85 3.00 ≤x < 3.10 24 240 3.10 ≤x < 3.20 19 190 3.20 ≤x < 3.40 8 a (i) Find the value of a. [1] (ii) Draw a histogram on graph paper to represent the data. [3] (iii) Find the probability that a randomly chosen car on the ferry is less than 3.20 m in length. [2] 3 When Andrea needs a taxi, she rings one of three taxi companies, A, B or C. 50% of her calls are to taxi company A, 30% to B and 20% to C. A taxi from company A arrives late 4% of the time, a taxi from company B arrives late 6% of the time and a taxi from company C arrives late 17% of the time. (i) Find the probability that, when Andrea rings for a taxi, it arrives late. [3] (ii) Given that Andrea’s taxi arrives late, find the conditional probability that she rang company B. [3] 4 The ages, x years, of 18 people attending an evening class are summarised by the following totals: Σx = 745, Σx2 = 33 951. (i) Calculate the mean and standard deviation of the ages of this group of people. [3] (ii) One person leaves the group and the mean age of the remaining 17 people is exactly 41 years. Find the age of the person who left and the standard deviation of the ages of the remaining 17 people. [4] 5 The length of Paulo’s lunch break follows a normal distribution with mean µ minutes and standard deviation 5 minutes. On one day in four, on average, his lunch break lasts for more than 52 minutes. (i) Find the value of µ. [3] (ii) Find the probability that Paulo’s lunch break lasts for between 40 and 46 minutes on every one of the next four days. [4] 9709/06/O/N/04

3 6 A box contains five balls numbered 1, 2, 3, 4, 5. Three balls are drawn randomly at the same time from the box. (i) By listing all possible outcomes (123, 124, etc.), find the probability that the sum of the three numbers drawn is an odd number. [2] The random variable L denotes the largest of the three numbers drawn. (ii) Find the probability that L is 4. [1] (iii) Draw up a table to show the probability distribution of L. [3] (iv) Calculate the expectation and variance of L. [3] 7 (i) State two conditions which must be satisfied for a situation to be modelled by a binomial distribution. [2] In a certain village 28% of all cars are made by Ford. (ii) 14 cars are chosen randomly in this village. Find the probability that fewer than 4 of these cars are made by Ford. [4] (iii) A random sample of 50 cars in the village is taken. Estimate, using a normal approximation, the probability that more than 18 cars are made by Ford. [4] 9709/06/O/N/04

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary, Advanced Level and AICE MARK SCHEME for the November 2004 question papers 9709/0390 MATHEMATICS 9709/06, 0390/06 Paper 6 (Probability and Statistics 1), maximum raw mark 50 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2004 question papers for most IGCSE and GCE Advanced Level syllabuses.

Grade thresholds taken for Syllabus 9709/0390 (Mathematics) in the November 2004 examination. minimum mark required for grade: maximum mark available A B E Component 6 50 41 37 22 The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.

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.

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.

November 2004 GCE A, AS LEVEL and AICE MARK SCHEME MAXIMUM MARK: 50 SYLLABUS/COMPONENT: 9709/06, 0390/06 MATHEMATICS (Probability and Statistics 1)

Page 1 Mark Scheme Syllabus Paper AICE, A AND AS LEVEL – NOVEMBER 2004 9709/0390 6 © University of Cambridge International Examinations 2005 1 (i) !2 !2 !9 = 90720 B1 For dividing by 2! or 2 once or twice, or 9P7 or 9C7 seen, can be implied B1 2 For correct answer (ii) !2 !2 !4 !5 = 720 OR could do by probs and multiply by their (i) B1 For 5! or 4! or 4P4 or 5P5 seen in num B1 For 5! × 4! × k in num of a term, k = 1 or 2 only B1 3 For correct final answer 2 (i) a = 8/0.2 (= 40) B1 1 (ii) M1 B1 B1 ft 3 Uniform linear scales from at least 2.8 to 3.4 on the x-axis and 0 to 240 on the y-axis, both axes labelled, accept m or length, on x-axis Correct widths, no 0.05s, no gaps Four bars correct, ft on their (i) consistent with their vertical labelling, heights within ½ small square (iii) 8 19 24 17 19 24 17 + + + + + M1 For three terms in num and 4 terms in denom (can be implied) NB fd.s ie 85 + 240 + 190/555 get M1 A0 = 60/68 or 0.882 A1 2 For correct answer, a.e.f . 3 (i) P(L) = 0.5 x 0.04 + 0.3 x 0.06 + 0.2 x 0.17 M1 For summing three relevant 2-factor terms A1 For correct expression = 0.072 (9/25) A1 3 For correct answer (ii) P (BL) = 072 . 0 06 . 0 3 . 0 × B1 ft For their 0.3 x 0.06 in numerator, must be divided by k ≠ 1 M1 For dividing by their P(L) = 0.25 A1 3 For correct answer

Page 2 Mark Scheme Syllabus Paper AICE, A AND AS LEVEL – NOVEMBER 2004 9709/0390 6 © University of Cambridge International Examinations 2005 4 (i) Mean = 745/18 = 41.4 B1 For correct answer, a.e.f. sd = 2 18 745 18 33951       − M1 For [33951/18 – (their 41.4)2] or ÷ 17 = 13.2 A1 3 For correct answer 13.1 gets A0 with PA # 1 (ii) 17 x 41 = 697 M1 For multiplying 17 by 41 745 – 697 = 48 A1 2 For correct answer sd = 2 2 41 17 48 33951 − − = 13.4 M1 For subtracting their 482 from 33951 (ignore anything else) A1 2 For correct answer 5 (i) z = 0.674 or 0.675 allow 0.67 to 0.675 B1 For correct z, can be + or - 5 52 µ − = 0.674 M1 For an equation relating 52, 5, µ and any z ≠ 0.5987 or 0.7734 ish µ = 48.6 A1 3 For correct answer (ii) z1 = 5 63 . 48 40 − = -1.726 M1 For standardising 40 or 46, 5 or 5 in denom or 52 with their mean, no cc z2 = 5 63 . 48 46 − = 0.526 prob = 0.9578 – 0.7005 = 0.2573 M1 For subtracting two probs consistent with their mean ie usually 1 Φ - 2 Φ or (1 - 1 Φ ) - (1 - 2 Φ ) but could be of type 1 Φ - (1 - 2 Φ ) if their mean is in between 40 and 46 (0.2573)4 M1 For raising their answer above to a power 4 = 0.00438 or 4.38 x 10-3 A1 ft 4 For correct answer accept 0.00449 x 10-3 NB 0.0045 gets A0 and RE #1

Page 3 Mark Scheme Syllabus Paper AICE, A AND AS LEVEL – NOVEMBER 2004 9709/0390 6 © University of Cambridge International Examinations 2005 6 (i) Options 123, 124, 125, 134, 135, 145, 234, 235, 245, 345 M1 For listings options, at least 4 different ones P (odd) = 0.4 B1 2 For correct answer, legit obtained (ii) P(largest is 4) = 0.3 B1 1 For correct answer OR 3 5 2 3 1 C C × SR if 9 options in (i) give B1 for 3/9 or 2/9 depending on their missing option (iii) l 3 4 5 P(L = l) 0.1 0.3 0.6 M1 For 3, 4, 5, in table or 1, 2 as well, no need for any probs but need to see an (uncompleted) second line M1 For evaluating another probability based on their list A1 3 For correct answer (iv) E(L) = ∑ = × + × + × = 5 . 4 6 . 0 5 4 . 0 4 1 . 0 3 p l B1 ft For correct answer, ft if their ∑p= 1 Var (L) = 32 × 0.1 + 42 × 0.3 + 52 × 0.6 – (their 4.52) M1 For evaluating their ∑ − ) 4.5 (their 2 2 p l (must see – their 4.52) each p < 1, in first numerical instance, ie can forget the sq rt subsequently = 0.45 A1 3 For correct answer

Page 4 Mark Scheme Syllabus Paper AICE, A AND AS LEVEL – NOVEMBER 2004 9709/0390 6 © University of Cambridge International Examinations 2005 7 (i) constant p, independent trials, fixed number of trials, only two outcomes B1 For an option B1 2 For a second option (ii) P(X<4) = 0.7214 + 14C1 × 0.28 × 0.7213 + 14C2 × 0.282 × 0.7212 + 14C3 × 0.283 × 0.7211 M1 For adding with some C in P(0 + 1 + 2 + 3) or P(1 + 2 + 3) or P(0 + 1 + 2 + 3 + 4) or P(1 + 2 + 3 + 4) M1 For 0.28 and 0.72 to powers which sum to 14 Need 2 or more terms (= 0.0101 + 0.0548 + 0.1385 + 0.2154) A1 For completely correct unsimplified form = 0.419 A1 4 For correct final answer NB 0.418 is A0 if PA # 1 or A1 if PA # 2 (iii) ) 14 ( 28 . 0 50 = = x µ B1 For 14 and 10.08 seen, can be implied σ 2 = 50 x 0.28 x 0.72 (=10.08) M1 For standardising with or without cc, must have sq root P(more than 18) = 1 -       − Φ 08 . 10 14 5 . 18 M1 For continuity correction 17.5 or 18.5 AND a final answer < 0.5 = 1 – ) 417 . 1 ( Φ = 1 – 0.9218 or 0.9217 = 0.0782 or 0.0783 A1 4 For correct answer NB 0.078 is A0 if RE # 1 or A1 if RE # 2