Cambridge A Level Mathematics 9709 — 2005 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 2005 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 2005 [Turn over

2 1 A study of the ages of car drivers in a certain country produced the results shown in the table. Percentage of drivers in each age group Young Middle-aged Elderly Males 40 35 25 Females 20 70 10 Illustrate these results diagrammatically. [4] 2 Boxes of sweets contain toffees and chocolates. Box A contains 6 toffees and 4 chocolates, box B contains 5 toffees and 3 chocolates, and box C contains 3 toffees and 7 chocolates. One of the boxes is chosen at random and two sweets are taken out, one after the other, and eaten. (i) Find the probability that they are both toffees. [3] (ii) Given that they are both toffees, find the probability that they both came from box A. [3] 3 A staff car park at a school has 13 parking spaces in a row. There are 9 cars to be parked. (i) How many different arrangements are there for parking the 9 cars and leaving 4 empty spaces? [2] (ii) How many different arrangements are there if the 4 empty spaces are next to each other? [3] (iii) If the parking is random, find the probability that there will not be 4 empty spaces next to each other. [2] 4 A group of 10 married couples and 3 single men found that the mean age xw of the 10 women was 41.2 years and the standard deviation of the women’s ages was 15.1 years. For the 13 men, the mean age xm was 46.3 years and the standard deviation was 12.7 years. (i) Find the mean age of the whole group of 23 people. [2] (ii) The individual women’s ages are denoted by xw and the individual men’s ages by xm. By first finding Σx2 w and Σx2 m, find the standard deviation for the whole group. [5] 5 A box contains 300 discs of different colours. There are 100 pink discs, 100 blue discs and 100 orange discs. The discs of each colour are numbered from 0 to 99. Five discs are selected at random, one at a time, with replacement. Find (i) the probability that no orange discs are selected, [1] (ii) the probability that exactly 2 discs with numbers ending in a 6 are selected, [3] (iii) the probability that exactly 2 orange discs with numbers ending in a 6 are selected, [2] (iv) the mean and variance of the number of pink discs selected. [2] © UCLES 2005 9709/06/O/N/05

3 6 In a competition, people pay $1 to throw a ball at a target. If they hit the target on the first throw they receive $5. If they hit it on the second or third throw they receive $3, and if they hit it on the fourth or fifth throw they receive $1. People stop throwing after the first hit, or after 5 throws if no hit is made. Mario has a constant probability of 1 5 of hitting the target on any throw, independently of the results of other throws. (i) Mario misses with his first and second throws and hits the target with his third throw. State how much profit he has made. [1] (ii) Show that the probability that Mario’s profit is $0 is 0.184, correct to 3 significant figures. [2] (iii) Draw up a probability distribution table for Mario’s profit. [3] (iv) Calculate his expected profit. [2] 7 In tests on a new type of light bulb it was found that the time they lasted followed a normal distribution with standard deviation 40.6 hours. 10% lasted longer than 5130 hours. (i) Find the mean lifetime, giving your answer to the nearest hour. [3] (ii) Find the probability that a light bulb fails to last for 5000 hours. [3] (iii) A hospital buys 600 of these light bulbs. Using a suitable approximation, find the probability that fewer than 65 light bulbs will last longer than 5130 hours. [4] © UCLES 2005 9709/06/O/N/05

4 BLANK PAGE Every reasonable effort has been made to trace all copyright holders where the publishers (i.e. UCLES) are aware that third-party material has been reproduced. The publishers would be pleased to hear from anyone whose rights they have unwittingly infringed. University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. 9709/06/O/N/05

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary and Advanced Level MARK SCHEME for the November 2005 question paper 9709/0390 MATHEMATICS 9709/06, 0390/06 Paper 6 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. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

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.

Page 1 Mark Scheme Syllabus Paper GCE A/AS LEVEL – November 2005 9709/0390 06 © University of Cambridge International Examinations 2005 1 two pie charts or 2 bars (m and f) 3 different age categories in each group correct height or angle labels m and f, percentage, drivers, y,m elderly M1 A1 B1 B1 4 3 lots of 2 or 2 lots of 3, bars, lines or sectors one category touching, not superimposed, one category not touching, bars equal width accept pie chart visually correct 2 (i) P(T, T) = 9 5 10 6 3 1 × × + 7 4 8 5 3 1 × × + 9 2 10 3 3 1 × × = 53/210 (0.252) B1 M1 A1 3 For one correct 3-factor term For summing three 3-factor or 2-factor probs For correct answer (ii) P(A TT) = 0.111/0.252 = 70/159 (0.440) M1 M1 A1 3 For choosing only their P(A T T) in num or denom For dividing by their (i) or what they think is P(T,T) For correct answer using either 2 or 3-term probs Constant prob B0M1A0M1M1A0 max 3 (i) 13P9 = 259,459,200 or 259,000,000 M1 A1 2 For using a permutation involving 13 For correct answer (ii) 10! or 10P9 = 3628800 M1 M1 A1 3 For using a 10 For using a 9! For correct answer (iii) 1 – (ii) / (i) = 0.986 M1 A1ft 2 For a subtraction of a suitable prob < 1, from 1 For correct answer, ft on their (i) and (ii) 4 (i) (41.2 × 10 + 46.3 × 13) / 23 = 44.1 M1 A1 2 For multiplying by 10 and 13 respctively and dividing by 23 For correct answer (ii) 2 2 2 2. 41 10 1. 15 − = ∑ w x Σxw 2 = 19254.5 2 2 2 3. 46 13 7. 12 − = ∑ m x Σxm 2 = 29964.74 Total Σ = 49219.24 sd =       − 2 1. 44 23 24 . 49219 = 14.0 M1 A1 A1 M1 A1 5 For correct substitution from recognisable formula with or without sq rt For correct Σxw 2 (can be rounded) For correct Σxm 2 (can be rounded) For using 23 and their answer to (i) in correct formula For correct answer

Page 2 Mark Scheme Syllabus Paper GCE A/AS LEVEL – November 2005 9709/0390 06 © University of Cambridge International Examinations 2005 5 (i) P(no orange) = (2/3)5 or 0.132 or 32/243 (ii) P(2 end in 6) = (1/10)2 × (9/10)3 × 5C2 = 0.0729 (iii) P(2 orange end in 6) = (1/30)2 × (29/30)3× 5C2 = 0.0100 accept 0.01 (iv) n = 5, p = 1/3, mean = 5/3, variance = 10/9 B1 1 B1 M1 A1 3 M1 A1 2 B1 B1 ft 2 For correct final answer either as a decimal or a fraction For using (1/10)k k>1 For using a binomial expression with their 1/10 or seeing some p2 *(1-p)3 For correct answer For their (1/10)/3 seen For correct answer For recognising n=5, p = 1/3 For correct mean and variance, ft their n and p, p<1 6 (i) $2 (ii) P(MMMH) +P(MMMMH) = 0.83 × 0.2 + 0.84 × 0.2 = 0.184 AG (iii) x 4 2 0 -1 P(X = x) 0.2 0.288 0.184 0.328 (iv) E(X) = 0.8 + 0.576 – 0.328 = $1.05 B1 1 M1 A1 2 B1 B1ft B1 3 M1 A1 2 For correct answer For attempting to sum P(MMMH) and P(MMMMH) For correct answer For one correct prob other than 0.184 For another correct prob other than 0.184, ft only if the –1 ignored and their 3rd prob is 1 – Σ the other 2 For correct table, can have separate 2s For attempt at Σ xp from their table, at least 2 non- zero terms For correct answer 7 (i) 1.282 = (5130-µ)/ 40.6 µ = 5080 (5078) rounding to 5080 (ii) P(<5000) = Φ[(5000-5078)/ 40.6] = Φ(-1.921) = 1 – 0.9727 = 0.0273 or 2.73% (iii) µ = 60, var = 54 P(fewer than 65) = Φ (64.5 – 60) / √ 54 = Φ(0.6123) = 0.730 accept 0.73 B1 M1 A1 3 M1 M1 A1 3 B1 M1 M1 A1 4 For ± 1.282 seen, or 1.28, 1.281, not 1.29 or 1.30 For standardising, with or without sq rt, squared, no cc For correct answer For standardising, criteria as above, can include cc For correct area found using tables ie < 0.5ft on wrong (i) For correct answer, accept 0.0274 For 60 and 54 seen (could be sd or variance) For using 64.5 or 65.5 in a standardising process For standardising, must have √( their 54) in denom For correct answer