Cambridge A Level Mathematics 9709 — 2008 May/June Paper 7 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level MATHEMATICS 9709/07 Paper 7 Probability & Statistics 2 (S2) May/June 2008 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 2008 [Turn over *6044709223*

2 1 A magazine conducted a survey about the sleeping time of adults. A random sample of 12 adults was chosen from the adults travelling to work on a train. (i) Give a reason why this is an unsatisfactory sample for the purposes of the survey. [1] (ii) State a population for which this sample would be satisfactory. [1] A satisfactory sample of 12 adults gave numbers of hours of sleep as shown below. 4.6 6.8 5.2 6.2 5.7 7.1 6.3 5.6 7.0 5.8 6.5 7.2 (iii) Calculate unbiased estimates of the mean and variance of the sleeping times of adults. [3] 2 The lengths of time people take to complete a certain type of puzzle are normally distributed with mean 48.8 minutes and standard deviation 15.6 minutes. The random variable X represents the time taken in minutes by a randomly chosen person to solve this type of puzzle. The times taken by random samples of 5 people are noted. The mean time X is calculated for each sample. (i) State the distribution of X, giving the values of any parameters. [2] (ii) Find P(X < 50). [3] 3 The lengths of red pencils are normally distributed with mean 6.5 cm and standard deviation 0.23 cm. (i) Two red pencils are chosen at random. Find the probability that their total length is greater than 12.5 cm. [3] The lengths of black pencils are normally distributed with mean 11.3 cm and standard deviation 0.46 cm. (ii) Find the probability that the total length of 3 red pencils is more than 6.7 cm greater than the length of 1 black pencil. [4] 4 People who diet can expect to lose an average of 3 kg in a month. In a book, the authors claim that people who follow a new diet will lose an average of more than 3 kg in a month. The weight losses of the 180 people in a random sample who had followed the new diet for a month were noted. The mean was 3.3 kg and the standard deviation was 2.8 kg. (i) Test the authors’ claim at the 5% significance level, stating your null and alternative hypotheses. [5] (ii) State what is meant by a Type II error in words relating to the context of the test in part (i). [2] © UCLES 2008 9709/07/M/J/08

3 5 When a guitar is played regularly, a string breaks on average once every 15 months. Broken strings occur at random times and independently of each other. (i) Show that the mean number of broken strings in a 5-year period is 4. [1] A guitar is fitted with a new type of string which, it is claimed, breaks less frequently. The number of broken strings of the new type was noted after a period of 5 years. (ii) The mean number of broken strings of the new type in a 5-year period is denoted by λ. Find the rejection region for a test at the 10% significance level when the null hypothesis λ = 4 is tested against the alternative hypothesis λ < 4. [4] (iii) Hence calculate the probability of making a Type I error. [1] The number of broken guitar strings of the new type, in a 5-year period, was in fact 1. (iv) State, with a reason, whether there is evidence at the 10% significance level that guitar strings of the new type break less frequently. [2] 6 People arrive randomly and independently at the elevator in a block of flats at an average rate of 4 people every 5 minutes. (i) Find the probability that exactly two people arrive in a 1-minute period. [2] (ii) Find the probability that nobody arrives in a 15-second period. [2] (iii) The probability that at least one person arrives in the next t minutes is 0.9. Find the value of t. [4] 7 If Usha is stung by a bee she always develops an allergic reaction. The time taken in minutes for Usha to develop the reaction can be modelled using the probability density function given by f(t) = ⎧⎪⎨ ⎪⎩ k t + 1 0 ≤t ≤4, 0 otherwise, where k is a constant. (i) Show that k = 1 ln 5. [4] (ii) Find the probability that it takes more than 3 minutes for Usha to develop a reaction. [3] (iii) Find the median time for Usha to develop a reaction. [3] © UCLES 2008 9709/07/M/J/08

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. 9709/07/M/J/08

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2008 question paper 9709 MATHEMATICS 9709/07 Paper 7, 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. 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 discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2008 9709 07 © UCLES 2008 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 Syllabus Paper GCE A/AS LEVEL – May/June 2008 9709 07 © UCLES 2008 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 Syllabus Paper GCE A/AS LEVEL – May/June 2008 9709 07 © UCLES 2008 1 (i) commuters are not representative of the whole population (ii) people who travel to work on (this) train (iii) mean = 6.17 o.e. variance =       −12 74 56 . 463 11 1 2 = 0.657 B1 [1] B1 [1] B1 M1 A1 [3] Any sensible answer Substituting in formula from tables Correct answer 2 (i) X ∼ N(48.8, 15.62 / 5) (ii) P(( ) ( )      − Φ = < 5 / 6. 15 8. 48 50 50 X o.e. = Φ(0.1720) = 0.568 B1 B1 [2] M1 M1 A1 [3] For normal Correct mean and variance/s.d. Standardising with sq root Correct area > 0.5 Correct answer 3 (i) 2R ~ N(13.0, 2 x 0.232) P(2R > 12.5) = 1 –       − Φ 1058 .0 13 5. 12 = Φ (1.537) = 0.938 (ii) 3R – B ~ N(8.2, 3 × 0.232 + 0.462) P((3R – B) > 6.7) = 1 – Φ       − 3703 .0 2.8 7.6 = Φ ( 2.465) = 0.993 B1 M1 A1 [3] B1 M1 M1 A1 [4] Correct mean and variance Standardising and area > 0.5 Correct answer Correct mean and variance Considering P((3R – B) > 6.7 o.e. Correct probability area > 0.5 Correct answer 4 (i) H0: µ = 3 H1: µ > 3 Test statistic z = 179 / 8.2 3 3.3 − = 1.43 critical value z = 1.645 not enough evidence to support the claim B1 M1 A1 M1 A1ft [5] Both hypotheses correct Standardising attempt with sq rt in denom Correct z value accept rounding to 1.44 from 180 (OR alt method finding crit value 3.344 M1 A1) Comparing with z = 1.645 (or z consistent with their H1) or eqiv comparison of areas Correct answer, ft their z. No contradictions. (OR compare C.V 3.344 with 3.3 M1 A1ft)

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2008 9709 07 © UCLES 2008 (ii) Say no extra weight loss when there is. B2 [2] Correct statement in context SR1 B1 for partially correct statement in context SR2 B1 for any true statement about a Type II error including accept Ho when it is false 5 (i) 1 in 15 months is equivalent to 4 in 60 months (ii) P(0) = e–4 = 0.01831 P(1) = e–4× 4 = 0.07326 P(2) = e–4× 42/2 = 0.147 too big P(0) + P(1) = 0.0916 Rejection region at 10% level is 0 or 1. (iii) P(type I error) = 0.0916 (iv) 1 is in rejection region there is evidence that the new guitar string lasts longer B1 [1] M1* M1* M1 A1*dep [4] B1 [1] B1 B1ft [2] Or equivalent Attempt to find P(0) and / or P(1) Comparing sum with 0.10 Considering and rejecting P(2) Correct answer. No errors seen. Correct answer identifying where 1 is correct conclusion ft their rejection region 6 (i)  λ = 0.8 P(2) = 2 8.0 2 8.0 − e = 0.144 (ii) λ = 0.2 P(0) = e–0.2 = 0.819 (iii) 1 – e–0.8t = 0.9 ln 0.1 = –0.8t t = 2.88 M1 A1 [2] M1 A1 [2] M1 A1 M1 A1 [4] Attempt at Poisson calculation with attempt at λ Correct answer Attempt at P(0) with attempt at λ Correct answer Equation containing at least 0.9 and e–k Correct equation with 0.8t attempt to solve equation by ln correct answer 7 (i) 1 4 0 1 = + ∫ dt t k ( ) 1 ] 1 ln [ 4 0 = + t k k = 1/ln 5 AG (ii) P(T > 3) = dt t k ∫+ 4 3 1 = [ ]4 3 )1 ln( + t k = 1 – ln4/ln5 = 0.139 M1 A1 M1 A1 [4] M1 A1 A1 [3] Equating to 1 and attempt to integrate ln(t + 1) seen (or ln (x + 1) Correct use of limits 0 and 4 Correct given answer legit obtained Attempt to integrate with one limit 3 Correct integration with correct limits seen (o.e.) Correct answer

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2008 9709 07 © UCLES 2008 (iii) dt m t k ∫+ 0 1 = 0.5 ( ) m t k 0] 1 ln [ + = 0.5 k ln(m + 1) = 0.5 m = 1.24 min M1 M1 A1 [3] Equating to 0.5 and attempt to integrate Attempt to solve equation with at least k, ln and m on LHS and 0.5 on RHS correct answer