Cambridge A Level Mathematics 9709 — 2012 May/June Paper 7 · Variant 3

*9878370743* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level MATHEMATICS 9709/73 Paper 7 Probability & Statistics 2 (S2) 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_73/RP © UCLES 2012 [Turn over

2 1 Leaves from a certain type of tree have lengths that are distributed with standard deviation 3.2 cm. A random sample of 250 of these leaves is taken and the mean length of this sample is found to be 12.5 cm. (i) Calculate a 99% confidence interval for the population mean length. [3] (ii) Write down the probability that the whole of a 99% confidence interval will lie below the population mean. [1] 2 The independent random variables X and Y have the distributions N(6.5, 14) and N(7.4, 15) respectively. Find P(3X −Y < 20). [5] 3 The lengths, x mm, of a random sample of 150 insects of a certain kind were found. The results are summarised by Σ x = 7520 and Σx2 = 413 540. (i) Calculate unbiased estimates of the population mean and variance of the lengths of insects of this kind. [3] (ii) Using the values found in part (i), calculate an estimate of the probability that the mean length of a further random sample of 80 insects of this kind is greater than 53 mm. [3] 4 The number of lions seen per day during a standard safari has the distribution Po(0.8). The number of lions seen per day during an off-road safari has the distribution Po(2.7). The two distributions are independent. (i) Susan goes on a standard safari for one day. Find the probability that she sees at least 2 lions. [2] (ii) Deena goes on a standard safari for 3 days and then on an off-road safari for 2 days. Find the probability that she sees a total of fewer than 5 lions. [3] (iii) Khaled goes on a standard safari for n days, where n is an integer. He wants to ensure that his chance of not seeing any lions is less than 10%. Find the smallest possible value of n. [3] 5 (i) Deng wishes to test whether a certain coin is biased so that it is more likely to show Heads than Tails. He throws it 12 times. If it shows Heads more than 9 times, he will conclude that the coin is biased. Calculate the significance level of the test. [3] (ii) Deng throws another coin 100 times in order to test, at the 5% significance level, whether it is biased towards Heads. Find the rejection region for this test. [5] © UCLES 2012 9709/73/M/J/12

3 6 Last year Samir found that the time for his journey to work had mean 45.7 minutes and standard deviation 3.2 minutes. Samir wishes to test whether his journey times have increased this year. He notes the times, in minutes, for a random sample of 8 journeys this year with the following results. 46.2 41.7 49.2 47.1 47.2 48.4 53.7 45.5 It may be assumed that the population of this year’s journey times is normally distributed with standard deviation 3.2 minutes. (i) State, with a reason, whether Samir should use a one-tail or a two-tail test. [2] (ii) Show that there is no evidence at the 5% significance level that Samir’s mean journey time has increased. [5] (iii) State, with a reason, which one of the errors, Type I or Type II, might have been made in carrying out the test in part (ii). [2] 7 x f( )x 0 p 2 3 A random variable X has probability density function given by f(x) = ( k sin x 0 ≤x ≤2 3π, 0 otherwise, where k is a constant, as shown in the diagram. (i) Show that k = 2 3. [2] (ii) Show that the median of X is 1.32, correct to 3 significant figures. [4] (iii) Find E(X). [4] © UCLES 2012 9709/73/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. 9709/73/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 9709/73 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, 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 73 © 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 73 © 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 73 © University of Cambridge International Examinations 2012 1 (i) z = 2.574 to 2.576 12.5 ± z 250 2.3 12.0 to 13.0 (3 sfs) B1 M1 A1 3 Any z Correct form Allow 12 to 13 (ii) 0.005 or 0.5% B1 1 Not just 0.5 [Total 4] 2 (i) E(3X - Y) = 12.1 Var(3X - Y) = 9 × 14 + 15 = (141) ' 141 ' 1. 12 20− (= 0.665) Φ(‘0.665’) = 0.747 (3 sfs) B1 B1 M1 M1 A1 5 Allow without √ (No Continuity Correction) Correct area consistent with their working [Total 5] 3 (i) x = 150 7520 = (50.1) (3 sfs) s2 = 2 150 413540 7520 149 150 150 ( ( ) ) − = 245 or 246 (3 sfs) B1 M1 A1 3 Attempt at unbiased variance (either formula) Allow s2 = 15.72 (3 sfs) (ii) 80 ' 217 . 245 ' 150 7520 53− (= 1.637 to 1.638) 1 – Φ(‘1.637’) = 0.0488 to 0.0509 M1 M1 A1 3 For Standardising (±) with their mean and their variance must have √80 (ignore cc) Correct area consistent with their working Correct working only [Total 6] 4 (i) 1 – e-0.8(1 + 0.8) = 0.191 (3 sfs) M1 A1 2 Allow one end error (ii) λ = 3 × 0.8 + 2 × 2.7 (= 7.8) e-‘7.8’(1 + 7.8 + !4 8.7 !3 8.7 2 8.7 4 3 2 + + ) = 0.112 (3 sfs) M1 M1 A1 3 Attempt find λ P(0, 1, 2, 3, 4) Using their λ. Allow one end error (iii) e-0.8n < 0.1 Allow ‘=’ -0.8n < ln0.1 Allow ‘=’ min n = 3 M1* M1* dep A1 3 or e-x < 0.1 -x < ln0.1 Correctly obtained e-1.6 = 0.20 } M1 for trial and improvement to find P(0) try n = 2 e-2.4 = 0.09 } M1*dep try n = 3 both correct n = 3 A1 [Total 8]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 73 © University of Cambridge International Examinations 2012 5 (i) P(> 9 Heads | unbiased) = 12C10 × 0.510 × 0.52 + 12 × 0.511 × 0.5 + 0.512 = 0.0193 Level is 1.93% or 1.9% M1 M1 A1 3 Allow Bin P(X = 9, 10, 11, 12) correct or 1 – P(X = (9),10,11,12)) any p/q Allow Bin P (X = 9, 10, 11, 12) correct p/q Allow 2% if correct working seen (ii) B(100, 0.5) ≈ N(50, 25) ' 25 ' ' 50 ' 5.0 − − x = z z = 1.645 x = 58.7 Rejection region is > 59 B1 M1 B1 A1 A1ft 5 Or proportion method N(0.5,0.0025) Allow with wrong or no cc or no √ (cc for proportion method 0.5/100) + only ( consistent with their standardisation) or > 58 (region and integer required) [Total 8] 6 (i) Test is for bias in one direction One-tail B1 B1 2 ‘Increased’ rather than ‘changed’ or statement that µ >45.7 dep 1st B1 (ii) H0: pop mean = 45.7 H1: pop mean > 45.7 x = 47.375 or 47.4 or 379/8 8 2.3 7. 45 ' 375 . 47 ' − (= 1.481 to 1.503)) z = 1.645 ‘1.481’ < 1.645 hence no evidence mean time increased (AG) B1 B1 M1 M1 A1 5 Allow µ, but not ‘mean’ (follow through their (i)) Allow without √ Explicit comparison with their z from table Comparison with 1.645 or probability (0.0664 to 0.0693) with 0.05 Correct conclusion – accept H0 No errors seen Not rejected H0 Type II possible B1 B1 2 dep 1st B1 No contradictions for either mark [Total 9]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 73 © University of Cambridge International Examinations 2012 7 (i) k 2 3 0 sin d x x π ∫ = 1 2 3 0 [ cos ] k x π − k[–cos 3 2π + cos 0] = 1 k[0.5 + 1] = 1 (k = 3 2 AG) M1 A1 2 Integ & = 1. Ignore limits Must see this line or next (ii) 3 2 0 sin d m x x ∫ = 0.5 0 2[ cos ] 0.5 3 m x − = 3 2 (–cosm + 1) = 0.5 cosm = 0.25 m = 1.32 (3 sfs) AG M1* M1* dep A1 A1 4 Integ & = 0.5. Ignore limits Correct integrand & limits 0 to unknown & = 0.5 But allow a cos m = b where b/a = 0.25 dep cosm = 0.25 seen NB accept full verification (iii) 3 2 2 3 0 sin d x x x π ∫ 2 3 0 2{[ ( cos )] 3 x x π = − – 2 3 0 ( cos ) x dx π − ∫ } 2 3 0 2{ 0 [ sin ] 3 3 x π π = − −− = 3 2 ( 3 π + sin 3 2π ) = 9 3 3 2 + π or 1.28 (3 sf) M1 M1* dep M1* dep A1 4 Integ xf(x). Ignore limits 1st step attempted ie x(-cosx) oe. Ignore limits 2nd step attempted including correct limits applied oe [Total 10]