Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 7 · Variant 2

*6370714324* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level MATHEMATICS 9709/72 Paper 7 Probability & Statistics 2 (S2) October/November 2013 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. JC13 11_9709_72/FP © UCLES 2013 [Turn over

2 1 Each computer made in a factory contains 1000 components. On average, 1 in 30 000 of these components is defective. Use a suitable approximate distribution to find the probability that a randomly chosen computer contains at least 1 faulty component. [4] 2 Heights of a certain species of animal are known to be normally distributed with standard deviation 0.17 m. A conservationist wishes to obtain a 99% confidence interval for the population mean, with total width less than 0.2 m. Find the smallest sample size required. [4] 3 Following a change in flight schedules, an airline pilot wished to test whether the mean distance that he flies in a week has changed. He noted the distances, x km, that he flew in 50 randomly chosen weeks and summarised the results as follows. n = 50 Σx = 143 300 Σx2 = 410 900 000 (i) Calculate unbiased estimates of the population mean and variance. [3] (ii) In the past, the mean distance that he flew in a week was 2850 km. Test, at the 5% significance level, whether the mean distance has changed. [5] 4 The number of radioactive particles emitted per 150-minute period by some material has a Poisson distribution with mean 0.7. (i) Find the probability that at most 2 particles will be emitted during a randomly chosen 10-hour period. [3] (ii) Find, in minutes, the longest time period for which the probability that no particles are emitted is at least 0.99. [5] 5 The volume, in cm3, of liquid left in a glass by people when they have finished drinking all they want is modelled by the random variable X with probability density function given by fx = T kx −22 0 ≤x ≤2, 0 otherwise, where k is a constant. (i) Show that k = 3 8. [2] (ii) 20% of people leave at least d cm3 of liquid in a glass. Find d. [3] (iii) Find EX. [3] © UCLES 2013 9709/72/O/N/13

3 6 At the last election, 70% of people in Apoli supported the president. Luigi believes that the same proportion support the president now. Maria believes that the proportion who support the president now is 35%. In order to test who is right, they agree on a hypothesis test, taking Luigi’s belief as the null hypothesis. They will ask 6 people from Apoli, chosen at random, and if more than 3 support the president they will accept Luigi’s belief. (i) Calculate the probability of a Type I error. [3] (ii) If Maria’s belief is true, calculate the probability of a Type II error. [3] (iii) In fact 2 of the 6 people say that they support the president. State which error, Type I or Type II, might be made. Explain your answer. [2] 7 Kieran and Andreas are long-jumpers. They model the lengths, in metres, that they jump by the independent random variables K ∼N5.64, 0.0576 and A ∼N4.97, 0.0441 respectively. They each make a jump and measure the length. Find the probability that (i) the sum of the lengths of their jumps is less than 11 m, [4] (ii) Kieran jumps more than 1.2 times as far as Andreas. [6] © UCLES 2013 9709/72/O/N/13

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. © UCLES 2013 9709/72/O/N/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2013 series 9709 MATHEMATICS 9709/72 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9709 72 © Cambridge International Examinations 2013 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 LEVEL – October/November 2013 9709 72 © Cambridge International Examinations 2013 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 LEVEL – October/November 2013 9709 72 © Cambridge International Examinations 2013 1 λ = 30 1 1 – 30 1 e − = 0.0328 (3 s.f.) B1 M1 M1 A1 [4] o.e 1 – P(X = 0) by Poisson, any λ allow 1 end error 1 – P(X = 0) by Poisson, correct λ no end errors S.R. Binomial with final answer 0.0328 B2 Correct answer, no working scores B2 2 z = 2.576 2 × z × n 17 .0 = 0.2 oe n = 2) ( 2.0 576 .2 17 .0 2 × × oe (= 19.2) Smallest n is 20 B1 M1 M1 A1 [4] Seen (accept 2.574 to 2.579) Allow without ‘2 ×’ OR with incorrect z Attempt to arrange equ of correct form (with correct z and ‘2×’ into the form n= or n = 3 (i) est (µ) = 2866 or 2870 (3 s.f.) est (σ2) = 49 1 (410900000 – 50 1433002 ) (= 4126.53) = 4130 (3 sf) B1 M1 A1 [3] Accept 143300/50 o.e. Correct subst in correct formula (ii) H0: Pop mean (or µ) = 2850 H1: Pop mean (or µ) ≠ 2850 50 ' 53 . 4126 ' 2850 50 143300 − = 1.761 ‘1.761’ < 1.96 No evidence mean distance changed B1 M1 A1 M1 A1f [5] Both. Not just ‘mean’ Allow ‘4126.53’ without , but must have all 50 Or correct c.v. (2867.81) for alt method For valid comparison of z values, areas or c.v. Dep 1.96; ft their 1.761 If H1: µ > 2850 and c.f. 1.645, max B0M1A1M1A0 (c.v. for 1 tail test 2864.94) 4 (i) λ = 2.8 ) 2 8.2 8.2 1( e 2 8.2 + + − = 0.469 (3 s.f.) or 0.47(0) B1 M1 A1 [3] seen any λ allow one end error As final answer (ii) e-0.7n [ 0.99 or e-λ [ 0.99 –0.7n [ ln 0.99 or –λ [ ln 0.99 n Y 0.01436 or λ Y 0.01005 ‘0.01436’ × 150 or ‘0.01005’ × 150 ÷ 0.7 Max period is 2.15 mins (3 sf) M1 M1 A1 M1 A1 [5] Allow ‘=’ throughout Attempt ln both sides Can be implied. Accept 3 s.f. Note n ) 150 / 7.0 ( e− [ 0.99 scores 1st and 3rd M1 T & I leading to ans 2.2 mins, SC: B2

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9709 72 © Cambridge International Examinations 2013 5 (i) ∫ − 2 0 2d ) 2 ( x x k = 1 ( 0 2 3 3 ) 2 (     − x k = 1) k [0 – ( 3 8 − )] = 1 k = 8 3 AG M1 A1 [2] Attempt to integrate f(x) with correct limits and = 1 Must see this line or better, e.g. k × 3 8 = 1 (ii) ∫ − 2 2 8 3 d ) 2 ( d x x = 0.2 ( d x 2 3 3 ) 2 ( 8 3       − = 0.2)      − − 3 3 ) 2 ( 8 3 0 d = 0.2 oe ( (d – 2)3 = –1.6) d = 0.83(0) (3 s.f.) M1 M1 A1 [3] ʃf(x)dx with limits d and 2 or 0 and d, and = 0.2 or =0.8 Condone missing ‘k’ Reasonable attempt to integrate from a correct expression, with limits substituted to give expression in d³. Condone missing ‘k’ (iii) ∫ − 2 0 2 d ) 2 ( 8 3 x x x (= ∫ + 2 0 2 3 d 4 4 - 8 3 x x x x ) = [ ]0 2 2 3 4 2 3 4 4 8 3 x x x + − = 2 1 M1 A1 A1 [3] Attempt integ xf(x); ignore limits, condone missing k ( 0 2 d 3 ) 2 ( 3 ) 2 ( 8 3 3 3    × ∫ − − − x x x x ) = 0 2 12 ) 2 ( 3 ) 2 ( 8 3 4 3     − − − × x x x Correct integration & limits, condone missing k 6 (i) P(Type I) = 1 - P(> 4 assuming p = 0.7) 1–(6C4 × 0.74×0.32 + 6C5 × 0.75 × 0.3 + 0.76) (= 1 – 0.744) = 0.256 (3 s.f.) M1 M1 A1 [3] or P(< 3 assuming p = 0.7) May be implied 6C3×0.73×0.33+6C2×0.72×0.34+6C1×0.7×0.35 + 0.36 Allow one end error = 0.256 (3 s.f.) SR if zero scored allow B1 for use of B(6, 0.7) in any two or more terms (ii) P(Type II) = P([ 4 assuming p = 0.35) = 6C4 × 0.354 × 0.652 + 6C5 × 0.355 × 0.65 + 0.356 = 0.117 M1 M1 A1 [3] May be implied Allow one end error SR if zero scored allow B1 for use of B(6, 0.35) in any two or more terms (iii) Type 1 They will reject Luigi’s belief, although it might be true. B1 B1 [2] In context

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9709 72 © Cambridge International Examinations 2013 7 (i) N(10.61, 0.1017) ' 1017 .0' ' 61 . 10 ' 11− (= 1.223) Φ(‘1.223’) = 0.889 (3 s.f.) B1 M1 M1 A1 [4] o.e. Stated or implied (accept in un-simplified form) Allow without √ For attempt to find correct area consistent with their working (ii) P(K – 1.2A > 0 ) Var = 0.0576 + 1.22 × 0.0441 (= 0.121104) N(–0.324, 0.121104) ' 121104 .0' ) 324 .0 ( 0 − − (= 0.931) 1 – Φ(‘0.931’) = 0.176 (3 s.f.) M1 B1 B1 M1 M1 A1 [6] Or similar stated or implied o.e. May be implied (accept in un-simplified form) Allow without √ For attempt to find correct area consistent with their working