Cambridge A Level Mathematics 9709 — 2004 Oct/Nov Paper 7 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level HIGHER MATHEMATICS MATHEMATICS 8719/07 9709/07 Paper 7 Probability & Statistics 2 (S2) 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 number of radioactive particles emitted per second by a certain metal is random and has mean 1.7. The radioactive metal is placed next to an object which independently emits particles at random such that the mean number of particles emitted per second is 0.6. Find the probability that the total number of particles emitted in the next 3 seconds is 6, 7 or 8. [4] 2 Over a long period of time it is found that the amount of sunshine on any day in a particular town in Spain has mean 6.7 hours and standard deviation 3.1 hours. (i) Find the probability that the mean amount of sunshine over a random sample of 300 days is between 6.5 and 6.8 hours. [4] (ii) Give a reason why it is not necessary to assume that the daily amount of sunshine is normally distributed in order to carry out the calculation in part (i). [1] 3 A random sample of 150 students attending a college is taken, and their travel times, t minutes, are measured. The data are summarised by Σt = 4080 and Σt2 = 159 252. (i) Calculate unbiased estimates of the population mean and variance. [3] (ii) Calculate a 94% confidence interval for the population mean travel time. [4] 4 The weights of men follow a normal distribution with mean 71 kg and standard deviation 7 kg. The weights of women follow a normal distribution with mean 57 kg and standard deviation 5 kg. The total weight of 5 men and 2 women chosen randomly is denoted by X kg. (i) Show that E(X) = 469 and Var(X) = 295. [2] (ii) The total weight of 4 men and 3 women chosen randomly is denoted by Y kg. Find the mean and standard deviation of X −Y and hence find P(X −Y > 22). [5] 5 Of people who wear contact lenses, 1 in 1500 on average have laser treatment for short sight. (i) Use a suitable approximation to find the probability that, of a random sample of 2700 contact lens wearers, more than 2 people have laser treatment. [4] (ii) In a random sample of n contact lens wearers the probability that no one has laser treatment is less than 0.01. Find the least possible value of n. [3] 6 A continuous random variable X has probability density function given by f(x) =
3(1 −x)2 0 ≤x ≤1, 0 otherwise. Find (i) P(X > 0.5), [3] (ii) the mean and variance of X. [6] © UCLES 2004 9709/07/O/N/04

3 7 In a research laboratory where plants are studied, the probability of a certain type of plant surviving was 0.35. The laboratory manager changed the growing conditions and wished to test whether the probability of a plant surviving had increased. (i) The plants were grown in rows, and when the manager requested a random sample of 8 plants to be taken, the technician took all 8 plants from the front row. Explain what was wrong with the technician’s sample. [1] (ii) A suitable sample of 8 plants was taken and 4 of these 8 plants survived. State whether the manager’s test is one-tailed or two-tailed and also state the null and alternative hypotheses. Using a 5% significance level, find the critical region and carry out the test. [7] (iii) State the meaning of a Type II error in the context of the test in part (ii). [1] (iv) Find the probability of a Type II error for the test in part (ii) if the probability of a plant surviving is now 0.4. [2] © UCLES 2004 9709/07/O/N/04

4 BLANK PAGE 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. 9709/07/O/N/04

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary and Advanced Level MARK SCHEME for the November 2004 question papers 9709 MATHEMATICS 8719 HIGHER MATHEMATICS 8719/07, 9709/07 – Paper 7 (Probability and Statistics 2) 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 8719 and 9709 (Mathematics and Higher Mathematics) in the November 2004 examination. minimum mark required for grade: maximum mark available A B E Component 7 50 41 38 23 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 AND AS LEVEL MARK SCHEME MAXIMUM MARK: 50 SYLLABUS/COMPONENT: 8719/07 AND 9709/07 MATHEMATICS AND HIGHER MATHEMATICS Paper 7 (Probability and Statistics 2)

Page 1 Mark Scheme Syllabus Paper GCE AS/A LEVEL EXAMINATIONS – NOVEMBER 2004 8719 and 9709 7 © University of Cambridge International Examinations 2005 1 9 . 6 3 3 . 2 = × = λ P(6, 7, 8) =       + + − !8 9 . 6 !7 9 . 6 !6 9 . 6 e 8 7 6 9 . 6 = ( ) 06 . 425 e 9 . 6 − = 0.428 M1 A1 A1ft A1 4 For attempt at Poisson, any mean For correct mean For correct expression with their mean For correct answer 2(i) X ~       300 1 . 3 ,7 . 6 2 N 5587 . 0 300 / 1 . 3 7 . 6 8 . 6 1 = − = z 117 . 1 300 / 1 . 3 7 . 6 5 . 6 2 − = − = z Prob = Φ(0.5587) – {1 – Φ(1.117)} = 0.7119 – (1 – 0.8679) = 0.580 (ii) 300 is large, so X is approx normal even if X is not i.e. CLT application M1 A1 M1 A1 4 B1 1 For standardising, (with or without 300 in denom) For two correct expressions for z For subtracting 2 probabilities For correct answer For reference to large n and/or CLT 3(i) 2 . 27 150 4080 = = x 324 150 4080 159252 149 1 2 2 =       − = s B1 M1 A1 3 For 4080/150 For correct expression, (from formulae sheet or equiv.) For correct answer (ii) 94% CI 150 324 882 . 1 2 . 27 × ± = = (24.4, 30.0) M1 B1 A1ft A1 4 For one of correct form n s z x n s z x × − × + or For z = 1.881 or 1.882 only For correct expression with their x z s and , 150 / Or equivalent statement (c.w.o.) 4(i) 5M + 2W ~ N(355 + 114, 72× 5 + 52 × 2) ~ N(469, 295) B1 B1 2 For mean = 5× 71 + 2× 57 For variance = 72× 5 + 52× 2 (ii) Y ~ 4M + 3W ~ N(455, 271) X – Y ~ (5M + 2W) – (4M + 3W) ~ N(14, 566) Mean = 14, s.d. = 8 . 23 566 = P (X – Y > 22) = 1 – Φ       − 566 14 22 = 1 – Φ(0.3363) = 1 – 0.631 or 1 – 0.632 = 0.368 or 0.369 B1 M1 A1ft M1 A1 5 For correct mean and variance of 4M + 3W For adding their two variances and subtracting their two means For both correct (must be s.d.), ft on wrong mean and var of Y For standardising and using tables, either end, need the sq rt For correct answer

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL EXAMINATIONS – NOVEMBER 2004 8719 and 9709 7 © University of Cambridge International Examinations 2005 5(i) 8 . 1 = λ P(X > 2) = 1 – [P(0) + P(1) + P(2)] =       + + − − !2 8 . 1 8 . 1 1 e 1 2 8 . 1 = 1 – 0.7306 = 0.269 M1 A1 M1 A1 4 For attempt at Poisson, any mean For correct mean For finding 1 – P(0) – P(1) – P(2) or 1 – P(0) – P(1) For correct answer SR1 Normal scores B1 for 2.5 – 1.8/ ) 7988 . 1 ( SR2 Binomial scores M1 for complete method leading to final answer of 0.269 A1 (ii) 1500 / n = λ or P(0) < 0.01 i.e. 1500 n e − < 0.01 1500 n − < In 0.01 n > 6907.7 n = 6908 OR (1499/1500)n < 0.01 n = 6906 B1 M1 A1 3 (B1) (M1) (A1) For correct Poisson mean For equation or inequality involving their P(0) and 0.01 For correct answer For correct Binomial p For correct equation/inequality involving their P(0) and 0.01 For correct answer 6(i) ( ) ( ) ∫         − − = − 1 5 . 0 1 5 . 0 3 2 3 1 3 d 1 3 x x x = [0] – [−1](0.5)3 = 0.125 M1 A1 A1 3 For attempt at integrating and using limits Or equivalent correct integration (missing factors of 3 can still gain A1) For correct answer (ii) ( ) ( ) ∫ ∫ + − = − = 1 0 1 0 3 2 2 d 3 6 3 d 1 3 E x x x x x x x X = 1 0 4 3 2 4 3 3 6 2 3         + − x x x = 25 . 0 4 3 2 2 3 = + − Var (X) = ( ) ( ) [ ] ∫ − − 1 0 2 2 2 E d 1 3 X x x x = ( ) ∫ − + − 1 0 2 4 3 2 25 . 0 d 3 6 3 x x x x = ( )2 1 0 5 4 3 25 . 0 5 3 4 6 3 3 −         + − x x x = 0.0375 M1 A1 A1 M1 B1 A1 6 For attempt at ( ) ∫ x x x d f with or without limits For 2 or 3 correct parts of the integral (missing factors of 3 can still gain A1) For correct answer For attempt at ( ) ( ) [ ] ( ) [ ] ∫ − − 2 2 2 E . i.e E d f X X x x x must be seen even if it is ignored in the next line For 2 or 3 correct parts of the integral (missing factors of 3 can still gain A1) For correct answer

Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL EXAMINATIONS – NOVEMBER 2004 8719 and 9709 7 © University of Cambridge International Examinations 2005 7(i) not random, could be more light etc. B1 1 Any sensible reason (ii) One-tailed test H0: p = 0.35 H1: p > 0.35 P(8) = 0.358 = 0.000225 P(7) = 0.357× 0.651× 8C7 = 0.0033456 P(6) = 0.356× 0.652× 8C6 = 0.02174 P(5) = 0.355× 0.653× 8C5 = 0.08077 Crit region is 6, 7, 8 survive 4 is not in CR (OR Pr( ≥4) = 0.294 and comparison 0.5/or equiv.) ⇒ no significant improvement in survival rate B1 B1 M1* M1 A1 M1*dep A1ft 7 For correct answer For H0 and H1 For attempt at any Bin expression P(0) – P(8) For summing probabilities starting at P(8) and working backwards until > 0.05 (or equiv.) For correct answer For deciding whether 4 is in their CR or not OR finding relevant prob and showing comparison For correct conclusion (ft from their critical region) (iii) Saying no improvement when there is B1 1 Or equivalent, relating to the question (iv) Need P(0, 1, 2, 3, 4, 5) or 1 – P(6, 7, 8) P(8) = 0.48 (= 0.0006554) P(7) = 0.47× 0.6× 8C7 (= 0.007864) P(6) = 0.46× 0.62× 8C6 (= 0.04128) 1 – (0.48 + 0.47× 0.6× 8C7 + 0.46× 0.62× 8C6) = 0.950 M1 A1 2 For identifying type II error For correct answer