Cambridge A Level Mathematics 9709 — 2005 May/June 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) May/June 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 Exam marks, X, have mean 70 and standard deviation 8.7. The marks need to be scaled using the formula Y = aX + b so that the scaled marks, Y, have mean 55 and standard deviation 6.96. Find the values of a and b. [4] 2 Jenny has to do a statistics project at school on how much pocket money, in dollars, is received by students in her year group. She plans to take a sample of 7 students from her year group, which contains 122 students. (i) Give a suitable method of taking this sample. [1] Her sample gives the following results. 13.40 10.60 26.50 20.00 14.50 15.00 16.50 (ii) Find unbiased estimates of the population mean and variance. [3] (iii) Is the estimated population variance more than, less than or the same as the sample variance? [1] (iv) Describe what you understand by ‘population’ in this question. [1] 3 A survey of a random sample of n people found that 61 of them read The Reporter newspaper. A symmetric confidence interval for the true population proportion, p, who read The Reporter is 0.1993 < p < 0.2887. (i) Find the mid-point of this confidence interval and use this to find the value of n. [3] (ii) Find the confidence level of this confidence interval. [4] 4 A study of a large sample of books by a particular author shows that the number of words per sentence can be modelled by a normal distribution with mean 21.2 and standard deviation 7.3. A researcher claims to have discovered a previously unknown book by this author. The mean length of 90 sentences chosen at random in this book is found to be 19.4 words. (i) Assuming the population standard deviation of sentence lengths in this book is also 7.3, test at the 5% level of significance whether the mean sentence length is the same as the author’s. State your null and alternative hypotheses. [5] (ii) State in words relating to the context of the test what is meant by a Type I error and state the probability of a Type I error in the test in part (i). [2] 5 A clock contains 4 new batteries each of which gives a voltage which is normally distributed with mean 1.54 volts and standard deviation 0.05 volts. The voltages of the batteries are independent. The clock will only work if the total voltage is greater than 5.95 volts. (i) Find the probability that the clock will work. [4] (ii) Find the probability that the average total voltage of the batteries of 20 clocks chosen at random exceeds 6.2 volts. [3] © UCLES 2005 9709/07/M/J/05

3 6 At a petrol station cars arrive independently and at random times at constant average rates of 8 cars per hour travelling east and 5 cars per hour travelling west. (i) Find the probability that, in a quarter-hour period, (a) one or more cars travelling east and one or more cars travelling west will arrive, [4] (b) a total of 2 or more cars will arrive. [2] (ii) Find the approximate probability that, in a 12-hour period, a total of more than 175 cars will arrive. [3] 7 The random variable X denotes the number of hours of cloud cover per day at a weather forecasting centre. The probability density function of X is given by f(x) =   (x −18)2 k 0 ≤x ≤24, 0 otherwise, where k is a constant. (i) Show that k = 2016. [3] (ii) On how many days in a year of 365 days can the centre expect to have less than 2 hours of cloud cover? [3] (iii) Find the mean number of hours of cloud cover per day. [4] © UCLES 2005 9709/07/M/J/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. 9709/07/M/J/05

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary and Advanced Level MARK SCHEME for the June 2005 question papers 9709 MATHEMATICS 8719 HIGHER MATHEMATICS 8719/07, 9709/07 - Paper 7, 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 June 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Grade thresholds for Syllabus 8719 and 9709 (Mathematics and Higher Mathematics) in the June 2005 examination. minimum mark required for grade: maximum mark available A B E Component 7 50 39 34 18 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.

JUNE 2005 GCE AS AND A LEVEL MARK SCHEME MAXIMUM MARK: 50 SYLLABUS/COMPONENT: 9709/07, 8719/07 MATHEMATICS AND HIGHER MATHEMATICS Paper 7

Page 1 Mark Scheme Syllabus Paper GCE AS/A LEVEL – JUNE 2005 8719 and 9709 7 © University of Cambridge International Examinations 2005 1 55 = 70a + b 6.96 = 8.7a or 6.962 = 8.72 a2 a = 0.8 b = −1 M1 M1 A1 A1 (4) For an equation relating to the means For an equation relating to the variance or sd, only a in it For correct a For correct b 2 (i) Put names in a hat and draw out, or assign a number to each person in year and generate 7 random numbers by calculator. (ii) est pop mean 116.5/7 (= 16.6) est pop var = 27.1 (iii) more (iv) (pocket money of) all pupils in Jenny’s year at school B1 (1) B1 M1 A1 (3) B1 (1) B1 (1) Or any equivalent method, could use systematic sampling For using a correct formula (can be implied) For correct answer Need to see all of this 3 (i) (0.1993 + 0.2887)/2 (= 0.244) = 61/n n = 250 (ii) 0.0447 = 250 ) 244 . 0 1 ( 244 . 0 − × z (or equiv. equ. leading to this) z = 1.646 90% confidence interval B1 M1 A1 (3) M1 M1 A1 A1ft (4) For correct mid-point For equating their mid-point with 61/n For correct answer For equating half-width with n pq z × or equiv. For solving for z from a reasonable looking equation For obtaining z = 1.64 or 1.65 For correct answer (nearest whole no.) 4 (i) H0: µ = 21.2 H1: µ ≠ 21.2 Test statistic z = ) 90 / 3 . 7 ( 2 . 21 4 . 19 − = −2.34 CV z = ± 1.96 In CR, reject H0. Sig evidence to say not the same author or Φ (−2.339) = 1 − 0.9903 = 0.0097/0.0096 Compare with 0.025 say sig evidence to say not the same sentence length or author or x = 21.2 ± 1.96x(7.3/√ 90) = 19.7(22.7) Compare with 19.4 etc. (ii) Say it is not the same sentence length or author when it is P (Type I error) = 5% B1 M1 A1 M1 A1ft (5) M1 A1ft M1 A1 M1 A1ft B1 B1 (2) For H0 and H1 correct must be ≠ For standardising must have √ 90 For correct z accept +/- For correct comparison with correct critical value, ft from their H1 For correct conclusion ft on their z and their CV For correct Φ and correct comparison (consistent with H1) For correct conclusion ft on their Φ and 0.025 For expression for x with correct (consistent) z For correct comparison and conclusion(ft) For correct statement For correct answer

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – JUNE 2005 8719 and 9709 7 © University of Cambridge International Examinations 2005 5 (i) T~ N(1.54 × 4, 0.052 × 4) [~ N(6.16, 0.01)] P(T>5.95) = 1 - Φ {(5.95-6.16)/√0.01} = Φ (2.1) = 0.982 (ii) Av ~ N(6.16, 0.01/20) P(Av > 6.2) = 1 - Φ {(6.2 - 6.16) /√(0.01/20)} = 1 - Φ (1.789) = 1 - 0.9633 = 0.0367 or 0.0368 or Tot~ N(123.2, 0.2) P(Tot>124) = 1-Φ {(123.2-124) /√0.2} etc. B1 M1 M1 A1 (4) B1ft M1 A1 (3) B1ft M1 For mult mean and variance by 4 For standardising must have √ For correct area i.e. > 0.5 For correct answer For dividing their variance by 20 For standardising (must use consistent values) For correct answer 6 (i) (a) East P(≥1) = 1 − e−2 = 0.8647 West P(≥1) = 1 − e−1.25 = 0.7135 P(Both) = 0.8647 × 0.7135 = 0.617 (b) P(total ≥2) = 1 − e-13/4(1+13/4) = 0.835 or P(total ≥2) = P(2) + P(3) +…P(13) etc. (ii) T ~ N(156, 156) P(>175) = 1-Φ {(175.5-156) /√156} = 1 - Φ (1.5612) = 1 - 0.9407 = 0.0593/0.0592 B1 M1 M1 A1 (4) M1 A1 (2) M1 B1 M1 A1 (3) Correct mean of 2 or 1.25 used in Poisson expression One Poisson expression P(≥1)=1 - P(0) or 1 - P(0) - P(1) any mean. For multiplying their 2 probs together For correct answer For attempt at summing their means and for 1 - their P(0, 1) or 1 - their P(0,1,2) or 1 - P(0E,0W) - P(1E,0W) - P(0E,1W) or equiv. expression incl.2 For correct answer For correct mean and variance For standardising, with or without cc or sq rt For correct answer

Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – JUNE 2005 8719 and 9709 7 © University of Cambridge International Examinations 2005 7 (i) 1 ) 18 ( 24 0 2 = − ∫ k x 1 3 ) 18 ( 24 0 3 =       − k x 2016 1 2016 = ⇒ = k k AG (ii) x x x p d 2016 ) 18 ( ) 2 ( 2 0 2 ∫ − = < 2 0 3 2016 3 ) 18 (       × − = x 6048 ) 18 ( ) 16 ( 3 3 − − − = = 0.2870(31/108) Number of days = 0.287 × 365 = 104 or 105 (iii) mean = ( ) x k x x d 18 24 0 2 ∫ − 24 0 2 3 4 2 324 3 36 4 1       + − = x x x k = 5.14       7 1 5 M1 A1 A1 (3) M1 A1 B1ft (3) M1 A1 A1 A1 (4) For equating to 1 and attempting to integrate For correct integration with correct limits seen For given answer legit obtained For integration attempt between 0 and 2 (condone missing k) For correct answer For multiplying their prob by 365 For attempting to integrate xf(x) (condone missing k) For one correct integrated term with correct limits(condone missing k) or For integration by parts correct first stage answer with limits seen(condone missing k) For fully correct integrated expression with limits(condone missing k) For correct answer