Cambridge A Level Mathematics 9709 — 2007 Oct/Nov Paper 6 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/06 Paper 6 Probability & Statistics 1 (S1) October/November 2007 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 2007 [Turn over *4005043363*

2 1 A summary of 24 observations of x gave the following information: Σ(x −a) = −73.2 and Σ(x −a)2 = 2115. The mean of these values of x is 8.95. (i) Find the value of the constant a. [2] (ii) Find the standard deviation of these values of x. [2] 2 The random variable X takes the values −2, 0 and 4 only. It is given that P(X = −2) = 2p, P(X = 0) = p and P(X = 4) = 3p. (i) Find p. [2] (ii) Find E(X) and Var(X). [4] 3 The six digits 4, 5, 6, 7, 7, 7 can be arranged to give many different 6-digit numbers. (i) How many different 6-digit numbers can be made? [2] (ii) How many of these 6-digit numbers start with an odd digit and end with an odd digit? [4] 4 The random variable X has a normal distribution with mean 4.5. It is given that P(X > 5.5) = 0.0465 (see diagram). (i) Find the standard deviation of X. [3] (ii) Find the probability that a random observation of X lies between 3.8 and 4.8. [4] © UCLES 2007 9709/06/O/N/07

3 5 The arrival times of 204 trains were noted and the number of minutes, t, that each train was late was recorded. The results are summarised in the table. Number of minutes late (t) −2 ≤t < 0 0 ≤t < 2 2 ≤t < 4 4 ≤t < 6 6 ≤t < 10 Number of trains 43 51 69 22 19 (i) Explain what −2 ≤t < 0 means about the arrival times of trains. [1] (ii) Draw a cumulative frequency graph, and from it estimate the median and the interquartile range of the number of minutes late of these trains. [7] 6 On any occasion when a particular gymnast performs a certain routine, the probability that she will perform it correctly is 0.65, independently of all other occasions. (i) Find the probability that she will perform the routine correctly on exactly 5 occasions out of 7. [2] (ii) On one day she performs the routine 50 times. Use a suitable approximation to estimate the probability that she will perform the routine correctly on fewer than 29 occasions. [5] (iii) On another day she performs the routine n times. Find the smallest value of n for which the expected number of correct performances is at least 8. [2] 7 Box A contains 5 red paper clips and 1 white paper clip. Box B contains 7 red paper clips and 2 white paper clips. One paper clip is taken at random from box A and transferred to box B. One paper clip is then taken at random from box B. (i) Find the probability of taking both a white paper clip from box A and a red paper clip from box B. [2] (ii) Find the probability that the paper clip taken from box B is red. [2] (iii) Find the probability that the paper clip taken from box A was red, given that the paper clip taken from box B is red. [2] (iv) The random variable X denotes the number of times that a red paper clip is taken. Draw up a table to show the probability distribution of X. [4] © UCLES 2007 9709/06/O/N/07

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. 9709/06/O/N/07

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2007 question paper 9709 MATHEMATICS 9709/06 Paper 6, 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 October/November 2007 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 – October/November 2007 9709 06 © UCLES 2007 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 – October/November 2007 9709 06 © UCLES 2007 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 – October/November 2007 9709 06 © UCLES 2007 1 (i) –73.2/24 (= -3.05) a = 8.95 + 3.05 = 12 OR 8.95×24 (= 214.8) 2. 73 − = Σ − Σ a x 12 288 = = Σ a a (ii) standard deviation = 2) 05 .3 ( 24 2115 − − = 8.88 OR sd = 2 95 .8 24 2. 3814 − = 8.88 M1 A1 M1 A1 2 M1 A1 M1 A1 2 Accept (-72.4 + anything)/ 24 Correct answer For 8.95×24 seen Correct answer obtained using Σx and Σa For 2) mean coded their ( 24 2115 ± − Correct answer For 2 2 95 .8 24 their − Σx where Σx2 is obtained from expanding 2 ) ( a x − Σ with 2aΣx seen Correct answer 2 (i) 2p +p +3p = 1 p = 1/6 (= 0.167) (ii) E( X) = -2× 2/6 + 0 + 4 × 3/6 = 4/3 (=1.33) Var (X) = 4 × 2/6 + 0 + 16 × 3/6 – (4/3)2 = 7.56 (68/9) M1 A1 2 M1 A1ft M1 A1 4 Equation involving ps and summing to 1 Correct answer Using correct formula for E(X), in terms of p or their p< 1 Correct expectation ft on their p if 3 / 1 ≤ p Substitution in their Σpx2 – their E2(X) need 2 terms Correct answer 3 (i) 120 !3 !6 = (ii) 5 ….7 = 12 !2 !4 = 7…. 5 = 12 !2 !4 = 7…. 7 = !4 = 24 total = 48 M1 A1 2 M1 B1 B1 A1 4 For dividing by 3! Correct answer For identifying different cases For 4!/2! seen For 4! alone seen or in a sum or product Correct final answer

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9709 06 © UCLES 2007 4 (i) z = ± 1.68 σ 5.4 5.5 − = z σ = 0.595 accept 25/42 (ii) 5952 .0 5.4 8.3 1 − = z = -1.176 5952 .0 5.4 8.4 2 − = z = 0.504 prob = Φ(0.504) – (1-Φ(1.176)) = 0.6929 – (1-0.8802) = 0.573 B1 M1 A1 3 M1 A1ft M1 A1 4 Number rounding to 1.68 seen Standardising and attempting to solve with their z, ; must be z value, no cc, no σ2 ,no σ Correct answer For standardising 3.8 or 4.8, mean 4.5 not 5.5, their σ or σ or σ2 in denom One correct z-value, ft on their σ Correct area ie Φ1 + Φ2 - 1 or Φ1 - Φ2 if µ taken to be 5.5 Correct answer only 5 (i) some trains were up to 2 minutes early (ii) cf table B1 1 Or sensible equivalent, must use the idea ‘early’ 2 not needed NB All M marks are independent. Min late, less than 0 2 4 6 10 C freq 43 94 163 185 204 M1 Attempt at C F table with upper limits no halves cf 200 150 100 or curve 50 2 0 2 4 6 8 10 mins Median = rounding to 2.1 to 2.4 min IQ range = rounding to 3.2 to 3.6 min M1 M1 M1 M1 B1 A1 7 Uniform linear scales from at least 0 to 10 and 0 to 204 and at least one axis labelled, CF or mins or t Attempt at graph their 5 points. (-2, 0) not nec (could be midpoints or lower bounds not f d) Attempt at median along 102 or 102.5 line Attempt at LQ along 51/52 line and UQ along 153/154 line from graph Correct median Correct IQ range allow from midpoints etc

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9709 06 © UCLES 2007 6 (i) P(X=5) = (0.65)5× (0.35)2× 7C5 = 0.298 allow 0.2985 (ii) µ = 65 .0 50× (= 32.5), σ2 = 35 .0 65 .0 50 × × (= 11.375) P(fewer than 29) =       − Φ 375 . 11 5. 32 5. 28 = 1 – Φ(1.186) = 1 – 0.8822 = 0.118 (iii) 0.65 n ≥8 smallest n = 13 M1 A1 2 B1 M1 M1 M1 A1 5 M1 A1 2 Expression with 3 terms, powers summing to 7 and a 7C term Correct answer 32.5 and 11.375 seen or implied standardising, with or without cc, must have sq rt for continuity correction 28.5 or 29.5 correct area ie < 0.5 must be from a normal approx correct answer equality or inequality with np and 8 correct answer 7 (i) P(W, R) = 1/6 × 7/10 =7/60 (0.117) (ii) P(R, R) = 5/6× 8/10 (=40/60) P(red) = 47/60 (= 0.783) (iii) P( ) ( ) ( ) R P R R P R R ∩ = = 47 40 (= 0.851) (iv) P(R, W) = 5/6× 2/10 = 10/60 P(W, W) = 1/6× 3/10 = 3/60 M1 A1 2 M1 A1 2 M1 A1 2 B1 B1 For a single product with 6 and 10 in denoms Correct answer For finding their P(R ,R) and adding it to their (i) Correct answer Their P(R ,R) / their P(R) ie something × 5/6 ÷ their (ii) Correct answer x = 0, 1, 2, only, seen, no probabilities needed one correct probability x 0 1 2 P(X=x) 3/60 17/60 40/60 B1 B1ft 4 another correct probability ft if only one prob correct and Σp = 1