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MATHS QUEST 12 Further Mathematics 4TH EDITION



MATHS QUEST 12 Further Mathematics 4TH EDITION VCE MATHEMATICS UNITS 3 & 4 ANTHONY NOVAK | RUTH BAKOGIANIS | KYLIE BOUCHER JENNIFER NOLAN | GEOFF PHILLIPS CONTRIBUTING AUTHORS ELENA IAMPOLSKY | MARK BARNES | STEPHEN HEAMES | ROBERT CAHN JANET HEFFERNAN | CHRIS LONGHURST | NICK SIMPSON SUPPORT MATERIAL JOHN DOWSEY | DENNIS FITZGERALD | EMILY HUI | CAROLINE MEWS | VINOD NARAYAN PETER SWAIN | DAVID TYNAN | IAN YOUNGER | WAYNE YOUNGS SIMONE RICHARDSON | DINA ANTONIOU | NORRENE HILL

Fourth edition published 2013 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Third edition published 2010 Second edition published 2006 First edition published 2000 Typeset in 10/12 pt Times LT Std © John Wiley & Sons Australia, Ltd 2000, 2006, 2010, 2013 The moral rights of the authors have been asserted. National Library of Australia Cataloguing-in-Publication data Author: No vak, Anthony. Title: Maths Quest 12 Further Mathematics/ Anthon y Novak. Edition: 4th ed. Publisher: Milton, Qld: John Wiley & Sons Australia, 2009. ISBN: 978 1 1183 1786 0 (student pbk.) 978 1 1183 1793 8 (fle xisaver) 978 1 1183 1787 7 (student eBook) Notes: Includes inde x. Target Audience: F or secondary school age. Subjects: Mathematics — Textbooks. Mathematics — Problems, e xercises, etc. Dewey Number: 510 Repr oduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Cover and internal design images: © vic&dd/Shutterstock.com Cartography by MAPgraphics Pty Ltd Brisbane Illustrated by Aptara and Wiley Composition Services Typeset in India by Aptara Printed in China by Shenzhen Donnelley Printing Co., Ltd. 10 9 8 7 6 5 4 3

Contents Introduction ix About eBookPLUS xi Acknowledgements xii Chapter 1 Univariate data 1 1A Types of data 1 Exercise 1A 2 1B Stem plots 3 Exercise 1B 5 1C Dot plots, frequency histograms and bar charts 7 Exercise 1C 10 1D Describing the shape of stem plots and histograms 11 Exercise 1D 12 1E The median, the interquartile range, the range and the mode 15 Exercise 1E 18 1F Boxplots 19 Exercise 1F 23 1G The mean 24 Exercise 1G 27 1H Standard deviation 28 Exercise 1H 30 1I The 68–95–99.7% rule and z-scores 31 Exercise 1I 36 1J Populations and simple random samples 40 Exercise 1J 42 ■Summary 43 ■ Chapter review 45 ■ ICT activities 52 ■ Answers 53 Chapter 2 Bivariate data 57 2A Dependent and independent variables 57 Exercise 2A 58 2B Back-to-back stem plots 59 Exercise 2B 61 2C Parallel boxplots 62 Exercise 2C 64 2D Two-way frequency tables and segmented bar charts 65 Exercise 2D 67 2E Scatterplots 69 Exercise 2E 72 2F Pearson’s product–moment correlation coeffi cient 73 Exercise 2F 75 2G Calculating r and the coeffi cient of determination 76 Exercise 2G 79 ■ Summary 82 ■ Chapter review 84 ■ ICT activities 89 ■ Answers 90 Chapter 3 Introduction to regression 95 Fitting straight lines to bivariate data 95 3A Fitting a straight line by eye 95 Exercise 3A 96 3B Fitting a straight line — the 3-median method 97 Exercise 3B 100 3C Fitting a straight line — least-squares regression 102 Exercise 3C 104 3D Interpretation, interpolation and extrapolation 106 Exercise 3D 108 3E Residual analysis 109 Exercise 3E 112 3F Transforming to linearity 114 Exercise 3F 117 ■ Summary 120 ■ Chapter review 122 ■ ICT activities 125 ■ Answers 126 Chapter 4 Time series 129 4A Time series and trend lines 129 Exercise 4A 132 4B Fitting trend lines and forecasting 133 Exercise 4B 135 4C Smoothing time series 137 Exercise 4C 140 4D Smoothing with an even number of points 141 Exercise 4D 144 4E Median smoothing 145 Exercise 4E 147 4F Seasonal adjustment 148 Exercise 4F 153 ■ Summary 156 ■ Chapter review 157 ■ ICT activities 163 ■ Answers 164 ExAm PrACtICE 1 Based on Chapters 1–4 169

vi Contents Chapter 5 Arithmetic and geometric sequences 171 Introduction 171 5A Recognition of arithmetic sequences 171 Exercise 5A 173 5B Finding the terms of an arithmetic sequence 175 Exercise 5B 177 5C The sum of a given number of terms of an arithmetic sequence 178 Exercise 5C 181 5D Recognition of geometric sequences 183 Exercise 5D 185 5E Finding the terms of a geometric sequence 186 Exercise 5E 189 5F The sum of a given number of terms of a geometric sequence 191 Exercise 5F 193 5G Applications of geometric sequences 194 Exercise 5G 197 5H Finding the sum of an infinite geometric sequence 198 Exercise 5H 201 5I Contrasting arithmetic and geometric sequences through graphs 202 Exercise 5I 204 ■Summary 206 ■ Chapter review 208 ■ ICT activities 212 ■ Answers 213 Chapter 6 Difference equations 215 Introduction 215 6A Generating the terms of a sequence defined by a first order difference equation 215 Exercise 6A 217 6B The relationship between arithmetic sequences and first order difference equations 218 Exercise 6B 220 6C The relationship between geometric sequences and first order difference equations 221 Exercise 6C 222 6D Setting up first order difference equations to represent practical situations 223 Exercise 6D 227 6E Graphical representation of a sequence defined by a first order difference equation 228 Exercise 6E 231 6F Interpretation of the graph of first order difference equations 232 Exercise 6F 234 6G Fibonacci sequences as second order difference equations 237 Exercise 6G 239 ■Summary 242 ■ Chapter review 244 ■ ICT activities 247 ■ Answers 248 ExA m PrACt ICE 2 Based on Chapters 1–6 251 Chapter 7 Geometry: similarity and mensuration 253 Geometry 253 7A Properties of angles, triangles and polygons 253 Exercise 7A 256 7B Area and perimeter 258 Exercise 7B 261 7C Total surface area 263 Exercise 7C 266 7D Volume of prisms, pyramids and spheres 268 Exercise 7D 272 7E Similar figures 275 Exercise 7E 277 7F Similar triangles 279 Exercise 7F 281 7G Area and volume scale factors 283 Exercise 7G 288 ■Summary 291 ■ Chapter review 293 ■ ICT activities 297 ■ Answers 298 Chapter 8 Trigonometry 301 Trigonometry 301 8A Pythagoras’ theorem 301 Exercise 8A 303 8B Pythagorean triads 305 Exercise 8B 306 8C Three-dimensional Pythagoras’ theorem 307 Exercise 8C 308 8D Trigonometric ratios 310 Exercise 8D 314 Introduction — sine and cosine rules 316 8E The sine rule 317 Exercise 8E 320 8F Ambiguous case of the sine rule 322 Exercise 8F 324 8G The cosine rule 324 Exercise 8G 326 8H Special triangles 328 Exercise 8H 330 8I Area of triangles 331 Exercise 8I 333 ■Summary 336 ■ Chapter review 338 ■ ICT activities 343 ■ Answers 344

Chapter 9 Applications of geometry and trigonometry 347 Introduction 347 9A Angles 347 Exercise 9A 350 9B Angles of elevation and depression 352 Exercise 9B 354 9C Bearings 356 Exercise 9C 360 9D Navigation and specification of locations 361 Exercise 9D 366 9E Triangulation — cosine and sine rules 368 Exercise 9E 372 9F Triangulation — similarity 375 Exercise 9F 376 9G Contour maps 378 Exercise 9G 382 ■Summary 385 ■ Chapter review 387 ■ ICT activities 394 ■ Answers 395 ExA m PrACt ICE 3 Based on Chapters 1–4, 7–9 399 Chapter 10 Construction and interpretation of graphs 401 10A Constructing and interpreting straight-line graphs 401 Exercise 10A 407 10B Line segments and step functions 409 Exercise 10B 411 10C Simultaneous equations and break-even point 414 Exercise 10C 417 10D Interpreting non-linear graphs 418 Exercise 10D 420 10E Constructing non-linear relations and graphs 422 Exercise 10E 424 ■Summary 427 ■ Chapter review 428 ■ ICT activities 432 ■ Answers 433 Chapter 11 Linear inequations and linear programming 439 11A Linear inequations 439 Exercise 11A 441 11B Simultaneous linear inequations 442 Exercise 11B 444 11C Linear programming 445 Exercise 11C 451 11D Applications 454 Exercise 11D 459 ■ Summary 462 ■ Chapter review 463 ■ ICT activities 467 ■ Answers 468 ExA m PrACt ICE 4 Based on Chapters 1–4, 10–11 473 Chapter 12 Loans and investments 475 Introduction to growth and decay in business 475 12A Simple interest 476 Exercise 12A 479 12B Bonds, debentures and term deposits 480 Exercise 12B 482 12C Compound interest 483 Exercise 12C 488 Introduction to annuities 490 12D Reducing balance loans — the annuities formula 491 Exercise 12D 493 12E Reducing balance loans — further calculations 495 Exercise 12E 505 12F Hire-purchase 508 Exercise 12F 510 12G Reducing balance and flat rate loan comparisons 512 Exercise 12G 514 12H Effective rate of interest 516 Exercise 12H 517 12I Perpetuities 518 Exercise 12I 520 12J Annuity investments 521 Exercise 12J 526 ■ Summary 529 ■ Chapter review 532 ■ ICT activities 537 ■ Answers 538 Chapter 13 Financial transactions and asset value 541 13A Bank accounts 541 Exercise 13A 545 13B Financial computations 548 Exercise 13B 552 Depreciation 554 13C Flat rate (straight line) depreciation 555 Exercise 13C 557 13D Reducing balance depreciation 558 Exercise 13D 561 Contents vii

13E Unit cost depreciation 563 Exercise 13E 564 13F Inflation 566 Exercise 13F 568 ■ Summary 569 ■ Chapter review 571 ■ ICT activities 575 ■ Answers 576 ExA m PrACt ICE 5 Based on Chapters 1–4, 12 and 13 579 Chapter 14 Undirected graphs and networks 581 14A Basic concepts of a network 581 Exercise 14A 585 14B Planar graphs and Euler’s formula 587 Exercise 14B 591 14C Paths and circuits 592 Exercise 14C 596 14D Trees and their applications 599 Exercise 14D 605 ■ Summary 609 ■ Chapter review 611 ■ ICT activities 616 ■ Answers 617 Chapter 15 Directed graphs and networks 621 15A Reachability and dominance in directed networks 621 Exercise 15A 624 15B Critical path analysis 626 Exercise 15B 633 15C Critical path analysis with backward scanning and crashing 634 Exercise 15C 643 15D Network flow 647 Exercise 15D 652 15E Assignment problems and bipartite graphs 654 Exercise 15E 659 ■ Summary 663 ■ Chapter review 665 ■ ICT activities 673 ■ Answers 674 ExA m PrACt ICE 6 Based on Chapters 1–4, 14 and 15 681 Chapter 16 Matrices 683 16A Matrix representation 683 Exercise 16A 687 16B Addition, subtraction and scalar operations with matrices 688 Exercise 16B 694 16C Multiplying matrices 696 Exercise 16C 701 16D Multiplicative inverse and solving matrix equations 704 Exercise 16D 707 16E Application of matrices to simultaneous equations 709 Exercise 16E 712 16F Transition matrices 715 Exercise 16F 721 ■ Summary 726 ■ Chapter review 728 ■ ICT activities 733 ■ Answers 734 ExA m PrACt ICE 7 Based on Chapters 1–4 and 16 739 ■Exam practice answers 743 Index 745 viii Contents

Maths Quest 12 Further Mathematics Fourth edition is specifi cally designed for the VCE Further Mathematics course and based on the award-winning Maths Quest series. The suite of resources for this title include: • a student textbook with accompanying eBookPLUS • a TI-Nspire CAS calculator companion • a Casio ClassPAD calculator companion • a Solutions Manual • Flexi-saver versions of all print products • teacher support material available on the eGuidePLUS. Student textbook studyON icons provide links to Concept screens, See mores and Do mores for online study, revision and exam practice. Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos and to make navigation through the text easier. Clear, concise theory sections contain worked examples and highlighted important text and remember boxes. Icons appear for the eBookPLUS to indicate that interactivities and eLessons are available online to help with the teaching and learning of particular concepts. Worked examples in a Think–Write format provide clear explanation of key steps and suggest presentation of solutions. Many worked examples have eBookPLUS icons to indicate that a Tutorial is available to elucidate the concepts being explained. Worked examples also have calculator icons that indicate support in the Calculator Companion books, which contain compre\ hensive step-by-step CAS calculator instructions, fully integrated into the examples, for the TI-Nspire CAS and Casio ClassPad calculators. Exercises contain many carefully graded skills and application problems, including multiple c\ hoice questions. Cross-references to relevant worked examples appear with the fi rst ‘matching’ question throughout the exercises. A selection of questions are tagged as technology-free to indicate to st\ udents that they should avoid using their calculators or other technologies to assist them in fi nding a solution. Exam practice sections contain exam-style questions, including time and mark allocations for each question. Fully worked solutions are available on the eBookPLUS for students. Each chapter concludes with a summary and chapter review exercise containing examination-style questions (multiple choice, short answer and extended response), which help consolidate students’ learning of new concepts. Also included are questions from past VCE exams along with relevant exam tips. Student website — eBookPLUS The accompanying eBookPLUS contains the entire student textbook in HTML plus additional exercises. Students may use the eBookPLUS on laptops, tablets, school or home compu\ ters, and cut and paste material for revision, assignments or the creation of notes for exams. Career profi les and History of mathematics place mathematical concepts in context. Investigations, often suggesting the use of technology, provide further discovery learning opportunities. WorkSHEET icons link to editable Word documents, and may be completed on screen or printed and completed by hand. SkillSHEET icons link to printable pages designed to help students revise required concepts, and contain additional examples and problems. Interactivity icons link to dynamic animations, which help students to understand dif\ fi cult concepts. Introduction Introduction ix

eLesson icons link to videos or animations designed to elucidate concepts in ways that are more than what the teacher can achieve in the classroom. Tutorial icons link to one-way engagement activities which explain the worked examples in detail for students to view at home or in the classroom. Test Yourself tests are also available. Answers are provided for students to receive instant feedback. teacher website — eGuidePLUS The accompanying eGuidePLUS contains everything in the eBookPLUS and more. Two tests per chapter, fully worked solutions to WorkSHEETs, the work program and other curriculum advice in editable Word format are provided. Maths Quest is a rich collection of teaching and learning resources within one package. Maths Quest 12 Further Mathematics provides ample NEW material, such as application tasks, analysis tasks and semester exams, from which teachers may set school-assessed coursework (SAC), as well as additional investigations, worksheets and technology files. x Introduction



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ChapTer 1 • Univariate data 1 ChapTer ConTenTS 1a Types of data 1B Stem plots 1C Dot plots, frequency histograms and bar charts 1d Describing the shape of stem plots and histograms 1e The median, the interquartile range, the range and the mode 1F Boxplots 1G The mean 1h Standard deviation 1i The 68–95–99.7% rule and z -scores 1J Populations and simple random samples ChapTer 1 Univariate data diGiTal doC doc-9399 10 Quick Questions 1a Types of data Univariate data are data that contain one variable. That is, the information deals with only one quantity that changes. Therefore, the number of cars sold by a car salesman during one week is an example of univariate data. Sets of data that contain two variables are called bivariate data and those that contain more than two variables are called multivariate data. You will learn more about bivariate data in chapter 2. Data can be numerical, categorical, discrete or continuous. The methods we use to display data depend on the type of information we are dealing with. numerical and categorical data Examples of numerical data are: 1. the heights of a group of teenagers 2. the marks for a maths test 3. the number of universities in a country 4. ages 5. salaries. As the name suggests, numerical data involve quantities which are, broadly speaking, measurable or countable. Examples of categorical data are: 1. genders 2. AFL football teams 3. religions 4. fi nishing positions in the Melbourne Cup 5. municipalities 6. ratings of 1–5 to indicate preferences for 5 different cars 7. age groups, for example 0–9, 10–19, 20–29 8. hair colours. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 1 Concept: 1 Do more Interact  with classif cation  of data. ConTenTS 1a Types of data 1 exercise 1a Types of data 2 1B Stem plots 3 exercise 1B Stem plots 5 1C dot plots, frequency histograms and bar charts 7 exercise 1C dot plots, frequency histograms and bar charts 10 1d describing the shape of stem plots and histograms 11 exercise 1d describing the shape of stem plots and histograms 12 1e The median, the interquartile range, the range and the mode 15 exercise 1e The median, the interquartile range, the range and the mode 18 1F Boxplots 19

2 Maths Quest 12 Further Mathematics Such categorical data, as the name suggests, have categories like masculine, feminine and neuter for gender, or Christian, Islamic, Buddhist and so on for religious denomination, \ or 1st, 2nd, 3rd for fi nishing position in the Melbourne Cup. Note: Some numbers may look like numerical data, but are actually names or titles (for example, ratings of 1 to 5 given to different samples of cake — ‘This one’s a 4’; the numbers on netball players’ uniforms — ‘she’s number 7’). These ‘titles’ are not countable; they place the subject in a category (with a name), and so they are categorical. discrete and continuous data Data are said to be discrete when a variable can take only certain fi xed values. For example, if we counted the number of children per household in a particular suburb, the data obtained would always be whole numbers starting from zero. A value in between, such as 2.5, would clearly not be possible. Other examples of variables that produce discrete data are the number of crayfi sh caught in a fi sherman’s pots each day and the number of people that attend a restaurant each d\ ay. If objects can be counted, then the data are discrete. Continuous data are obtained when a variable takes any value between two values. If the heights of students in a school were obtained, then the data could consist of an\ y values between the smallest and largest heights. The values recorded would be restricted only by the precision of the measuring instrument. Other examples of variables that produce continuous data are weight, length and the time to\ complete a certain task. If variables can be measured, then the data are continuous. exercise 1a Types of data 1 Write whether each of the following represents numerical or categorical data. a The heights, in centimetres, of a group of children b The diameters, in millimetres, of a collection of ball-bearings c The numbers of visitors to an exibit each day d The modes of transport that students in Year 12 take to school e The 10 most-watched television programs in a week f The occupations of a group of 30-year-olds g The numbers of subjects offered to VCE students at various schools h Life expectancies i Species of fi sh j Blood groups k Years of birth l Countries of birth m Tax brackets 2 For each set of numerical data identified in question 1 above, state whether the data are discrete or continuous. 3 mC An example of a numerical variable is: a attitude to 4-yearly elections (for or against) B year level of students C the total attendance at Carlton football matches d position in a queue at the pie stall e television channel numbers shown on a dial 4 mC The weight of each truck-load of woodchips delivered to the wharf during a one-month period was recorded. This is an example of: a categorical and discrete data B discrete data C continuous and numerical data d continuous and categorical data e numerical and discrete data diGiTal doC doc-9400 WorkSHEET 1.1

ChapTer 1 • Univariate data 3 1B Stem plots A stem-and-leaf plot, or stem plot for short, is a way of displaying a set of data. It is best suited to data which contain up to about 50 obser\ vations (or records). The stem plot below shows the ages of people attending an advanced computer class. The ages of the members of the class are 16, 22, 22, 23, 30, 32, 34, 36,\ 42, 43, 46, 47, 53, 57 and 61. A stem plot is constructed by splitting the numerals of a record into two parts — the stem, which in this case is the fi rst digit, and the leaf, which is always the last digit. Worked example 1 The number of cars sold in a week at a large car dealership over a 20-week period is given below. 16 12 8 7 26 32 15 51 29 45 19 11 6 15 32 18 43 31 23 23 Construct a stem plot to display the number of cars sold in a week at th\ e dealership. Think WriTe 1In this example the observations are one- or two-digit numbers and so the stems will be the digits referring to the ‘tens’, and the leaves will be the digits referring to the units. Work out the lowest and highest numbers in the data in order to determine what the stems will be. Lowest number = 6 Highest number = 51 Use stems from 0 to 5. 2Before we construct an ordered stem plot, construct an unordered stem plot by listing the leaf digits in the order they appear in the data. Stem 0 1 2 3 4 5Leaf 8 7 6 6 2 5 9 1 5 8 6 9 3 3 2 2 1 5 3 1 3Now rearrange the leaf digits in numerical order to create an ordered stem plot. Include a key so that the data can be understood by anyone viewing the stem plot. Stem 0 1 2 3 4 5Leaf 6 7 8 1 2 5 5 6 8 9 3 3 6 9 1 2 2 3 5 1 Key: 2|3 = 23 cars Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 2 Concept: 4 Stem 1 2 3 4 5 6Leaf 6 2 2 3 0 2 4 6 2 3 6 7 3 7 1 Key: 2|2 = 22 years old

4 Maths Quest 12 Further Mathematics Worked example 2 The masses (in kilograms) of the members of an Under-17 football squad are given below. 70.3 65.1 72.9 66.9 68.6 69.6 70.8 72.4 74.1 75.3 75.6 69.7 66.2 71.2 68.3 69.7 71.3 68.3 70.5 72.4 71.8 Display the data in a stem plot. Think WriT e 1In this case the observations contain 3 digits. The last digit always becomes the leaf and so in this case the digit referring to the tenths becomes the leaf and the two preceding digits become the stem. Work out the lowest and highest numbers in the data in order to determine what the stems will be. Lowest number = 65.1 Highest number = 75.6 Use stems from 65 to 75. 2Construct an unordered stem plot. Note that the decimal points are omitted since we are aiming to present a quick visual summary of data. Stem 65 66 67 68 69 70 71 72 73 74 75 Leaf 1 9 2 6 3 3 6 7 7 3 8 5 2 3 8 9 4 4 1 3 6 3Construct an ordered stem plot. Provide a key. Stem 65 66 67 68 69 70 71 72 73 74 75 Leaf 1 2 9 3 3 6 6 7 7 3 5 8 2 3 8 4 4 9 1 3 6 Key: 74|1 = 74.1 kg Sometimes data which are very bunched make it difficult to get a clear idea about the data variation. To overcome the problem, we can split the stems. Stems can be split into halv\ es or fifths. Worked example 3 A set of golf scores for a group of professional golfers trialling a new 18-hole golf course is shown on the following stem plot. Stem 6 7Leaf 1 6 6 7 8 9 9 9 0 1 1 2 2 3 7 Key: 6|1 = 61 Produce another stem plot for these data by splitting the stems into: a halves b fifths.

ChapTer 1 • Univariate data 5 Think WriT e a By splitting the stem 6 into halves, any leaf digits in the range 0–4 appear next to the 6, and any leaf digits in the range 5–9 appear next to the 6*. Likewise for the stem 7. a Stem 6* 6* 7* 7* Leaf 1 6 6 7 8 9 9 9 0 1 1 2 2 3 7 Key: 6|1 = 61 b Alternatively, to split the stems into fifths, each stem would appear 5 times. Any 0s or 1s are recorded next to the first 6. Any 2s or 3s are recorded next to the second 6. Any 4s or 5s are recorded next to the third  6. Any 6s or 7s are recorded next to the fourth 6 and, finally, any 8s or 9s are recorded next to the fifth 6. This process would be repeated for those observations with a stem of 7. b Stem 6 6 6 6 6 7 7 7 7 7Leaf 1 6 6 7 8 9 9 9 0 1 1 2 2 3 7 Key: 6|1 = 61 e xercise 1B Stem plots 1 In each of the following, write down all the pieces of data shown on the stem plot. a Stem 0* 0* 1* 1* 2* 2* 3* Leaf 1 2 5 8 2 3 3 6 6 7 1 3 4 5 5 6 7 0 2 b Stem 1 2 3 4 5 6Leaf 0 1 3 3 0 5 9 1 2 7 5 2 c Stem 10 11 12 13 14 15 Leaf 1 2 5 8 2 3 3 6 6 7 1 3 4 5 5 6 7 d Stem 5 5 5 5 5Leaf 0 1 3 3 4 5 5 6 6 7 9 e Stem 0* 0* 1* 1* 2* 2* Leaf 1 4 5 8 0 2 6 9 9 1 1 5 9 The key used for each stem plot is 3|2 = 32. 2 We1 The money (to the nearest dollar) earned each week by a busker over an 18-week period is shown below. Construct a stem plot for the busker’s weekly earnings. What can you say about the busker’s earnings? 5 31 19 5211 4327 3723 4135 3918 4542 3229 36

6 Maths Quest 12 Further Mathematics 3 The ages of those attending an embroidery class are given below. Construct a stem plot for these data and draw a conclusion from it. 39 6368 4951 5257 6163 5851 5937 4942 53 4 mC The observations shown on the stem plot at right are: a 4 10 27 28 29 31 34 36 41 B 14 10 27 28 29 29 31 34 36 41 41 C 4 22 27 28 29 29 30 31 34 36 41 41 d 14 22 27 28 29 30 30 31 34 36 41 41 e 4 2 27 28 29 29 30 31 34 36 41 Stem 0 1 2 3 4 Leaf 4 2 7 8 9 9 0 1 4 6 1 1 Key: 2|5 = 25 5 The ages of the mothers of a class of children attending an inner-city kindergarten are given below. Construct a stem plot for these data. Based on your display, comment on the statement ‘Parents of kindergarten children are very young’. 32 28 3730 29 3319 34 2928 32 3525 35 3829 39 3332 30 6 The number of hit outs made by each of the principal ruckmen in each of \ the AFL teams for Round 11 is recorded below. Construct a stem plot to display these data. Which teams had the three highest scoring ruckmen? Team Number of hit outs Team Number of hit outs Collingwood Bulldogs Kangaroos Port Adelaide Geelong Sydney Melbourne Brisbane 19 41 29 24 21 31 40 25Adelaide St Kilda Essendon Carlton West Coast Fremantle Hawthorn Richmond 32 34 31 26 29 22 33 28 7 We2 The heights of members of a squad of basketballers are given below in metres. Construct a stem plot for these data. 1.96 1.951.85 2.03 2.03 2.092.21 2.052.17 2.01 1.89 1.96 1.99 1.971.87 1.91 8 The 2008 median house price of a number of Melbourne suburbs is given below. Construct a stem plot for these data and comment on it. Suburb Price (× $1000) Suburb Price (× $1000) Ashburton Ashwood Blackburn Bulleen Burwood Caulfield East Chadstone Chettenham Clayton Cobury 670 600 670 628 652 653 608 576 525 526Collingwood Dancaster Essendon Highett Huntingdale Ivanhoe Moonee Ponds Newport Oakleigh Preston 583 620 670 600 517 633 638 536 548 515

ChapTer 1 • Univariate data 7 9 We3 The data below give the head circumference (to the nearest cm) of 16 four-year-old girls. 48 5049 5047 5352 5251 4350 4749 4948 50 Construct a stem plot for head circumference, using: a the stems 4 and 5 b the stems 4 and 5 split into halves c the stems 4 and 5 split into fi fths. 10 A random sample of 20 screws is taken and the length of each is recorded to the nearest millimetre below. 23 19 1715 20 1918 16 2117 20 2317 21 2019 19 2122 23 Construct a stem plot for screw length using: a the stems 1 and 2 b the stems 1 and 2 split into halves c the stems 1 and 2 split into fi fths. Use your plots to help you comment on the screw lengths. 1C dot plots, frequency histograms and bar charts Dot plots, frequency histograms and bar charts display data in graphical form. dot plots In picture graphs, a single picture represents each data value. Similarly, in dot plots, a single dot represents each data value. Dot plots are used to display discrete data where values are not spread out very much. They are also used to display categorical data. Dot plots have a scaled horizontal axis and each data value is indicated by a dot above this scale. The end result is a set of vertical ‘lines’ of evenly-spaced dots. 2 4 6 8 10 12 Score Worked example 4 The number of hours per week spent on art by 18 students is given below. 4 0 3 1 3 4 2 2 3 4 1 3 2 5 3 2 1 0 Display the data as a dot plot. Think draW 1Determine the lowest and highest scores and then draw a suitable scale. 0 1 2 3 4 5 • • • • • • • • •• • • • • • • • • Hours/week 2Represent each score by a dot on the scale. Frequency histograms A histogram is a useful way of displaying large data sets (say, over 50 observations). The vertical axis on the histogram displays the frequency and the horizontal axis displays class intervals of the variable (for example, height or income). When data are given in raw form — that is, just as a list of fi gures in no particular order — it is helpful to fi rst construct a frequency table. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 2 Concept: 3 Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 2 Concept: 5

8 Maths Quest 12 Further Mathematics Worked example 5 The data below show the distribution of masses (in kilograms) of 60 students in Year 7 at Northwood Secondary College. Construct a frequency histogram to display the data more clearly. 45.7 45.8 45.9 48.2 48.3 48.4 34.2 52.4 52.3 51.8 45.7 56.8 56.3 60.2 44\ .2 53.8 43.5 57.2 38.7 48.5 49.6 56.9 43.8 58.3 52.4 54.3 48.6 53.7 58.7 57\ .6 45.7 39.8 42.5 42.9 59.2 53.2 48.2 36.2 47.2 46.7 58.7 53.1 52.1 54.3 51\ .3 51.9 54.6 58.7 58.7 39.7 43.1 56.2 43.0 56.3 62.3 46.3 52.4 61.2 48.2 58\ .3 Think WriTe/draW 1First construct a frequency table. The lowest data value is 34.2 and the highest is 62.3. Divide the data into class intervals. If we started the fi rst class interval at, say, 30 kg and ended the last class interval at 65 kg, we would have a range of 35. If each interval was 5 kg, we would then have 7 intervals which is a reasonable number of class intervals. While there are no set rules about how many intervals there should be, somewhere between about 5 and 15 class intervals is usual. So, in this example, we would have class intervals of 30–34.9 kg, 35–39.9 kg, 40–44.9 kg and so on. Complete a tally column using one mark for each value in the appropriate interval. Add up the tally marks and write them in the frequency column. Class interval Tally Frequency 30–34.9 35–39.9 40–44.9 45–49.9 50–54.9 55–59.9 60–64.9 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 1 4 7 16 15 14 3 Total 60 Mass (kg) Frequency 0 2 4 6 8 10 12 14 30 35 40 45 50 55 60 65 16 2Check that the frequency column totals 60. The data are in a much clearer form now. 3A histogram can be constructed. Worked example 6 The marks out of 20 received by 30 students for a book-review assignment are given in the frequency table below. Mark 12 13 14 15 16 17 18 19 20 Frequency 2 7 6 5 4 2 3 0 1 Display these data on a histogram. Think draW In this case we are dealing with integer values (discrete data). Since the horizontal axis should show a class interval, we extend the base of each of the columns on the histogram halfway either side of each score. Mark out of 20 Frequency 0 1 2 3 4 5 6 7 12 13 14 15 16 17 18 19 20

ChapTer 1 • Univariate data 9 Bar charts A bar chart is similar to a histogram. However, it consists of bars of equal width separated by small, equal spaces and may be arranged either horizontally or vertically. Bar charts are often used to display categorical data. 2 4 6 Number of students Student pet preferences 8 10 12 Dog Cat Rabbit Bird Snake Goldfsh 0 5 10 15 20 25 1 2 Number of children in family Number of families families 3 4 5 In bar charts, the frequency is graphed against a variable as shown in both fi gures above. The variable may or may not be numerical. However, if it is, the variable should represent discrete data because the scale is broken by the gaps between the bars. The numerical values are generally close together and have little spread, for example, consecutive years. The bar chart at right represents the data presented in Worked example 6. It could also have been drawn with vertical bars (columns). Segmented bar charts A segmented (divided) bar chart is a single bar which is used to represent all the dat\ a being studied. It is divided into segments, each segment representing a particular group of the data. Generally, the information is presented as percentages and so the total bar length repr\ esents 100% of the data. Consider the following table, showing fatal road accidents in Australia. Road traffi c accidents involving fatalities Accidents involving fatalities Year NSW Vic. Qld SA WATas. NT ACTAust. 2003 483 294 284 136 155 39 44 10 1445 2004 471 313 288 128 162 52 34 10 1458 2005 469 316 294 127 151 48 51 25 1481 2006 453 309 314 104 183 42 39 12 1456 2007 405 289 338 107 214 39 49 14 1453 2008 376 278 293 87 189 38 67 14 1342 Persons killed Year NSW Vic. Qld SA WATas. NT ACTAust. 2003 539 330 310 157 180 41 53 11 1621 2004 522 343 311 139 178 58 35 10 1596 2005 518 348 328 148 163 50 55 26 1636 2006 500 337 336 117 202 54 42 13 1601 2007 435 332 360 124 235 45 58 14 1603 2008 397 303 327 99 209 40 75 14 1464 Source: Australian Bureau of Statistics 2010, Year book Australia 2009–10, cat. no. 1301.0, ABS, Canberra, table 24.20, p. 638. It is appropriate to represent the number of accidents involving fatalities in all states and territories during 2008 as a segmented bar chart. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 2 Concept: 1 Do more Interact  with bar charts. 1 2 3 Frequency or number of students Mark out of 20 4 5 6 7 17 16 15 19 20 18 14 13 12 Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 2 Concept: 2 See more Watch  a video about  segmented graphs. diGiTal doC doc-9401 Spreadsheet Segmented bar charts

10 Maths Quest 12 Further Mathematics First, we convert each state’s proportion of accidents out of the total to a percentage. State Number of accidents Percentage NSW 376376 ÷ 1342 × 100% = 28.0% Vic. 278278 ÷ 1342 × 100% = 20.7% Qld 293293 ÷ 1342 × 100% = 21.8% SA 8787 ÷ 1342 × 100% = 6.5% WA 189189 ÷ 1342 × 100% = 14.1% Tas. 3838 ÷ 1342 × 100% = 2.8% NT 6767 ÷ 1342 × 100% = 5.0% ACT 1414 ÷ 1342 × 100% = 1.0% The segmented bar chart is drawn to scale. An appropriate scale would be constructed by drawing the total bar 100 mm long, so that 1 mm represents 1%. That is, accidents in NSW would be represented by a segment of 28 mm, those in Victoria by a segment of 20.7 mm and so on. Each segment is then labelled directly, or a key may be used. NSW 28% Vic. 20.7% WA 14.1% Tas. 2.8% QLD 21.8% SA 6.5% NT 5.0% ACT 1.0% exercise 1C dot plots, frequency histograms and bar charts 1 We5 Construct a frequency table for each of the following sets of data. a 4.3 4.5 4.7 4.9 5.1 5.3 5.5 5.6 5.2 3.6 2.5 4.3 2.5 3.7 4.5 6.3 1.3 b 11 13 15 15 16 18 20 21 22 21 18 19 20 16 18 20 16 10 23 24 25 27 28 30 \ 35 28 27 26 29 30 31 24 28 29 20 30 32 33 29 30 31 33 34 c 0.4 0.5 0.7 0.8 0.8 0.9 1.0 1.1 1.2 1.0 1.3 0.4 0.3 0.9 0.6 2 Using the frequency tables from question 1, construct a histogram for each set of data. 3 Using a CAS calculator, construct a histogram for each of the sets of data given in question 1. Compare this histogram with the one drawn for question 2. 4 We4 The data below represent the number of hours each week that 40 teenagers spent on hou\ sehold chores. Display these data on a bar chart and a dot plot. 2 5 2 0 8 7 8 5 1 0 2 1 8 0 4 2 2 9 8 5 7 5 4 2 1 2 9 8 1 2 8 5 8 10 0 3 4 5 2 8 5 Using the information provided in the table below: a calculate the proportion of residents who travelled in 2005 to each of the countries listed b draw a segmented bar graph showing the major destinations of Australians travelling abroad in 2005. Short-term resident departures by major destinations 2004 (× 1000) 2005 (× 1000) 2006 (× 1000) 2007 (× 1000) 2008 (× 1000) New Zealand 815.8 835.4 864.7 902.1 921.1 United States of America 376.1 426.3 440.3 479.1 492.3 United Kingdom 375.1 404.2 412.8 428.5 420.3 Indonesia 335.1 319.7 194.9 282.6 380.7 China (excluding Special Administrative Regions (SARs)) 182.0 235.1 251.0 284.3 277.3 diGiTal doC doc-9402 S killSHEET  1.1 Converting a fraction to a percentage diGiTal doC doc-9403 Spreadsheet Frequency histograms

ChapTer 1 • Univariate data 11 2004 (× 1000) 2005 (× 1000) 2006 (× 1000) 2007 (× 1000) 2008 (× 1000) Thailand 188.2 202.7 288.0 374.4 404.1 Fiji 175.4 196.9 202.4 200.3 236.2 Singapore 159.0 188.5 210.9 221.5 217.8 Hong Kong (SAR of China) 152.6 185.7 196.3 206.5 213.1 Malaysia 144.4 159.8 168.0 181.3 191.0 Source: Australian Bureau of Statistics 2010, Year book Australia 2009–10, cat. no. 1301.0, ABS, Canberra, table 23.12, p. 621. 6 Presented below is information about adult participation in sport and physical activities in 2005–06. Draw a segmented bar graph to compare the participation of all persons from various age groups. Comment on the statement, ‘Only young people participate in sport and\ physical activities’. Participation in sport and physical activities (a) — 2005–06 Males Females Persons Age group (years) Number (× 1000) Participation rate (%) Number (× 1000) Participation rate (%) Number (× 1000) Participation rate (%) 18–24 735.2 73.3 671.3 71.8 1406.4 72.6 25–34 1054.5 76.3 1033.9 74.0 2088.3 75.1 35–44 975.4 66.7 1035.9 69.1 2011.2 68.0 45–54 871.8 63.5 923.4 65.7 1795.2 64.6 55–64 670.1 60.4 716.3 64.6 1386.5 62.5 65 and over 591.0 50.8 652.9 48.2 1243.9 49.4 Total 4898 64.6 5033.7 64.4 9931.5 64.5 (a) Relates to persons aged 18 years and over who participated in sport or physical activity as a player during the 12 months prior to interview. Source: Participation in Sport and Physical Activities, Australia, 2005–06 (4177.0). Viewed 10 October 2008 1d describing the shape of stem plots and histograms Symmetric distributions The data shown in the histogram at right can be described as symmetric. There is a single peak and the data trail off on both sides of this peak in roughly the same fashion. Similarly in the stem plot at right, the distribution of the data could be described as symmetric. The single peak for these data occur at the stem 3. On either side of the peak, the number of observations reduces in approximately matching fashion. Frequency Stem 0 1 2 3 4 5 6 Leaf 7 2 3 2 4 5 7 9 0 2 3 6 8 8 4 7 8 9 9 2 7 8 1 3 Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 2 Concept: 6

12 Maths Quest 12 Further Mathematics Skewed distributions Each of the histograms shown below are examples of skewed distributions. The fi gure below left shows data which are negatively skewed. The data in this case peak to the right and trail off to the left. The fi gure below right shows positively skewed data. The data in this case peak to the left and trail off to the right. Negatively skewed distribution Positively skewed distribution Worked example 7 The ages of a group of people who were taking out their fi rst home loan is shown below. Stem 1 2 3 4 5 6Leaf 9 9 1 2 4 6 7 8 8 9 0 1 1 2 3 4 7 1 3 5 6 2 3 7 Key: 1|9 = 19 years old Describe the shape of the distribution of these data. Think WriTe Check whether the distribution is symmetric or skewed. The peak of the data occurs at the stem 2. The data trail off as the stems increase in value. This seems reasonable since most people would take out a home loan early in life to give themselves time to pay it off. The data are positively skewed. exercise 1d describing the shape of stem plots and histograms 1 We7 For each of the following stem plots, describe the shape of the distribution of the data. a Stem 0 1 2 3 4 5 6 7Leaf 1 3 2 4 7 3 4 4 7 8 2 5 7 9 9 9 9 1 3 6 7 0 4 4 7 1 Key: 1| 2 = 12 b Stem 1 2 3 4 5 6Leaf 3 6 3 8 2 6 8 8 9 4 7 7 7 8 9 9 0 2 2 4 5 Key: 2|6 = 2.6 c Stem 2 3 4 5 6 7 8 9 10 Leaf 3 5 5 6 7 8 9 9 0 2 2 3 4 6 6 7 8 8 2 2 4 5 6 6 6 7 9 0 3 3 5 6 2 4 5 9 2 7 Key: 10|4 = 104 TUTorial eles-1254 Worked example 7

ChapTer 1 • Univariate data 13 d Stem 1 * 1*  2 * 2*  3 * 3* 4 * 4* Leaf 5 1 4 5 7 8 8 9 1 2 2 3 3 3 4 4 5 5 5 6 3 4 Key: 2|4 = 24 e Stem 3 3 4 4 4 4 4Leaf 8 9 0 0 1 1 1 2 3 3 3 3 3 4 5 5 5 6 7 8 Key: 4|3 = 0.43 f Stem 60 61 62 63 64 65 66 67 Leaf 2 5 8 1 3 3 6 7 8 9 0 1 2 4 6 7 8 8 9 2 2 4 5 7 8 3 6 7 4 5 8 3 5 4 Key: 62|3 = 623 2 For each of the following histograms, describe the shape of the distribution of the data and comment on the existence of any outliers. a Frequency b Frequency c Frequency d Frequency e Frequency f Frequency 3 mC The distribution of the data shown in this stem plot could be described as: a negatively skewed B negatively skewed and symmetric C positively skewed d positively skewed and symmetric e symmetric Stem 0 0 0 0 0 1 1 1 1 1Leaf 1 2 4 4 5 6 6 6 7 8 8 8 8 9 9 0 0 0 1 1 1 1 2 2 2 3 3 3 4 4 5 5 6 7 7 8 9 Key: 1|8 = 18 4 mC The distribution of the data shown in the histogram at right could be described as: a negatively skewed B negatively skewed and symmetric C positively skewed d positively skewed and symmetric e symmetric Frequency 5 The average number of product enquiries per day received by a group of small businesses who advertised in the Yellow Pages telephone directory is given at right. Describe the shape of the distribution of these data. Frequency Number of enquiries 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

14 Maths Quest 12 Further Mathematics 6 The number of nights per month spent interstate by a group of flight attendants is shown on the stem plot at right. Describe the shape of distribution of these data and explain what this tells us about the number of nights per month spent interstate by this group of flight attendants. Stem 0 0 0 0 0 1 1 1 1Leaf 0 0 1 1 2 2 3 3 3 3 3 3 3 3 4 4 5 5 5 5 5 6 6 6 6 7 8 8 8 9 0 0 1 4 4 5 5 7 Key: 1|4 = 14 nights 7 The mass (to the nearest kilogram) of each dog at a dog obedience school is shown on the stem plot below. a Describe the shape of the distribution of these data. b What does this information tell us about this group of dogs? Stem 0* 0* 1* 1* 2* 2* Leaf 4 5 7 9 1 2 4 4 5 6 6 7 8 9 1 2 2 3 6 7 Key: 0 |4 = 4 kg 8 The amount of pocket money (to the nearest 50 cents) received each week by students in a Grade 6 class is illustrated in the histogram below. a Describe the shape of the distribution of these data. b What conclusions can you reach about the amount of pocket money received weekly by this group of students? Frequency Pocket money ($) 0 1 2 3 4 5 6 7 8 5.566.5 77.5 88.5 99.5 1010.5 9 Statistics were collected over 3 AFL games on the number of goals kicked by forwards over 3 weeks. This is displayed in the histogram below. a Describe the shape of the histogram. b Use the histogram to determine: i the number of players who kicked 3 or more goals over the 3 weeks ii the percentage of players who kicked between 2 and 6 goals over the 3 weeks. Number of goals kicked by players over 3 weeks 0 Frequency 3 2 1 4 5 Number of goals 1 2 3 4 5 6 7 diGiTal doC doc-9404 WorkSHEET 1.2

ChapTer 1 • Univariate data 15 1e The median, the interquartile range, the range and the mode After displaying data using a histogram or stem plot, we can make even more sense of the data by calculating what are called summary statistics. Summary statistics are used because they give us an idea about: 1. where the centre of the distribution is 2. how the distribution is spread out. We will look fi rst at four summary statistics — the median, the interquartile range,\ the range and the mode — which require that the data be in ordered form before they can be calculated. The median The median is the midpoint of an ordered set of data. Half the data are less than or equal to the median. Consider the set of data: 2 5 6 8 11 12 15. These data are in ordered form (that is, from lowest to highest). There are 7 observations. The median in this case is the middle or fourth score; that is, 8. Consider the set of data: 1 3 5 6 7 8 8 9 10 12. These data are in ordered form also; however, in this case there is an even number of scores. The median of this set lies halfway between the 5th score (7) and the 6th score (8). So the median is 7.5. (Alternatively, median = 78+78+ 2 = 7.5.) When there are n records in a set of ordered data, the median can be located at the +       n 1 2 th pos ition. Checking this against our previous example, we have n = 10; that is, there were 10 observations in the set. The median was located at the +   10 1 2 = 5.5th position; that is, halfway between the 5th and the 6th terms. A stem plot provides a quick way of locating a median since the data in a stem plot are already ordered. Worked example 8 Consider the stem plot below which contains 22 observations. What is the median? Stem 2* 2* 3* 3* 4* 4* Leaf 3 3 5 7 9 1 3 3 4 4 5 8 9 9 0 2 2 6 8 8 8 9 Key: 3| 4 = 34 Think WriTe 1Find the median position, where n = 22. Median = +                       n 1 2 th position = +                       22 1 2 th position = 11.5th position 2Find the 11th and 12th terms. 11th term = 35 12th term = 38 3The median is halfway between the 11th and 12th terms. Median = 36.5 Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 3 Concept: 1 inTeraCTiViTY int-0084 The median, the interquartile range, the range and the mode

16 Maths Quest 12 Further Mathematics The interquartile range We have seen that the median divides a set of data in half. Similarly, quartiles divide a set of data in quarters. The symbols used to refer to these quartiles are Q 1, Q 2 and Q 3. The middle quartile, Q 2, is the median. The interquartile range IQR = Q 3 − Q 1. The interquartile range gives us the range of the middle 50% of values in a set of data. There are four steps to locating Q 1 and Q 3. Step 1. Write down the data in ordered form from lowest to highest. Step 2. Locate the median; that is, locate Q 2. Step 3. Now consider just the lower half of the set of data. Find the middle score. This score is Q 1. Step 4. Now consider just the upper half of the set of data. Find the middle score\ . This score is Q 3. The four cases given below illustrate this method. Case 1 Consider data containing the 6 observations: 3 6 10 12 15 21. The data are already ordered. The median is 11. Consider the lower half of the set, which is 3 6 10. The middle score is 6, so Q 1 = 6. Consider the upper half of the set, which is 12 15 21. The middle score is 15, so Q 3 = 15. Case 2 Consider a set of data containing the 7 observations: 4 9 11 13 17 23 30. The data are already ordered. The median is 13. Consider the lower half of the set, which is 4 9 11. The middle score is 9, so Q 1 = 9. Consider the upper half of the set, which is 17 23 30. The middle score is 23, so Q 3 = 23. Case 3 Consider a set of data containing the 8 observations: 1 3 9 10 15 17 21 26. The data are already ordered. The median is 12.5. Consider the lower half of the set, which is 1 3 9 10. The middle score is 6, so Q 1 = 6. Consider the upper half of the set, which is 15 17 21 26. The middle score is 19, so Q 3 = 19. Case 4 Consider a set of data containing the 9 observations: 2 7 13 14 17 19 21 25 29. The data are already ordered. The median is 17. Consider the lower half of the set, which is 2 7 13 14. The middle score is 10, so Q 1 = 10. Consider the upper half of the set, which is 19 21 25 29. The middle score is 23, so Q 3 = 23. Worked example 9 The ages of the patients who attended the casualty department of an inner-suburban hospital on one particular afternoon are shown below. 14 3 27 42 19 17 73 60 62 21 23 2 5 58 33 19 81 59 25 17 69 Find the interquartile range of these data. Think WriTe 1Order the data. 2 3 5 14 17 17 19 19 21 23 25 27 33 42 58 59 60 62 69 73 81 2Find the median.The median is 25 since ten scores lie below it and ten lie above it. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 3 Concept: 3 TUTorial eles-1255 Worked example 9

ChapTer 1 • Univariate data 17 3Find the middle score of the lower half of the data. For the scores 2 3 5 14 17 17 19 19 21 23, the middle score is 17. So, Q 1 = 17. 4Find the middle score of the upper half of the data.For the scores 27 33 42 58 59 60 62 69 73 81, the middle score is 59.5. So, Q 3 = 59.5. 5Calculate the interquartile range. IQR = Q 3 − Q 1 = 59.5 − 17 = 42.5 A CAS or graphics calculator can be a fast way of locating quartiles and hence fi nding the value of the interquartile range. Worked example 10 Parents are often shocked at the amount of money their children spend. The data below give the amount spent (to the nearest whole dollar) by each child in a group that was taken on an excursion to the Royal Melbourne Show. 15 12 17 23 21 19 16 11 17 18 23 24 25 21 20 37 17 25 22 21 19 Calculate the interquartile range for these data. Think WriTe 1Enter the data into a calculator. (There is no need to order it.) Use the calculator to generate one-variable statistics. Copy down the values of the fi rst and third quartiles. Q 1 = 17 and Q 3 = 23 2Calculate the interquartile range. So, IQR = Q 3 − Q 1 = 23 − 17 = 6 The range The range of a set of data is the difference between the highest and lowest values in that set. It is usually not too diffi cult to locate the highest and lowest values in a set of data. Only when there is a very large number of observations might the job be made more diffi cult. In the previous worked example, the minimum and maximum values were 11 and 37, respectively. The range, therefore, can be calculated as: Range = maxX − minX= 37 − 11 = 26. While the range gives us some idea about the spread of the data, it is not very informative since it gives us no idea of how the data are distributed between the highest and lowest values. Now let us look at another measure of the centre of a set of data: the mod\ e.

18 Maths Quest 12 Further Mathematics The mode The mode is the score that occurs most often; that is, it is the score w\ ith the highest frequency. If there is more than one score with the highest frequency, then all scores with that frequency are the modes. The mode is a weak measure of the centre of data because it may be a value that is close to the extremes of the data. If we consider the set of data in Worked example 8, the mode is 48 since it occurs three times and hence is the score with the highest frequency. In Worked example 9 there are two modes, 17 and 19, because they equally occur most frequently. exercise 1e The median, the interquartile range, the range and the mode 1 We8 Write the median, the range and the mode of the sets of data shown in the following stem plots. The key for each stem plot is 3|4 = 34. a Stem 0 1 2 3 4 5 6Leaf 7 2 3 2 4 5 7 9 0 2 3 6 8 8 4 7 8 9 9 2 7 8 1 3 b Stem 0 0 0 0 0 1 1 1 1 1 Leaf 0 0 1 1 2 2 3 3 4 4 5 5 5 5 5 5 5 5 6 6 6 6 7 8 8 8 9 0 0 1 3 3 5 5 7 c Stem 0 0 0 0 0 1 1 1 1 1Leaf 1 2 4 4 5 6 6 6 7 8 8 8 8 9 9 0 0 0 1 1 1 1 2 2 2 3 3 3 4 4 5 5 6 7 7 8 9 d Stem 3 3 3 3 3 4 4 4 4 4Leaf 1 6 8 9 0 0 1 1 1 2 2 3 3 3 3 4 5 5 5 6 7 9 e Stem 60 61 62 63 64 65 66 67 Leaf 2 5 8 1 3 3 6 7 8 9 0 1 2 4 6 7 8 8 9 2 2 4 5 7 8 3 6 7 4 5 8 3 5 4 2 For each of the following sets of data, write the median and the range. a 2 4 6 7 9 b 12 15 17 19 21 c 3 4 5 6 7 8 9 d 3 5 7 8 12 13 15 16 e 12 13 15 16 18 19 21 23 24 26 f 3 8 4 2 1 6 5 g 16 21 14 28 23 15 11 19 25 h 7 4 3 4 9 5 10 4 2 11 i 29 23 22 33 26 18 37 22 16 3 a We9 The number of cars that used the drive-in at a McBurger restaurant during each hour, from 7.00 am until 10.00 pm on a particular day, is shown below. 14 18 8 9 12 24 25 15 18 25 24 21 25 24 14 Find the interquartile range of this set of data. b On the same day, the number of cars stopping during each hour that the nearby Kenny’s Fried Chicken restaurant was open is shown below. 7 9 13 16 19 12 11 18 20 19 21 20 18 10 14 Find the interquartile range of these data. c What do these values suggest about the two restaurants? diGiTal doC doc-9405 Spreadsheet one-variable statistics

ChapTer 1 • Univariate data 19 4 Write down a set of data for which n = 5, the median is 6 and the range is 7. Is this the only set of data with these parameters? 5 Is it possible to have a set of data in which the: a lower quartile equals the lowest score? b IQR is zero? Give an example of each. 6 mC The quartiles for a set of data are calculated and found to be Q 1 = 13, Q 2 = 18 and Q 3 = 25. Which of the following statements is true? a The interquartile range of the data is 5. B The interquartile range of the data is 7. C The interquartile range of the data is 12. d The median is 12. e The median is 19. 7 We10 For each of the following sets of data find the median, the interquartile range, the range and the mode. a 16 12 8 726 32 15 51 29 45 19 11 615 32 18 43 31 23 23 b 22 25 27 36 31 32 39 29 20 30 23 25 21 19 29 28 31 27 22 29 c 1.2 2.3 4.1 2.4 1.5 3.7 6.1 2.4 3.6 1.2 6.1 3.7 5.4 3.7 5.2 3.8 6.3 7.1 4.9 8 For each set of data shown on the stem plots, find the median, the interquartile range, the range and the mode. Compare these values for both data sets. a Stem 2 3 4 5 6 7 8 9 10 11 Leaf 3 5 5 6 7 8 9 9 0 2 2 3 4 6 6 7 8 8 2 2 4 5 6 6 6 7 9 0 3 3 5 6 2 4 5 9 2 7 4 Key: 4|2 = 42 b Stem 1* 1* 2* 2* 3* 3* 4* 4* Leaf 4 1 4 5 7 8 8 9 1 2 2 2 4 4 4 4 5 5 5 6 3 4 Key: 2|1 = 21 2*|5 = 25 1F Boxplots The fi ve number summary statistics that we looked at in the previous section can be illustrated very neatly in a special diagram known as a boxplot (or box-and-whisker diagram). The diagram is made up of a box with straight lines (whiskers) extending from opposite sides of the box. A boxplot displays the minimum and maximum values of the data together with the quartiles and is drawn with a labelled scale. The length of the box is given by the interquartile range. A boxplot gives us a very clear visual display of how the data are spread out. Q1 Q2 Median Q 3 25% of data 25% of data 25% of data 25% of data Minimum value Maximum value Whisker Whisker Box A boxplot Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 3 Concept: 5 See more Watch a  video about how to  construct boxplots.

20 Maths Quest 12 Further Mathematics Boxplots can be drawn horizontally or vertically. Vertical boxplot Horizontal boxplot Worked example 11 The boxplot at right shows the distribution of the part-time weekly earnings of a group of Year 12 students. Write down the range, the median and the interquartile range for these data. Think WriT e 1Range = Maximum value − Minimum value. The minimum value is 20 and the maximum value is 90. Range = 90 − 20 = 70 2The median is located at the bar inside the box. Median = 50 3The ends of the box are at 40 and 80. IQR = Q 3 − Q 1 Q1 = 40 and Q 3 = 80 IQR = 80 − 40 = 40 Earlier, we noted three general types of shape for histograms and stem plots: s\ ymmetric, negatively skewed and positively skewed. It is useful to compare the corresponding boxplots of distributions with such shapes. In the figures below, a symmetric distribution is represented in the histogram and in the boxplot. The characteristics of this boxplot are that the whiskers are about the same length and the median is located about halfway along the box. Symmetric histogram Symmetric boxplot The figures below show a negatively skewed distribution. In such a distribution, the data peak to the right on the histogram and trail off to the left. In corresponding fashion on the boxplot, the bunching of the data to the right means that the left-hand whisker is longer and the right-hand whisker is shorter; that is, the lower 25% of data are sparse and spread out whereas the top 25% of data are bunched up. The median occurs further towards the right end of the box. Negatively skewed histogram Negatively skewed boxplot Part-time weekly earnings ($) 10 20 30 40 50 60 70 80 90 100

ChapTer 1 • Univariate data 21 In the fi gures below, we have a positively skewed distribution. In such a distribution, the data peak to the left on the histogram and trail off to the right. In corresponding fashion on the boxplot, the bunching of the data to the left means that the left-hand whisker is shorter and the right-hand whisker is longer; that is, the upper 25% of data are sparse and spread out whereas the lower 25% of data are bunched up. The median occurs further towards the left end of the box. Positively skewed histogram Positively skewed boxplot Worked example 12 Explain whether or not the histogram and the boxplot shown below could represent the same data. Think WriTe The histogram shows a distribution which is positively skewed. The boxplot shows a distribution which is approximately symmetric. The histogram and the boxplot could not represent the same data since the histogram shows a distribution that is positively skewed and the boxplot shows a distribution that is approximately symmetric. Worked example 13 The results (out of 20) of oral tests in a Year 12 Indonesian class are: 15 12 17 8 13 18 14 16 17 13 11 12 Display these data using a boxplot and discuss the shape obtained. Think WriTe/draW 1Find the lowest and highest scores, Q 1, the median (Q 2) and Q 3 by fi rst ordering the data. 8 11 12 12 13 13 14 15 16 17 17 18 The median score is 13.5. The lower half of the scores are 8 11 12 12 13 13. So, Q 1 = 12 The upper half of the scores are 14 15 16 17 17 18. So, Q 3 = 16.5 The lowest score is 8. The highest score is 18. 2Using these fi ve number summary statistics, draw the boxplot. 7 8 9 10 11 12 13 14 15 16 17 18 19 Results 3Consider the spread of each quarter of the data. The scores are grouped around 12 and 13, as well as around 17 and 18 with 25% of the data in each section. The scores are more spread elsewhere.  CAS calculators can also be used to draw boxplots. TUTorial eles-1256 Worked example 13

22 Maths Quest 12 Further Mathematics outliers When one observation lies well away from other observations in a set, we call it an outlier. Sometimes an outlier occurs because data have been incorrectly obtained or misread. For example, at right we see a histogram showing the weights of a group of 5-year-old boys. The outlier, 33, may have occurred because a weight was incorrectly recorded as 33 rather than 23 or perhaps there was a boy in this group who, for some medical reason, weighed a lot more than his counterparts. When an outlier occurs, the reasons for its occurrence should be checked. To identify possible outliers, we can apply a simple rule. An outlier is a score, x, which lies outside the interval: Q 1 − 1.5 × IQR ≤ x ≤ Q 3 + 1.5 × IQR. An outlier is not included in the boxplot but simply plotted as a point beyond the end of the whisker. Worked example 14 The times (in seconds) achieved by the 12 fastest runners in the 100-m sprint at a school athletics meeting are listed below. 11.2 12.3 11.5 11.0 11.6 11.4 11.9 11.2 12.7 11.3 11.2 11.3 Draw a boxplot to represent the data, describe the shape of the distribution and comment on the existence of any outliers. Think WriTe/draW 1Determine the fi ve number summary statistics by fi rst ordering the data and obtain the interquartile range. 11.0  11.2  11.2  11.2  11.3  11.3  \ 11.4  11.5 11.6  11.9  12.3  12.7 Lowest score = 11.0 Highest s core = 12.7 Median = Q 2 = 11.35 Q 1 = 11.2 Q 3 = 11.75 IQR = 11.75 − 11.2 = 0.55 2Identify any outliers by applying the outlier rule. Q 1 − 1.5 × IQR = 11.2 − 1.5 × 0.55 = 10.375 The lowest score lies above 10.375, so there is no outlier below. Q 3 + 1.5 × IQR = 11.75 + 1.5 × 0.55 = 12.575 The score 12.7 lies above 12.575, so it is an outlier and 12.3 becomes the end of the upper whisker. 3Draw the boxplot with the outlier. 11.0 12.0 13.0 Time (s) 4Describe the shape of the distribution. Data peak to the left and trail off to the right with one outlier. The data are positively skewed with 12.7 seconds being an outlier. This may be due to incorrect timing or recording but more likely the top eleven runners were signifi cantly faster than the other competitors in the event. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 3 Concept: 6 Weight (kg) Frequency 0 5 10 15 20 25 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

ChapTer 1 • Univariate data 23 exercise 1F Boxplots 1 We11 For the boxplots shown, write down the range, the interquartile range and the median of the distributions which each one represents. a 2 4 6 8 10 12 14 x b 1 2 3 4 5 6 7 8 x c 100 200 300 400 500 x d 20 30 40 50 60 70 80 90 100 110120 130 140 x e 10 15 20 25 30 35 x 2 We12 Match each histogram below with the boxplot which could show the same distribution. a b c d i ii iii iv 3 We13 For each of the following sets of data, construct a boxplot. a 3 5 6 8 8 9 12 14 17 18 b 3 4 4 5 5 6 7 7 7 8 8 9 9 10 10 12 c 4.3 4.5 4.7 4.9 5.1 5.3 5.5 5.6 d 11 13 15 15 16 18 20 21 22 21 18 19 20 16 18 20 e 0.4 0.5 0.7 0.8 0.8 0.9 1.0 1.1 1.2 1.0 1.3 4 mC For the distribution shown in the boxplot below, it is true to say that: a the median is 30 B the median is 45 C the interquartile range is 10 d the interquartile range is 30 e the interquartile range is 60 5 The number of clients seen each day over a 15-day period by a tax consultant is: 3 5 2 7 5 6 4 3 4 5 6 6 4 3 4 Represent these data on a boxplot. 6 The maximum daily temperatures (in ° C) for the month of October in Melbourne are: 18 26 28 23 16 19 21 27 31 23 24 26 21 18 26 27 23 21 24 20 19 25 27 32 29 21 16 19 23 25 27 Represent these data on a boxplot. 7 We14 The number of rides that 16 children had at the annual show are listed below. 8 5 9 4 9 0 8 7 9 2 8 7 9 6 7 8 a Draw a boxplot to represent the data, describe the shape of the distribution and comment on the existence of any outliers. b Use a CAS calculator to draw a boxplot for these data. diGiTal doC doc-9406 Spreadsheet Boxplots 10 20 30 40 50 60 70 x

24 Maths Quest 12 Further Mathematics 8 A concentration test was carried out on 40 students in Year 12 across Australia. The test involved the use of a computer mouse and the ability to recognise multiple images\ . The less time required to complete the activity, the better the student’s ability to concentrate. The data are shown by the parallel boxplots below. 20 40 60 Time (s) Females Males 100 a Identify two similar properties of the concentration spans for boys and girls. b Find the interquartile range for boys and girls. c Comment on the existence of an outlier in the boys’ data. 1G The mean The mean of a set of data is what is referred to in everyday language as the average.For the set of data {4, 7, 9, 12, 18}: mean = ++ ++ 47++47++ 91++91++ 21 ++ 21 ++ 8 5 = 10. The symbol we use to represent the mean is x, that is, a lower-case x with a bar on top. So, in this case, x = 10. The formal defi nition of the mean is: x x n =∑ where Σ x represents the sum of all of the observations in the data set and n represents the number of observations in the data set. Note that the symbol, Σ, is the Greek letter, sigma, which represents ‘the sum of ’. The mean is also referred to as a summary statistic and is a measure of the centre of a distribution. The mean is the point about which the distribution ‘balances’. Consider the masses of 7 potatoes, given in grams, in the photograph below. 170 g 145 g 100 g 190 g 160 g 120 g 130 g The mean is 145 g. The observations 130 and 160 ‘balance’ each other since they are each 15 g from the mean. Similarly, the observations 120 and 170 ‘balance’ each other since they are each 25 g from the mean, as do the observations 100 and 190. Note that the median is also 145 g. That is, for this set of data the mean and the median give the same value for the centre. This is because the distribution is symmetric. Now consider two cases in which the distribution of data is not symmetric. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 3 Concept: 1

ChapTer 1 • Univariate data 25 Case 1 Consider the masses of a different set of 7 potatoes, given in grams below. 100 105 110 115 120 160 200 The median of this distribution is 115 g and the mean is 130 g. There are 5 observations that are less than the mean and only 2 that are more. In other words, the mean does not give us a good indication of the centre of the distribution. However, there is still a ‘balance’ between observations below the mean and those above, in terms of the spread of all the observations from the mean. Therefore, the mean is still useful to give a measure of the central tendency of the distribution but in cases where the distribution is skewed, the median gives a better indication of the centre. For a positively skewed distribution, as in the previous case, the mean will be greater than the median. For a negatively skewed distribution the mean will be less than the median. Case 2 Consider the data below, showing the weekly income (to the nearest $10) of 10 families living in a suburban street. $600 $1340 $1360 $1380 $1400 $1420 $1420 $1440 $1460 $1500 In this case, =x 13 320 10 = $1332, and the median is $1410. One of the values in this set, $600, is clearly an outlier. As a result, the value of the mean is below the weekly income of the other 9 households. In such a case the mean is not very useful in establishing the centre; however, the ‘balance’ still remains for this negatively skewed distribution. The mean is calculated by using the values of the observations and because of this it becomes a less reliable measure of the centre of the distribution when the distribution is skewed or contains an outlier. Because the median is based on the order of the observations rather than their value, it is a better measure of the centre of such distributions. Worked example 15 Calculate the mean of the set of data below. 10, 12, 15, 16, 18, 19, 22, 25, 27, 29 Think WriTe 1Write the formula for calculating the mean, where ∑ x is the sum of all scores; n is the number of scores in the set. =∑ x x n = 10 + 12 + 15 + 16 + 18 + 19 + 22 + 25 + 27 + 29 10 x = 19.3 2Substitute the values into the formula and evaluate. The mean, x, is 19.3.

26 Maths Quest 12 Further Mathematics When data are presented in a frequency table with class intervals and we don’t know what the raw data are, we employ another method to fi nd the mean of these grouped data. This other method is shown in the example that follows and uses the midpoints of the class intervals to represent the raw data. Recall that the Greek letter sigma, ∑, represents ‘the sum of ’. So, ∑ f means the sum of the frequencies and is the total of all the numbers in the frequency column. To fi nd the mean for grouped data, =∑× ∑ x fm ∑× fm ∑× f∑f∑ ()∑×()∑× fm()fm ∑× fm ∑×()∑× fm ∑× where f represents the frequency of the data and m represents the midpoint of the class interval of the grouped data. Worked example 16 The ages of a group of 30 people attending a superannuation seminar are recorded in the frequency table below. Age (class intervals) Frequency f Age (class intervals) Frequency f 20–29 30–39 40–49 1 6 13 50–59 60–69 70–796 3 1 Calculate the mean age of those attending the seminar. Think WriTe 1Since we don’t have individual raw ages, but rather a class interval, we need to decide on one particular age to represent each interval. We use the midpoint, m, of the class interval. Add an extra column to the table to display these. The midpoint of the fi rst interval is +20 29 2 = 24.5, the midpoint of the second interval is 34.5 and so on. Age (class intervals) Frequency f Midpoint of class interval m f × m 20–29 30–39 40–49 50–59 60–69 70–79 1 6 13 6 3 1 24.5 34.5 44.5 54.5 64.5 74.5 24.5 207 578.5 327 193.5 74.5 ∑ f = 30 ∑(f × m) = 1405 2Multiply each of the midpoints by the frequency and display these values in another column headed f × m. For the fi rst interval we have 24.5 × 1 = 24.5. For the second interval we have 34.5 × 6 = 207 and so on. 3Sum the product of the midpoints and the frequencies in the f × m column. 24.5 + 207 + 578.5 + 327 + 193.5 + 74.5 = 1405 4Divide this sum by the total number of people attending the seminar (given by the sum of the frequency column). So, x = 1405 30 ≈ 46.8 (correct to 1 decimal place). TUTorial eles-1257 Worked example 16

ChapTer 1 • Univariate data 27 exercise 1G The mean 1 We 15 Find the mean of each of the following sets of data. a 5 6 8 8 9 b 3 4 4 5 5 6 7 7 7 8 8 9 9 10 10 12 c 4.3 4.5 4.7 4.9 5.1 5.3 5.5 5.6 d 11 13 15 15 16 18 20 21 22 e 0.4 0.5 0.7 0.8 0.8 0.9 1.0 1.1 1.2 1.0 1.3 2 Calculate the mean of each of the following and explain whether or not it gives us a good indication of the centre of the data. a 0.7 0.8 0.85 0.9 0.92 2.3 b 14 16 16 17 17 17 19 20 c 23 24 28 29 33 34 37 39 d 2 15 17 18 18 19 20 3 The number of people attending sculpture classes at the local TAFE college for each week during the first semester is given below. 15 12 15 11 14   8 14 15 11 10   7 11 12 14 15 14 15   9 10 11 What is the mean number of people attending each week? (Express your an\ swer to the nearest whole number.) 4 mC The ages of a group of junior pilots joining an international airline ar\ e indicated on the stem plot at right. The mean age of this group of pilots is: a 20 B 28 C 29 d 29.15 e 29.5 5 mC The number of people present each week at a 15-week horticultural course is given by the stem plot at right. The mean number of people attending each week was closest to: a 17.7 B 18 C 19.5 d 20 e 21.2 6 For each of the following, write down whether the mean or the median would provide a better indication of the centre of the distribution. a A positively skewed distribution b A symmetric distribution c A distribution with an outlier d A negatively skewed distribution 7 We 16 Find the mean of each set of data given below. a Class interval Frequency, f b Class interval Frequency, f 0–9 10–19 20–29 30–39 40–49 50–59 1 3 6 17 12 5 0–4 5–9 10–14 15–19 20–24 25–29 2 5 7 13 8 6 Key: 2|1 = 21 years Stem 2 2 2 2 2 3 3 3 3 3Leaf 1 2 4 5 6 6 7 8 8 8 9 0 1 1 2 3 4 4 6 8 Key: 2|4 = 24 people Stem 0* 0* 1* 1* 2* 2* Leaf 4 7 2 4 5 5 6 7 8 1 2 4 7 7 7

28 Maths Quest 12 Further Mathematics c Class interval Frequency, f d Class interval Frequency, f 0–49 50–99 100–149 150–199 200–249 250–299 2 7 8 14 12 5 1–6 7–12 13–18 19–24 25–30 31–36 14 19 23 22 20 14 8 The ages of people attending a beginner’s course in karate are indicated in the following frequency table. a What is the mean age of those attending the course? (Express your answer correct to 1 decimal place.) b Calculate the median. What does this value, compared to the mean, suggest about the shape of the distribution? Age Frequency, f 10–14 15–19 20–24 25–29 30–34 35–39 40–44 45–49 5 5 7 4 3 2 2 1 1h Standard deviation The standard deviation gives us a measure of how data are spread around the mean. For the set of data {8, 10, 11, 12, 12, 13}, the mean, x = 11. The amount that each observation ‘deviates’ (that is, differs) from the mean is calculated and shown in the table below. Particular observation, x Deviation from the mean, (x − x) 8 10 11 12 12 13 8 − 11 = −3 10 − 11 = −1 11 − 11 = 0 12 − 11 = 1 12 − 11 = 1 13 − 11 = 2 The deviations from the mean are either positive or negative depending on whether the particular observation is lower or higher in value than the mean. If we were to add all the deviations from the mean we would obtain zero. If we square the deviations from the mean we will overcome the problem of positive and negative deviations cancelling each other out. With this in mind, a quantity known as sample variance (s 2) is defi ned: =∑− − s n ()()∑−()∑− xx()xx ∑− xx ∑−()∑− xx ∑− 1 2 2 . Technically, this formula for variance is used when the data set is a sub-set of a larger population. Variance gives the average of the squared deviations and is also a measure of spread. A far more useful measure of spread, however, is the standard deviation, which is the square root of variance (s). One reason for it being more useful is that it takes the same unit as the observations (for example, cm or number of people). Variance would square the units, for example, cm 2 or number of people squared, which is not very practical. Other advantages of the standard deviation will be dealt with later in the chapter. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 3 Concept: 4 Do more Interact  with standard  deviations.

ChapTer 1 • Univariate data 29 In summary, = ∑− − s n ()()∑−()∑− xx()xx ∑− xx ∑−()∑− xx ∑− 1 2 where s represents sample standard deviation ∑ represents ‘the sum of ’ x represents an observation x represents the mean n represents the number of observations. While some of the theory or formulas associated with standard deviation may look complex, the calculation of this measure of spread is straightforward using a statistical, graphics or CAS calculator. Manual computation of standard deviation is therefore rarely necessary. Worked example 17 The price (in cents) per litre of petrol at a service station was recorded each Friday over a 15-week period. The data are given below. 152.4 160.2 159.6 168.6 161.4 156.6 164.8 162.6 161.0 156.4 159.0 160.2 162.6 168.4 166.8 Calculate the standard deviation for this set of data, correct to 2 decimal places. Think WriTe 1On a CAS calculator, enter the data into the fi rst list in the spreadsheet and label it price. Select ‘One-Variable Statistics’ option and choose ‘price’ for the X1 List. Press OK to see all statistics. 2The entry, SX: = S n − 1 , gives us the standard deviation. Round the value correct to 2 decimal places. S x = 4.515 92 s = 4.52 cents/L Worked example 18 The number of students attending SRC meetings during the term is given in the stem plot at right. Calculate the standard deviation for this set of data, correct to 3 decimal places. Stem 0* 0* 1* 1* 2* 2* Leaf 4 8 8 1 3 4 5 6 8 3 5 Key: 1|4 = 14 students

30 Maths Quest 12 Further Mathematics Think WriTe 1On your calculator, enter the data from the stem plot into a spreadsheet. Using the given key, the scores are: 4, 8, 8, 11, 13, 14, 15, 16, 18, 23 and 25. To calculate the summary statistics, repeat the instructions for Worked example 17. 2The entry, SX, gives us the standard deviation. Round the value correct to 3 decimal places. SX = 6.363 25 s = 6.363 students The standard deviation is a measure of the spread of data from the mean. Consider the t\ wo sets of data shown below. Frequency 3 4 5 6 7 8 9 10 11 12 13 14 15 Score 16 17 18 Frequency 6 7 8 9 10 Score 11 12 13 14 Each set of data has a mean of 10. The set of data above left has a standard deviation of 1 and the set of data above right has a standard deviation of 3. As we can see, the larger the standard deviation, the more spread are the data from the mean. exercise 1h Standard deviation 1 We17 For each of the following sets of data, calculate the standard deviation correct to 2 decimal places. a 3 4 4.7 5.1 6 6.2 b 7 9 10 10 11 13 13 14 c 12.9 17.2 17.9 20.2 26.4 28.9 d 41 43 44 45 45 46 47 49 e 0.30 0.32 0.37 0.39 0.41 0.43 0.45 2 First-quarter profit increases for 8 leading companies are given below as percentages. 2.3 0.8 1.6 2.1 1.7 1.3 1.4 1.9 Calculate the standard deviation for this set of data and express your answer correct to 2 decimal places. 3 The heights in metres of a group of army recruits are given below. 1.8 1.95 1.87 1.77 1.75 1.79 1.81 1.83 1.76 1.80 1.92 1.87 1.85 1.83 Calculate the standard deviation for this set of data and express your answer correct to 2 decimal places. 4 We18 Times (to the nearest tenth of a second) for the heats in the 100 m sprint at the school sports carnival are given at right. Calculate the standard deviation for this set of data and express your answer correct to 2 decimal places. Stem 11 11 11 11 11 12 12 12 12 12 Leaf 0 2 3 4 4 5 6 6 8 8 9 0 1 2 2 3 4 4 6 9 Key: 11|0 = 11.0 s

ChapTer 1 • Univariate data 31 5 The number of outgoing phone calls from an office each day over a 4-week period is shown on the stem plot below. Stem 0 1 2 3 4 5Leaf 8 9 3 4 7 9 0 1 3 7 7 3 4 1 5 6 7 8 3 8 Key: 2|1 = 21 calls Calculate the standard deviation for this set of data and express your answer correct to 2 decimal places. 6 mC A new legal aid service has been operational for only 5 weeks. The number of people who have made use of the service each day during this period is set out at right. The standard deviation (to 2 decimal places) of these data is: a 6.00 B 6.34 C 6.47 d 15.44 e 16.00 1i The 68–95–99.7% rule and z -scores The 68 –95 –99.7% rule The heights of a large number of students at a graduation ceremony were recorded and are shown in the histogram at right. This set of data is approximately symmetric and has what is termed a bell shape. Many sets of data fall into this category and are often referred to as normal distributions. Examples are birth weights and people’s heights. Data which are normally distributed have their symmetrical, bell-shaped distribution centred on the mean value, x. An astounding feature of this type of distribution is that we can predict what percentage of the data lie 1, 2 or 3 standard deviations either side of the mean using what is termed the 68–95–99.7% rule. The 68–95–99.7% rule for a bell-shaped curve states that approximately: 1. 68% of data lie within 1 standard deviation either side of the mean 2. 95% of data lie within 2 standard deviations either side of the mean 3. 99.7% of data lie within 3 standard deviations either side of the mean. Stem 0* 0* 1* 1* 2* 2* Leaf 2 4 7 7 9 0 1 4 4 4 4 5 6 6 7 8 8 9 1 2 2 3 3 3 7 Key: 1|0 = 10 people 1*|6 = 16 people Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 4 Concept: 1 See more Watch a  video about normal  distributions. Frequency Height (cm) 150 160 170 180 190 200 210 220

32 Maths Quest 12 Further Mathematics Figure 1 68% _x+ s – s _x _x Figure 2 _ x _ x – 2s _ x + 2s 95% Figure 3 _ x _ x _x – 3s + 3s 99.7% In fi gure 1 above, 68% of the data shown lie between the value which is 1 standard deviation below the mean, that is x − s , and the value which is 1 standard deviation above the mean, that is, x + s . In fi gure 2 above, 95% of the data shown lie between the value which is 2 standard deviations below the mean, that is, x − 2 s, and the value which is 2 standard deviations above the mean, that is x + 2 s. In fi gure 3 above, 99.7% of the data shown lie between the value which is 3 standard deviations below the mean, that is, x − 3 s, and the value which is 3 standard deviations above the mean, that is, x + 3 s. Worked example 19 The wrist circumferences of a group of people were recorded and the results are shown in the histogram at right. The mean of the set of data is 17.7 and the standard deviation is 0.9. Write down the wrist circumferences between which we would expect approximately: a 68% of the group to lie b 95% of the group to lie c 99.7% of the group to lie. Think WriTe aThe distribution can be described as approximately bell-shaped and therefore the 68–95–99.7% rule can be applied. Approximately 68% of the people have a wrist circumference between x − s and x + s (or one standard deviation either side of the mean). ax − s = 17.7 − 0.9 = 16.8 x + s = 17.7 + 0.9 = 18.6 So approximately 68% of the people have a wrist size between 16.8 and 18.6 cm. bSimilarly, approximately 95% of the people have a wrist size between x − 2s and x + 2s. bx − 2s = 17.7 − 1.8 = 15.9 x + 2s = 17.7 + 1.8 = 19.5 Approximately 95% of people have a wrist size between 15.9 cm and 19.5 cm. cSimilarly, approximately 99.7% of the people have a wrist size between x − 3s and x + 3s. cx − 3s = 17.7 − 2.7 = 15.0 x + 3s = 17.7 + 2.7 = 20.4 Approximately 99.7% of people have a wrist size between 15.0 cm and 20.4 cm. Using the 68–95–99.7% rule, we can work out the various percentages of the distribution which lie between the mean and 1 standard deviation from the mean and between the mean and 2 standard deviations from the mean and so on. The diagram at right summarises this. Note that 50% of the data lie below the mean and 50% lie above the mean due to the symmetry of the distribution about the mean. inTeraCTiViTY int-0182 The 68–95–99.7% rule and z -scores Frequency Wrist circumference (cm) 0 10 20 30 40 50 60 15 15.5 16 16.5 17 17.5 18 18.5 19 19.5 20.5 20 _ x _ x + 3s _ x + 2s _ x + s _ x – s _ x – 2s _ x – 3s 34% 34% 13.5% 2.35% 2.35% 0.15% 0.15% 13.5% 68% 95% 99.7%

ChapTer 1 • Univariate data 33 Worked example 20 The distribution of the masses of packets of ‘Fibre-fi ll’ breakfast cereal is known to be bell-shaped with a mean of 250 g and a standard deviation of 5 g. Find the percentage of Fibre-fi ll packets with a mass which is: a less than 260 g b less than 245 g c more than 240 g d between 240 g and 255 g. Think WriTe/draW 1Draw the bell-shaped curve. Label the axis. x = 250, x + s = 255, x + 2s = 260 etc. 235 240 245 250 255 260 265 34% 34% 13.5% 2.35% 2.35% 0.15% 0.15% 13.5% a 260 g is 2 standard deviations above the mean. Using the summary diagram, we can fi nd the percentage of data which is less than 260 g. aMass of 260 g is 2 standard deviations above the mean. Percentage of distribution less than 260 g is 13.5% + 34% + 34% + 13.5% + 2.35% + 0.15% = 97.5% or 13.5% + 34% + 50% = 97.5% b 245 g is 1 standard deviation below the mean. bMass of 245 g is 1 standard deviation below the mean. Percentage of distribution less than 245 g is 13.5% + 2.35% + 0.15% = 16% or 50% − 34% = 16% c 240 g is 2 standard deviations below the mean. cMass of 240 g is 2 standard deviations below the mean. Percentage of distribution more than 240 g is 13.5% + 34% + 34% + 13.5% + 2.35% + 0.15%= 97.5% or 13.5% + 34% + 50% = 97.5% d Now, 240 g is 2 standard deviations below the mean while 255 g is 1 standard deviation above the mean. dMass of 240 g is 2 standard deviations below the mean. Mass of 255 g is 1 standard deviation above the mean. Percentage of distribution between 240 g and 255 g is 13.5% + 34% + 34% = 81.5% Worked example 21 The number of matches in a box is not always the same. When a sample of boxes was studied it was found that the number of matches in a box approximated a normal (bell-shaped) distribution with a mean number of matches of 50 and a standard deviation of 2. In a sample of 200 boxes, how many would be expected to have more than 48 matches? Think WriTe 1Find the percentage of boxes with more than 48 matches. Since 48 = 50 − 2, the score of 48 is 1 standard deviation below the mean. 48 matches is 1 standard deviation below the mean. Percentage of boxes with more than 48 matches = 34% + 50% = 84% 2Find 84% of the total sample. Number of boxes = 84% of 200 = 168 boxes TUTorial eles-1258 Worked example 20

34 Maths Quest 12 Further Mathematics Standard z -scores To fi nd a comparison between scores in a particular distribution or in different distributions, we use the z-score. The z-score (also called the standardised score) indicates the position of a certain score in relation to the mean. A z-score of 0 indicates that the score obtained is equal to the mean, a ne\ gative z -score indicates that the score is below the mean and a positive z -score indicates a score above the mean. The z-score measures the distance from the mean in terms of the standard deviation. A score that is exactly one standard deviation above the mean has a z-score of 1. A score that is exactly one standard deviation below the mean has a z-score of −1. To calculate a z-score we use the formula: =z xx−xx− s where x = the score, x = the mean and s = the standard deviation. Worked example 22 In an IQ test, the mean IQ is 100 and the standard deviation is 15. Dale’s test results give an IQ of 130. Calculate this as a z-score. Think WriTe 1Write the formula. =z xx−xx− s 2Substitute for x, x and s. = − 130 100 15 3Calculate the z-score. = 2 Dale’s z-score is 2, meaning that his IQ is exactly two standard deviations above the mean. Not all z-scores will be whole numbers; in fact most will not be. A whole number indicates only that the score is an exact number of standard deviations above or below the mean. Using the previous example, an IQ of 88 would be represented by a z-score of −0.8, as shown below. = = − =− z xx−xx− s 88 100 15 0.8 The negative value indicates that the IQ of 88 is below the mean but by less than one standard deviation. Worked example 23 To obtain the average number of hours of study done by Year 12 students per week, Kate surveys 20 students and obtains the following results. 12 18 15 14 9 10 13 12 18 25 15 10 3 21 11 12 14 16 17 20 a Calculate the mean and standard deviation (correct to 2 decimal places). b Robert studies for 16 hours each week. Express this as a z-score based on the above results. (Give your answer correct to 2 decimal places.) Think WriTe a 1 Enter the data into your calculator. a 2 Obtain the mean from your calculator. =x 14.25 3 Obtain the standard deviation from your calculator. s = 4.88 Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 4 Concept: 2

ChapTer 1 • Univariate data 35 b 1 Write the formula for z-score. b =z xx−xx− s 2 Substitute for x, x and s into the formula and evaluate. = − 16 14.25 4.88 = 0.36 Comparing data An important use of z-scores is to compare scores from different data sets. Suppose that in your maths exam your result was 74 and in English your result was 63. In which subject did you achieve the better result? At fi rst glance, it may appear that the maths result is better, but this does not take into account the diffi culty of the test. A mark of 63 on a diffi cult English test may in fact be a better result than 74 if it was an easy maths test. The only way that we can fairly compare the results is by comparing each result with its mean and \ standard deviation. This is done by converting each result to a z-score. If, for maths, x = 60 and s = 12, then z = xx−xx− s = − 74 60 12 = 1.17 And if, for English, x = 50 and s = 8, then z = xx−xx− s = − 63 50 8 = 1.625 The English result is better because the higher z-score shows that the 63 is higher in comparison to the mean of each subject. Worked example 24 Janine scored 82 in her physics exam and 78 in her chemistry exam. In physics, x 62= and s = 10, while in chemistry, x 66= and s = 5. a Write both results as a standardised score. b Which is the better result? Explain your answer. Think WriTe a 1 Write the formula for each subject. aPhysics: z = xx−xx− s Chemistry: z = xx−xx− s 2 Substitute for x, xand s. = − 82 62 10 = − 78 66 5 3 Calculate each z-score. = 2 = 2.4 bExplain that the subject with the highest z-score is the better result. bThe chemistry result is better because of the higher z-score. In each example the circumstances must be analysed carefully to see whether a hig\ her or lower z-score is better. For example, if we were comparing times for runners over different distances, the lower z-score would be the better one. Concept summary Read a summary  of this concept. Units: 3 & 4 AOS: DA Topic: 4 Concept: 3 Do more Interact  with comparisons  of data values.

36 Maths Quest 12 Further Mathematics exercise 1i The 68–95–99.7% rule and z -scores 1 In each of the following, decide whether or not the distribution is approximately bell-shaped. a Frequency b Frequency c Frequency d Frequency e Frequency f Frequency 2 Copy and complete the entries on the horizontal scale of the following distributions, given that x = 10 and s = 2. a 10 68% b 10 95% c 10 99.7% 3 Copy and complete the entries on the horizontal scale of the following distributions, given that x = 5 and s = 1.3. a 68% 5 b 5 95% c 5 99.7% 4 We19 The concentration ability of a randomly selected group of adults is test\ ed during a short task which they are asked to complete. The length of the concentration span of those involved during the task is shown at right. The mean, x, is 49 seconds and the standard deviation, s, is 14 seconds. Write down the values between which we would expect approximately: a 68% of the group’s concentration spans to fall b 95% of the group’s concentration spans to fall c 99.7% of the group’s concentration spans to fall. Frequency Concentration span (seconds) 100 20 30 40 50 60 70 80 90

ChapTer 1 • Univariate data 37 5 A research scientist measured the rate of hair growth in a group of hamsters. The findings are shown in the histogram below. The mean growth per week was 1.9 mm and the standard deviation was 0.6 mm. Write down the hair growth rates between which approximately: a 68% of the values fall b 95% of the values fall c 99.7% of the values fall. Frequency Growth per week (mm) 1 2 3 4 6 The force required to break metal fasteners has a distribution which is bell-shaped. A large sample of metal fasteners was tested and the mean breaking force required was 12 newtons with a standard deviation of 0.3 newtons. Write down the values between which approximately: a 68% of the breaking forces would lie b 95% of the breaking forces would lie c 99.7% of the breaking forces would lie. 7 The heights of the seedlings sold in a nursery have a bell-shaped distribution. The mean height is 7 cm and the standard deviation is 2.Write down the values between which approximately: a 68% of seedling heights will lie b 95% of seedling heights will lie c 99.7% of seedling heights will lie. 8 mC A set of scores in a competition has a mean of 15 and a standard deviation of 3. The distribution of the scores is known to be bell-shaped. Which one of the following could be true? a 68% of the scores lie between 3 and 15. B 68% of the scores lie between 15 and 18. C 68% of the scores lie between 12 and 15. d 68% of the scores lie between 13.5 and 16.5. e 68% of the scores lie between 12 and 18. 9 mC A distribution of scores is bell-shaped and the mean score is 26. It is known that 95% of scores lie between 21 and 31. It is true to say that: a 68% of the scores lie between 23 and 28. B 97.5% of the scores lie between 23.5 and 28.5. C The standard deviation is 2.5. d 99.7% of the scores lie between 16 and 36. e The standard deviation is 5. 10 We 20 The distribution of heights of a group of Melbourne-based employees who work for a large international company is bell-shaped. The data have a mean of 160 cm and a standard deviation of 10 cm. Find the percentage of this group of employees who are: a less than 170 cm tall b less than 140 cm tall c greater than 150 cm tall d between 130 cm and 180 cm in height. 11 The number of days taken off in a year by employees of a large company has a distribution which is approximately bell-shaped. The mean and standard deviation of this data are shown below. Mean = 9 days Standard deviation = 2 days Find the percentage of employees of this company who, in a year, take off: a more than 15 days b fewer than 5 days c more than 7 days d between 3 and 11 days e between 7 and 13 days.

38 Maths Quest 12 Further Mathematics 12 mC The mean number of Drool-mints in a packet is 48. The data have a standard deviation of 2. If the number of mints in a packet can be approximated by normal distribution. The percentage of packets which contain more than 50 Drool-mints is: a 0.15% B 2.5% C 16% d 50% e 84% 13 We21 The volume of fruit juice in a certain type of container is not always the same. When a sample of these containers was studied it was found that the volume of juice they contained approximated a normal distribution with a mean of 250 mL and a standard deviation of 5 mL. In a sample of 400 containers, how many would be expected to have a volume of: a more than 245 mL? b less than 240 mL? c between 240 and 260 mL? 14 A particular bolt is manufactured such that the length is not always the same. The distribution of the lengths of the bolts is approximately bell-shaped with a mean length of \ 2.5 cm and a standard deviation of 1 mm. a In a sample of 2000 bolts, how many would be expected to have a length: i between 2.4 cm and 2.6 cm? ii less than 2.7 cm? iii between 2.6 cm and 2.8 cm? b The manufacturer rejects bolts which have a length of less than 2.3 cm or a length of greater than 2.7 cm. In a sample of 2000 bolts, how many would the manufacturer expect to reject? 15 We22 In a maths exam, the mean score is 60 and the standard deviation is 12. Chifune’s mark is 96. Calculate her mark as a z -score. 16 In an English test, the mean score was 55 with a standard deviation of 5. Adrian scored 45 on the English test. Calculate Adrian’s mark as a z -score. 17 IQ tests have a mean of 100 and a standard deviation of 15. Calculate the z -score for a person with an IQ of 96. (Give your answer correct to 2 decimal places.) 18 The mean time taken for a racehorse to run 1 km is 57.69 s, with a standard deviation of 0.36 s. Calculate the z -score of a racehorse that runs 1 km in 58.23 s. 19 In a major exam, every subject has a mean score of 60 and a standard deviation of 12.5. Clarissa obtains the following marks on her exams. Express each as a z -score. a English 54 b Maths 78 c Biology 61 d Geography 32 e Art 95 20 We23 The length of bolts being produced by a machine needs to be measured. To do this, a sample of 20 bolts are taken and measured. The results (in mm) are given below. 20 19 18 21 20 17 19 21 22 21 17 17 21 20 17 19 18 22 22 20 a Calculate the mean and standard deviation of the distribution. b A bolt produced by the machine is 22.5 mm long. Express this result as a\ z-score. (Give your answer correct to 2 decimal places.) 21 mC In a normal distribution, the mean is 21.7 and the standard deviation is 1.9. A score of 20.75 corresponds to a z -score of: a −1 B −0.5 C 0.5 d 1 e 0.75 22 mC In a normal distribution the mean is 58. A score of 70 corresponds to a standardised score of 1.5. The standard deviation of the distribution is: a 6 B 8 C 10 d 12 e 9 diGiTal doC doc-9407 SkillSHEET 1.2 percentages

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