Mathspace VCE Units 1 & 2 General Mathematics

VCE

General Mathematics Units 1 & 2 Accreditation period from 2023

© 2025 Mathspace Group Holdings Ltd Copyright Notice This Work is copyright. All rights are reserved. Reproduction 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. For permission to use material from this text or product, please email hello@mathspace.com.au For our full digital offering, visit mathspace.co The Victorian Curriculum F-10 content elements are © VCAA, reproduced by permission. The VCAA does not endorse or make any warranties regarding this resource. The Victorian Curriculum F-10 and related content can be accessed directly at the VCAA website. Title: Mathspace VCE General Mathematics Units 1 & 2 Print run 1 - January 2026 ISBN: 978-1-963022-96-4 Editors and lead authors: Erin Gallagher, Jeremy Hunter, Elisha Templin, Nazeel Ali, Katrina Du, Neeti Meghani, Jessica Purton, Isha Tiwari Writing and development team: Adam Humphreys, Ma Lourdes Ona, Sheramie Castillo, Regine Jara, Lesley Ann Lopez, Adriane Abunda, Jady Amarillas, Joy Ann Antonio, Macy Castillo, Christian Dinglasan, Adrian Doctolero, Dharent Fernandez, Dhave Fernandez, Rona Josue, Jessa Lambinicio, Eric Malabag, Julie Ann Manzano, Kyra Joy Manzano, and Mary Joy Tacud, Albert Jara Images and design team: Scott Nolan, Chastine Marquez, Keith Gimeno, Jasper Jumawan, Mary Matheu, Jemark Orevillo, Jessa Ortega

Contents 1

Data distributions

2

1.01

Types of data

4

1.02

Categorical data displays

13

1.03

Numerical data displays

26

1.04

Measures of centre

42

Investigation: The problem with average 1.05

Data distribution

56

1.06

Quartiles and spread

72

1.07

Box plots

85

1.08

Outliers

102

1.09

Standard deviation

118

1.10

Compare sets of data

138

Chapter 1 review

2

Linear relationships

160

2.01

Solve linear equations

162

2.02

Linear equation applications

168

2.03

Graphs from tables

175

2.04

Slopes and intercepts

187

2.05

Sketch linear functions

200

2.06

Equations of straight lines

208

Investigation: Filling a pool Chapter 2 review

3

Linear applications

224

Linear models

226

3.01

Investigation: Intersection of lines 3.02

Graphical simultaneous equations

237

3.03

Algebraic simultaneous equations

247

3.04

Applications of simultaneous equations

256

3.05

Piecewise functions

262

3.06

Step graphs

276

Chapter 3 review

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iii

4

Financial arithmetic

288

4.01

Percentages

290

4.02

Rates and ratios

301

4.03

Unit cost comparisons

309

4.04

Percentage change and GST

315

4.05

Simple interest

326

4.06

Compound interest

337

Investigation: Spreadsheets and interest 4.07

Financial applications

346

4.08

Purchase options

359

Chapter 4 review

5

Number patterns and recurrence relations

374

5.01

Number patterns and sequences

376

5.02

First-order linear recurrence relations

387

5.03

Arithmetic sequences

400

5.04

Applications of arithmetic sequences

410

5.05

Geometric sequences

421

5.06

Applications of geometric sequences

431

Investigation: Saving money Chapter 5 review

6

Matrices

440

6.01

Introduction to matrices

442

6.02

Represent information in matrices

449

6.03

Addition and subtraction of matrices

460

6.04

Scalar multiplication

467

6.05

Matrix multiplication

473

6.06

Powers of matrices

486

6.07

Applications of matrices

498

6.08

Inverse matrices

510

6.09

Simultaneous equations with matrices

518

Investigation: Is there always a solution? 6.10

Transition matrices Chapter 6 review

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528

7

Graphs and networks

542

7.01

Introduction to graphs

544

7.02

Degree and adjacency matrices

556

7.03

Planar graphs and Euler’s formula

569

7.04

Walks and paths

583

7.05

Traverse connected graphs

600

7.06

Weighted graphs and networks

612

7.07

Trees

627

Investigation: Networks and coding Chapter 7 review

8

Compare numerical variables

642

8.01

Explanatory and response variables

644

8.02

Scatterplots and correlation

652

8.03

Association and causation

666

8.04

Line of good fit

675

Investigation: Answer questions with data Chapter 8 review

9

Variation 694 9.01

Direct and inverse variation

696

9.02

Orders of magnitude and logarithms

707

Investigation: Represent Australian populations 9.03

Log scales

717

9.04

Data transformations

725

9.05

Data models

746

Chapter 9 review

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10

Shape and measurement

760

10.01

Significant figures

762

10.02

Measurement units

770

10.03

Pythagoras’ theorem

781

10.04

Perimeter

793

Investigation: Circumference 10.05

Arc length and perimeter of sectors

803

10.06

Area

814

10.07

Area of composite shapes

823

10.08

Total surface area

835

10.09

Volume

851

10.10

Composite solids

863

10.11

Similarity and scale factors

877

Investigation: Scale diagrams 10.12

Area and volume in similar figures

890

Chapter 10 review

Chapter 10 review

11

Applications of trigonometry

902

11.01

Trigonometric ratios

904

11.02

2D Applications of trigonometry

920

11.03

Trigonometry in 3D

933

11.04

Trigonometry of obtuse angles

944

11.05

Sine rule

950

11.06

Cosine rule

966

11.07

Area of non-right angled triangles

977

11.08

Mixed applications

990

Investigation: Triangle inequality Chapter 11 review

Answers

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“Mathematicians enjoy thinking about the simplest possible things, and the simplest possible things are imaginary.” – Paul Lockhart

Big ideas • Effective data analysis begins with classifying data by type (categorical or numerical) and organising it through appropriate visual displays to reveal initial patterns and frequencies. • Key statistical measures are used to summarise and interpret the essential features of a dataset, including its central tendency, spread, and overall shape. • Visual summaries, such as box plots, and comparative statistics enable the analysis of multiple datasets, allowing for clear interpretation of their similarities and differences in centre, spread, and distribution.

1 Data distributions Chapter outline 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10

Types of data Categorical data displays Numerical data displays Measures of centre Investigation: The problem with average Data distribution Quartiles and spread Box plots Outliers Standard deviation Compare sets of data Chapter 1 review

4 13 26 42 56 72 85 102 118 138

Box plots help detect outliers! An outlier can be a surprising piece of data, like finding penguins in a grassy, rocky area– totally unexpected but real!

1.01   Types of data After this lesson, you will be able to… • distinguish between nominal, ordinal, discrete, and continuous data. • classify real-world data sets into appropriate data types. • compare and contrast the characteristics of nominal, ordinal, ratio, and interval scales.

Types of data A statistical variable is a measurable or observable characteristic or attribute of an item or person. Statistical data are the actual values or observations collected for these variables. Data is collected to observe and analyse changes in these variables. For example: Variable

Data

Height of students

162 cm, 168 cm, 173 cm, 183 cm

Type of animal

Dog, Cat, Bird, Fish

Univariate data Data relating to a single variable. Statistical variables are classified as either numerical or categorical: Numerical data Data where each data point is represented by a number. Also known as quantitative data. • Examples of numerical data include: the number of items sold each month, daily temperatures, heights of people, and ages of a population. • The data can be further defined as either discrete (associated with counting) or continuous (associated with measuring). Categorical data Data where each data point is represented by a word or label. Also known as qualitative data. • Examples of categorical data include: brand names, types of animals, favourite colours, and names of countries. • The data can be further defined as either ordinal (it can be ordered) or nominal (unordered).

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Data

Numerical

Discrete

Continuous

Categorical

Ordinal

Nominal

Discrete data Numerical data with distinct, separate values. There is a definite ‘gap’ separating one data point from the next. Discrete data usually, but not always, consists of whole numbers, and is often collected by some form of counting. Examples of discrete data Number of goals scored per match

0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 4, . . .

Number of products sold each day

386, 397, 401, 410, 411, 422, 437, . . .

Shoe size

5, 5.5, 6, 7, 8.5, 9, 10, 10, 12.5

In each of these cases, there are no in-between values. For example, there is no way to score 2.5 goals. Discrete data can involve decimals as long as there is a definite gap between possible values. For example, a shoe will not ever come in size 8.145, so it is discrete. Continuous data Numerical data that can take any value within a range. Continuous data often involves the use of decimal numbers, and is often collected using some form of measurement. Examples of continuous data Height of trees in a forest (in metres)

12.359, 14.022, 14.951, 18.276, 11.032, . . .

Times taken to run a 10 km race (minutes)

55.34, 58.03, 57.25, 61.49, 66.11, 59.87, . . .

Daily temperature (°C)

24.4, 23.0, 22.5, 21.6, 20.7, 20.2, 19.7, . . .

Since continuous data is typically measured, it is limited by the accuracy of the measuring device being used. So continuous data may appear discrete when it is rounded. For instance, it makes sense that time is continuous, even though it can only be measured to the nearest second (or millisecond, and so on). Ordinal data Categorical data with a natural order or rank.

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The word ‘ordinal’ means ‘ordered’. Examples of ordinal data Product rating on a survey

satisfactory, good, good, good, good, excellent, excellent, . . .

Exam grades

C, C, C, C, B, B, B, B, B, B, B, B, A, A, A, A, A, A, A, . . .

Size of fish in a lake

small, small, small, medium, medium, medium, large, large, . . .

Nominal data Categorical data with no inherent order. The word ‘nominal’ basically means ‘name’. Examples of nominal data Nationalities in a team

German, Austrian, Italian, Spanish, Dutch, Italian, . . .

Make of car driving through an intersection

Toyota, Holden, Mazda, Toyota, Ford, Toyota, Mazda, . . .

Hair colour of students in a class blonde, red, brown, blonde, black, brown, black, red, . . .

Example 1 Which one of these variables is discrete? A Height

B

The time it takes to swim 200 metres

C Daily temperature

D

The number of family pets

Create a strategy

Apply the idea

Choose the option that can be counted but is distinct and separate from one another.

The correct answer is option D: The number of family pets.

Example 2 Classify this data into its correct category: Weights of kittens A Numerical discrete

B

Categorical nominal

C Numerical continuous

D

Categorical ordinal

Create a strategy

Apply the idea

Determine if the data is represented by a number, or is separated into categories.

The weight of a kitten can be measured. So it is numerical or quantitative. Weight is a measurement that can have any number of decimal places, so it is continuous. The correct answer is option C.

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Idea summary A statistical variable is a measurable characteristic being studied. Data collected for one variable is called univariate data. Variables can be classified as either numerical or categorical. • Numerical data (quantitative) consists of numbers. It is further divided into: • Discrete data: Values are distinct and separate, often found by counting (e.g., shoe size). • Continuous data: Can take any value within a range, often found by measuring (e.g., height). • Categorical data (qualitative) consists of words or labels. It is further divided into: • Ordinal data: Categories have a natural, meaningful order (e.g., product ratings). • Nominal data: Categories have no inherent order (e.g., hair colour).

Level of measurement To perform statistical analysis of data, it is important to understand what summary statistics can be meaningfully calculated, interpreted, and compared from a given set of data. Different levels of measurement can be applied depending on the properties of the given data. The four widely applied levels of measurement are: Nominal scale

True zero

A level of measurement with categories that do not have a natural order or ranking.

A point on a scale representing the complete absence of the measured quantity.

Example: Employment status, blood type, or eye colour Ordinal scale A level of measurement with categories together with an explicit ranking. Example: Customer satisfaction ratings (very low, low, average, high, very high)

Interval scale A numerical scale for numerical data with equal intervals but no true zero. Example: Temperature in Celsius, as 30°C is not ‘double as warm’ as 10°C, but is 20°C warmer

Other examples of interval scale include: • pH scale - 0 does not mean an absence of pH, it means as acidic as possible • Dates - for the year 2025, 0 is not the ‘beginning of time’ • Time during the day using a 12-hour clock

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Ratio scale A numerical scale for numerical data with equal intervals and a true zero. Example: Mass, as 5 g is half as heavy as 10 g and 0 g means there is no mass Other examples of ratio scale include: • Temperature measured in Kelvin as 0 K has no molecular movement, so is a true zero • Speed, because 40 km/h is indeed twice as fast as 20 km/h • Time using a stopwatch The significant difference between an interval scale and a ratio scale is the inclusion of a ‘true zero’ or ‘absolute zero’.

RATIO Named + Ordered + Equal scale + Can accommodate absolute zero

INTERVAL Named + Ordered + Equal scale

ORDINAL Named + Ordered variables

NOMINAL Named variables A C D B

0 0.5 1 1.5 2 2.5 3

2 2.5 3 3.5 4 4.5 5

A<B<C<D

The properties of each scale can be compared in terms of how they lend themselves to being used to calculate and compare statistics:

Categorises the values

Nominal

Ordinal

Interval

Ratio

Y

Y

Y

Y

Ranks values in order

Y

Y

Y

Frequency distribution

Y

Y

Y

Y

Mode

Y

Y

Y

Y

Y

Y

Y

Mean

Y

Y

Can say one value is _ units greater or smaller than the other

Y

Y

Median

Can say one value is _ times greater or smaller than the other

Y

Note: The mean of ordinal data can sometimes be calculated by assigning numbers to the outcomes. For example, for a survey asking for a customer satisfaction rating of very low, low, average, high, and very high, the numbers 1 to 5 can be assigned to the outcomes, and the mean level of satisfaction reported can be calculated. However, it is debatable whether this can be meaningfully interpreted, as the original categories are not separated into equal intervals.

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Example 3 Classify this data into its correct category: Population of a town A Nominal

B

Ordinal

C

Interval

Ratio

D

Create a strategy Use the table:

Categorises the values

Nominal

Ordinal

Interval

Ratio

Y

Y

Y

Y

Y

Y

Y

Ranks values in order Frequency distribution

Y

Y

Y

Y

Mode

Y

Y

Y

Y

Y

Y

Y

Mean

Y

Y

Can say one value is _ units greater or smaller than the other

Y

Y

Median

Can say one value is _ times greater or smaller than the other

Y

Apply the idea The population can be compared through a ratio of values. For example, a population of 2000 is considered twice as large as a population of 1000. The correct answer is option D.

Idea summary The four levels of measurement classify data based on their properties, which determines the types of statistical analysis that can be performed. • Nominal scale: Used for categorical data where categories have no natural order or rank (e.g., blood type). • Ordinal scale: Used for categorical data where categories have a meaningful, explicit ranking (e.g., satisfaction ratings). • Interval scale: A numerical scale with equal intervals between values, allowing for meaningful differences. It lacks a true zero (e.g., temperature in Celsius). • Ratio scale: A numerical scale that has all the properties of an interval scale but includes a true zero, allowing for meaningful ratios (e.g., height, mass). The critical distinction between interval and ratio scales is the concept of a true zero (also known as an absolute zero), which represents the complete absence of the quantity being measured.

1.01 Types of data mathspace.co

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1.01 Practice questions What do you remember? 1

Fill in the blanks with the terms from this list: • a b c d

2

3

Interval

•

Numerical

•

Categorical

•

Variable

Univariate data can be numerical or ⬚ data. We can find the mean of ⬚ data. Univariate means one ⬚.

There is no true zero on an ⬚ scale.

Determine whether each statement is true or false: a

A ratio scale has a true zero.

b

An interval scale is a special type of ratio scale.

c

Discrete data is a special type of numerical data.

d

Categorical data can be broken down into ordinal and interval data.

e

The mode of categorical data can be determined.

Identify whether these examples are numerical or categorical data: a

Favourite flavours

b

Maximum temperature

c

Daily temperature

d

Types of horses

e

Amount owed on lay-by

f

Types of vegetables

g

Daily UV index

h

Maximum rainfall

i

Favourite colours

j

Brands of phones

k

Types of dogs at the park

l

Maximum snowfall

Practice 4 Ex 1

5

10

Identify whether these data variables are discrete or continuous: a

The number of classrooms in a school

b

Daily humidity

c

The time taken to run 200 metres

d

Lengths of cats’ whiskers

e

The distance from the Earth to nearby galaxies

f

The number of people taller than you in each class

g

Marks on the most recent Physics test

h

Time it takes to fall asleep at night

i

The amount of water consumed in a day

Identify whether these are nominal or ordinal categorical data: a

Eye colour

b

Birthdate

c

Temperature

d

Favourite song

e

State/territory of birth

f

Year level

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Classify these data examples as one of these types: • Numerical discrete • Numerical continuous • Categorical ordinal • Categorical nominal

Ex 2

7

8 Ex 3

9

a

Types of trailers in a shopping centre car park

b

Heights of people at an athletics carnival

c

Weights of dogs

d

The languages spoken in a class

e

The number of languages students in a town speak

f

The number of people at an athletics carnival

g

The time spent playing games each day

h

The number of sports people play

Classify these data examples as one of these types:

a b c d

• Numerical continuous • Numerical discrete • Categorical The workplaces of people living in Sydney The body temperature of a hospital patient taken over a 21-hour period The brands of breakfast foods in a supermarket The weights of club members

e

The number of people attending a netball game

Identify the level of measurement that can be applied to these data as nominal, ordinal, interval or ratio: a

Number of pets in a household

b

pH of liquid

c

Make of car

d

Weights of eggs

e

Daily rainfall in a city

f

Military rank

g

Temperature measured in degrees Celsius

h

Letter grades on an assignment (A to F)

i

Year of birth

Identify the level of measurement that can be applied to the data generated in each situation as nominal, ordinal, interval or ratio: a

The time taken for a team of swimmers to complete one lap of a swimming pool is recorded by their coach.

b

The shoe sizes of shoe prints left at the scene of a crime are recorded by a detective.

c

The religion practised by a person is recorded during a census.

d

A driver is being rated by customers on a scale of 1 star to 5 stars.

1.01 Types of data mathspace.co

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10

Determine whether these statistics can be meaningfully computed, interpreted, and compared, from the data in each situation: i

Mode

iii a

c

Differences between outcomes iv Ratios between outcomes The temperature, in degrees Celsius, of the ocean at a particular location is monitored and fed back to a weather station. A group of people is surveyed and asked to indicate their attitude on a recent government policy by selecting a number from 1 to 5, where 1 = strongly disagree, 2 = disagree, 3 = neutral, 4 = agree and 5 = strongly agree. The flavour of ice cream selected by customers at a market stall is recorded over a day.

d

The weight of puppies at a puppy school is recorded each week.

b

Median

The graph shows the height of each student in Kate’s class:

Number of students

11

ii

10 9 8 7 6 5 4 3 2 1 0

155 160 165 170 175 180 185 190 195 200 Height (cm)

a

Is the variable ‘height of student’ continuous or discrete? Explain your answer.

b

Is the variable ‘number of students in a given height range’ continuous or discrete? Explain your answer.

Extend your thinking 12

Patients at a hospital are asked to give their pain a rating from 1 to 10. Vera says: “The pain ratings are interval data because they are numerical and have a scale going up in equal amounts.” Critique Vera’s reasoning and give a correct explanation for the level of measurement that can be applied to the pain rating data.

13

Recording the distance, in kilometres, between towns is supposed to produce continuous data. However, distances listed on road signs are always represented as whole numbers. Explain how this can be.

14

Explain in your own words why the amount of money spent on groceries is a discrete variable even though it involves decimal values.

15

Explain in your own words why postcode is a categorical variable while the number of houses in a postcode is a numerical variable.

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1.02   Categorical data displays After this lesson, you will be able to… • construct frequency tables and bar charts for categorical data. • interpret graphical displays of categorical data distributions. • identify and interpret the mode in context for categorical data distributions.

Categorical data and frequency tables Categorical data uses labels or words to describe observations. When the same category appears multiple times, data displays should be used to organise the data with the number of occurrences of each response. Frequency table Lists the frequency (number of occurrences) of observations. Frequency distribution The division of a set of observations into a number of classes, together with a listing of the number of observations (the frequency) in that class.

Tally

Frequency

12

Black

llll

9

10

12

8

White

II

Blue Green

ll llll

Silver

7 4

l

Frequency

Vehicle colour

6 4

6

2 0

Yellow

II

2

Red

Illl

4

Colours of vehicles passing through intersection

k

ac

Bl

te

W

hi

ue

Bl

en

e Gr

r

lve

Si

w

lo

l Ye

d Re

Vehicle colour Notice that the frequency table has three columns: • The first column shows the categories or subgroups within the data. In this case, the colours. • The tally column (optional) uses tally marks to record the frequency of each category. • The final column sums the tally marks and records the frequency as a number. The sum of the frequencies is equal to the total number of data values. In this case, the colours of 44 vehicles were recorded.

1.02 Categorical data displays mathspace.co

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Example 1 Mr. Rodriguez recorded the number of pets owned by each of the 45 students in band. He found that 15 people had no pets, 19 people had one pet, 3 people had two pets and 8 people had three pets. Write Mr. Rodriguez's results in the frequency table: Number of pets

Frequency

0

⬚

1 2 3

Create a strategy

⬚ ⬚ ⬚

‘Frequency’ means the number of people. Use the given information to fill this column in.

Apply the idea Number of pets

Frequency

0

15

1

19

2

3

3

8

Reflect and check Total students surveyed: 15 + 19 + 3 + 8 = 45 The sum of frequencies (45) matches the total number of students.

Idea summary Frequency tables organise categorical data and their frequencies, showing the frequency distribution. The sum of frequencies equals the total observations.

Column graphs and bar charts

14

Column graph

Mode

A graph used in statistics for organising and displaying categorical data.

The category with the highest frequency in a distribution.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Exploration

1. Which sport is the most popular? Explain how you know. 2. Assuming that 20 students preferred hockey, how many students preferred tennis? Explain how you know.

Number of students

A group of students were surveyed about their preferred sport, and the results are shown in this column graph: Preferred sport

Football Tennis

Rugby Basketball Hockey Sport

3. If 12 students preferred football, 14 preferred rugby, and 14 preferred basketball, how many students were surveyed in total?

Product F 0 1 2 3 4 5 6 7 8 9 10 Sales (thousands of units) Bar chart

Product F

Product E

Product E

Product D

Product D

Product C

Product C

Product B

10 9 8 7 6 5 4 3 2 1 0

Product B

Product A

Product A

Sales (thousands of units)

Bar charts and column graphs are both used to display the frequency of categorical data. Each category is represented by a separate bar. Column and bar graphs have: • Equal-width bars with gaps. • One axis represents the categories and the other axis represents the frequencies. • The length (height) of a bar corresponds to the frequency of that category.

Column graph

Column graph features to remember: • Title the graph clearly. • Label each axis. • Label each category and give the vertical axis a clear scale. • Make sure the columns are equal widths and equally spaced. • Make sure that the height of the column matches the value it represents, which is often the frequency of the category (how often that outcome occurred). The heights of the bars can be used to interpret and make conclusions from the data set.

1.02 Categorical data displays mathspace.co

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Example 2

Number of students

Miss Yen recorded her students’ favourite pets in this column graph: 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Students’ favourite pets

Hamster

Dog

Cat

Fish

Bird

Pets a Identify the mode and interpret it in context.

Create a strategy

Apply the idea

Find the tallest bar, which represents the highest frequency.

The bird category has the tallest column in the column graph, indicating that birds are the most popular pet.

b Determine the number of students who selected fish as their favourite pet.

Apply the idea The height of the fish column is 8. So, 8 students picked a fish as their favourite pet.

c How many students were surveyed in total?

Apply the idea Number of students = 6 + 7 + 4 + 8 + 11 = 36 The sum of the frequencies is 6 + 7 + 4 + 8 + 11 = 36. Therefore, 36 students were surveyed in total.

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Write the equation Evaluate

Example 3 This table represents the sizes of t-shirts ordered from an online store over the past week: T-shirt size

Frequency

Small

10

Medium

8

Large

10

Extra Large

12

Total

40

a Classify this categorical variable as ordinal or nominal.

Create a strategy If the categories can be ordered or ranked in a meaningful way, the variable is ordinal. Otherwise, the variable is nominal.

Apply the idea T-shirt sizes have a meaningful order because they reflect increasing measurements. Each size represents a different range of body measurements, allowing for a natural ranking from smallest to largest. Therefore, t-shirt size is an ordinal variable.

b Construct a column graph to represent the data.

Create a strategy Place the categories on the horizontal axis and the frequencies on the vertical axis. Draw columns above each category to represent the corresponding frequency. Add labels for both axes and include a descriptive title.

Frequency

Apply the idea 12 10 8 6 4 2 0

Size breakdown of shirts ordered

Small

Medium

Large Extra large

T-shirt size

1.02 Categorical data displays mathspace.co

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Idea summary Column graphs and bar charts are used for organising and displaying categorical data. The category with the highest frequency in a distribution is the mode, represented by the tallest bar. When creating a column graph: • Title the graph clearly. • Label each axis. • Label each category and give the vertical axis a clear scale. • Make sure the columns are equal widths and equally spaced. • Make sure that the height of the column matches the value it represents, which is often the frequency of the category (how often that outcome occurred). The heights of the bars can be used to interpret and make conclusions from the data set.

1.02 Practice questions What do you remember? 1

2

Is each statement true or false: a

Column graphs are used for continuous data.

b

Frequencies can be read off column graphs.

c

The same data set can be represented with a frequency table or a column graph.

d

The sum of the frequencies can be found from a frequency table.

e

The vertical axis of column graphs is unlabelled.

For each of these scenarios, identify what categories A, B, C and D represent in the corresponding frequency tables: a

b

18

Sophia asked 28 of her students to pick their favourite sport. 8 picked football, 15 picked volleyball, 5 picked basketball, and 5 picked tennis. Rochelle asked her students how they spent their weekend. 11 said beach, 13 said skiing, 6 said fun park, and 10 said camping.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Sport

A

B

C

D

Frequency

8

15

5

5

Activity

A

B

C

D

Frequency

11

13

6

10

Match the frequency tables with the bar charts.

iii

ii

Animal Frequency

Bird

1

Bird

2

Cat

3

Cat

3

Dog

2

Dog

1

Fish

4

Fish

4

Animal Frequency

iv

Animal Frequency

Bird

1

Bird

2

Cat

2

Cat

1

Dog

3

Dog

3

Fish

4

Fish

4

Animal

a

Animal Frequency

b

Fish

Fish

Dog

Dog

Animal

i

Cat

0

1

2 3 Frequency

4

0

1

2 3 Frequency

4

0

1

2 3 Frequency

4

d

Fish

Fish

Dog

Dog

Animal

c

Cat Bird

Bird

Animal

3

Cat

Cat Bird

Bird 0

1

2 3 Frequency

4

1.02 Categorical data displays mathspace.co

19

Practice 4

5

6 Ex 2

For each scenario, construct a frequency table for the results: a

Katrina asked 34 people to choose their favourite fruit. 10 people picked apples, 15 people picked bananas, 6 people picked grapes and 3 people picked pears.

b

Ben asked 35 people about how many siblings they have. He found that 12 people had no siblings, 15 people had one sibling, 3 people had two siblings and 5 people had three siblings.

c

Bob asked 41 of his students to choose their favourite toy. 7 students picked cars, 13 people picked planes, 6 people picked balls and 15 people picked video games.

d

Noah asked his students to choose their favourite method of travelling. 9 students picked car, 13 people picked plane, and 5 people picked boat.

In a survey, some people were asked to name their favourite sport. The frequency table containing the results is given: Sport

Soccer

Basketball

Football

Tennis

Volleyball

Other

Frequency

17

14

26

9

4

12

a

Determine the number of people who took part in the survey.

b

Determine the sport that most people rate as their favourite.

A survey of the preferred sport was done for a group of boys and the results are shown in the given column graph:

Number of boys

Ex 1

Preferred sport

10 9 8 7 6 5 4 3 2 1 0 Football

Tennis

Rugby

Basketball

Hockey

Sport

20

a

Identify the mode and interpret it in context.

b

Determine the number of students who selected football as their preferred sport.

c

How many students participated in the survey?

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

7

The column graph shows the number of cars by brand that are fixed by the local mechanic: 30

Frequency

25 20 15 10 5 0

BMW

Holden Car brand

Mercedes

a

Identify and interpret the mode of this data set.

b

Which car brands does the mechanic fix equally often?

Mitsubishi

Kathleen recorded the number of phone calls she received over four days. The results are displayed in the given column graph:

Frequency

8

Toyota

10 9 8 7 6 5 4 3 2 1 0

Tuesday

Wednesday

Thursday

Friday

Day a

What day did Kathleen receive the most calls?

b

Which days did she receive an equal number of phone calls?

c

Determine the fraction of total calls received on Wednesday.

1.02 Categorical data displays mathspace.co

21

Edward asked his students to choose their favourite animals. He plotted the results in this column graph:

Frequency

9

10 9 8 7 6 5 4 3 2 1 0

Cats

Birds

Dogs

Hamster

Animal

10 Ex 3

a

Identify and interpret the mode of this data set.

b

Which animals were equally popular?

c

What fraction of the total students prefer cats?

A company conducted a customer satisfaction survey to understand how clients feel about their new product: a b

11

12

22

Satisfaction level

Frequency

Very unsatisfied

3

Classify this categorical variable as ordinal or nominal.

Unsatisfied

7

Neutral

12

Construct a bar chart to represent the data.

Satisfied

15

Very satisfied

8

Total

45

Amelia counted the cookbooks on her grandparents’ shelf and recorded the results in this table: a

Classify this categorical variable as ordinal or nominal.

b

Construct a bar chart to represent the data.

Type of cookbook

Number of cookbooks

Baking

2

Italian

3

Quick meals

2

Gourmet

4

Total

11

Laura asked some of her friends about their favourite colours and presented the results in the frequency table: Colour

Black

Brown

Blue

Red

Yellow

Frequency

3

10

12

5

15

a

Determine the number of people who took part in the survey.

b

Determine the least favourite colour.

c

Construct a column graph using the results from the table.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

13

A restaurant owner recorded how many employees arrived early, on time or late. The frequency table for the results is given: Early

On-time

Late

10

22

7

Frequency

14

15

a

Determine the number of employees that the restaurant owner surveyed in total.

b

When did the majority of employees arrive?

c

Construct a column graph using the results from the table.

In a survey some people were asked approximately how many minutes they take to decide between brands of a particular product. The frequency table with the results is given: Minutes taken

1

2

3

Frequency

13

17

12

a

Determine the number of people that took part in the survey.

b

Determine the number of minutes that most people take to decide between brands.

c

Construct the column graph using the results from the table.

For each of these scenarios: i

Construct a frequency table of the results.

ii

Create a column graph that represents the information in the frequency table.

a

Sandy asked 12 of her friends where they went on holiday last year. Their responses were: beach, camping, caravan, beach, snow, caravan, caravan, camping, beach, beach, caravan, caravan

b

In a survey, 20 people were asked how many languages they spoke. Their responses were: 1, 1, 1, 2, 3, 2, 3, 1, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1

Extend your thinking 16

The set of marks for a class of students is given: 81, 81, 71, 81, 81, 61, 71, 93, 71, 58, 71, 58, 61, 93, 93, 71, 61, 61, 81, 58, 93 a

Organise the data into a frequency table.

b

Determine the number of students in the class.

c

Determine the number of students that will get a Distinction grade (80 < Score ≤ 90).

d

Determine the number of students that will get a High Distinction grade (Score > 90).

e

Determine the percentage of students getting a High Distinction grade. Round your answer to two decimal places.

1.02 Categorical data displays mathspace.co

23

17

The table shows the number of people who visited Disneyland between 2008 and 2012: Year

2008

2009

2010

2011

2012

Number of people (in hundred thousands)

158

155

155

157

160

House

Yellow

Blue

Red

Green

Points

15

25

40

60

Construct a column graph for the data. 18

19

The table shows the house points earned by four different houses at their swimming carnival: a

Construct a column graph for the data.

b

Which house earned the most points?

c

Which house earned as many points as the Yellow house and Blue house combined?

The sales of different products are shown in this horizontal bar chart: Product A Product B Product C Product D Product E Product F 0

24

2

4 6 8 Sales (thousands of units)

10

a

What was the best-selling product?

b

Determine the total number of products sold.

c

Determine the revenue generated by Product B alone, given that Product B was sold at $50 each.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

For these three column graphs:

0

Tom

John Name

Kate

Is the tallest person the oldest? Explain your reasoning.

b

Is the heaviest person the oldest? Explain your reasoning.

c

Is the heaviest person the tallest? Explain your reasoning.

d

Is the youngest person the lightest? Explain your reasoning.

e

Is it possible for a 10-year old to be lighter than an 8-year old based on the data displayed? Give your reasoning.

50 45 40 35 30 25 20 15 10 5 0

Mass

Tom

John Name

Kate

Height

140 120 Centimetres

a

Age

10 9 8 7 6 5 4 3 2 1

Kilograms

Years

20

100 80 60 40 20 0

Tom

John Name

Kate

1.02 Categorical data displays mathspace.co

25

1.03   Numerical data displays After this lesson, you will be able to… • construct histograms, stem plots, and dot plots for numerical data. • select the most appropriate graphical representation (histograms, stem plots, dot plots) based on context and purpose. • interpret numerical data distributions using various graphical displays.

Histograms and grouped data Histograms display the frequency distribution of continuous numerical data, like time or height, using adjacent bars. Histogram A graph with no gaps between bars, showing the frequency of numerical data in class intervals.

Although a histogram looks similar to a bar chart, there are several important differences between them: Histogram

Column graph

Shows distribution of data values

Show and compare frequencies of categories

Used for numerical data

Used for primarily categorical data

Numerical scales on both axes

Numerical scale only on vertical axis

Columns cannot be reordered

Can reorder columns and retain meaning

Distribution of running times for a 10 km race

12

35

10 Frequency

25 20 15 10

6 4 2

5

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

hi te Bl ue G re en Si lv e Ye r llo w Re d

W

re

45 50 55 60 65 70 Running time (minutes)

en

0

Histogram

26

8

G

Frequency

30

0

Colours of vehicles passing through intersection

Vehicle colour Column graph

Class interval

Frequency

45 ≤ Time < 50

9

50 ≤ Time < 55

7

55 ≤ Time < 60

20

60 ≤ Time < 65

30

65 ≤ Time < 70

6

Total

72

This frequency table organises the data that is represented in the “Distribution of running times for a 10 kilometres race” histogram. Notice that the histogram has only five columns or grouped intervals, even though it represents 72 different data values.

Key features of a frequency histogram: • Continuous numerical scale on the horizontal axis (class intervals). • Data is grouped into equal-width class intervals. • Frequency on the vertical axis. • No gaps between bars and about half-column gap at the start. For class intervals with frequency zero, the gap should remain. • Technology can be used to verify or create histograms. There are several different ways that class intervals are defined. Here are some examples with two adjacent class intervals: Class interval formats

Frequency

45 ≤ Time < 50 0 ≤ Time < 55

Lower endpoint included, upper endpoint excluded. This is the most common approach.

45 < Time ≤ 50 0 < Time ≤ 55

Upper endpoint included, lower endpoint excluded.

45 to < 50

45 to < 50

Lower endpoint included, upper endpoint excluded.

45 - 49

50 - 54

Suitable for data rounded to the nearest whole number, or discrete data.

45 → 50

50 → 55

Not clear which endpoints are included or excluded. Assume the lower endpoint is included.

Regardless of the format used, each class interval for a given set of data should be consistent across all class intervals to avoid double counting any values in two intervals.

1.03 Numerical data displays mathspace.co

27

Exploration This histogram summarises numerical univariate data. 6 5 4 3 2 1 0

0

200

400

600

800

1000

1. The modal class can be said to be 100 – < 200. What is meant by this statement? 2. One data value is 842, what could that value be called? 3. One of the measures of centre is 199. Which measure of centre would this be? Explain. 4. One of the measures of centre is 114. Which measure of centre would this be? Explain.

Example 1 As part of a fuel watch initiative, the price of petrol at a service station was recorded each day for 21 days. The frequency table shows the findings: Price (cents per litre)

Class centre

Frequency

130.0 - 134.9

132.5

6

135.0 - 139.9

137.5

5

140.0 - 144.9

142.5

3

145.0 - 149.9

147.5

6

a What was the highest price that could have been recorded?

Create a strategy Look for the maximum value in the highest class interval.

Apply the idea Based on the table, the highest price that could have been recorded is 149.9 cents as this is the maximum value in the highest class interval.

28

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

b How many days was the price at least 140.0 cents?

Create a strategy Find the sum of the frequencies from the appropriate classes.

Apply the idea The prices were observed to be above 140.0 in the last two classes: 140.0 - 144.9 and 145.0 - 149.9. So, the frequencies of those two classes should be added. Days = 3 + 6 =9

Add the frequencies Evaluate

c Construct a histogram to show the data distribution.

Create a strategy The data is already in a grouped frequency table, so the class intervals (bins) are already determined. The horizontal axis should go from 130 to 150 going up by 5s. The vertical axis can go up by 1s up to 7.

Apply the idea

Reflect and check If class centres are used, it would look like this: Distribution of gas prices 6

5

5 Frequency

Frequency

Distribution of gas prices 6

4 3 2

3 2 1

1 0

4

130 135 140 145 150 Price of gas (cents per litre)

0

132.5 137.5 142.5 147.5 Price of gas (cents per litre)

1.03 Numerical data displays mathspace.co

29

Idea summary For grouped data: • Every data value must go into exactly one and only one class interval. Class intervals should be equal in width. • Each class interval must be the same size, for example: 1 – < 5, 5 – < 10, 10 – < 15, . . . Key features of a frequency histogram: • • • •

Continuous numerical scale on the horizontal axis (class intervals). Data is grouped into equal-width class intervals. Frequency on the vertical axis. No gaps between bars and about half-column gap at the start. For class intervals with frequency zero, the gap should remain. • Technology can be used to verify or create histograms.

Dot plots Dot plot A data display for numerical or categorical data with a horizontal axis with dots representing the frequency of data values.

Dot plots are best used for small to mediumsized sets of data and are good for visually highlighting how the data is spread and whether there are any gaps in the data or outliers. Each dot represents one occurrence (e.g., 5 dots at 2 days). Ideal for small data sets to show spread and gaps. 0 1 2 3 4 5 6 7 Days per week spent exercising Key features: • Labelled and evenly spaced horizontal axis • No vertical axis • Evenly spaced dots stacked above the numerical scale to show the frequency

30

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Example 2 Here is a dot plot of the number of goals scored in each of Bob’s soccer games:

0

1

2 3 4 Number of goals

5

a How many times was one goal scored?

Create a strategy

Apply the idea

Count the number of dots above the number 1.

There are 7 dots above the number 1, so 1 goal was scored 7 times.

b Identify the number of goals that were scored with the highest and equal frequency.

Create a strategy Look for the scores that have the same number of dots, and the most dots.

Apply the idea On the dot plot, there are two scores that have the same number of dots and also the most dots. These are 0 and 3. So, the correct answers are 0 and 3.

c How many games were played in total?

Create a strategy Count all the dots in the graph.

Apply the idea Add the number of dots in each column. Number of games = 10 + 10 + 5 + 7 + 6 + 3 = 41

Write the equation Evaluate

1.03 Numerical data displays mathspace.co

31

Idea summary Dot plots show numerical data with dots for each value, best for small sets to highlight distribution and gaps. Note: Technology can be used to construct dot plots of numerical data.

Stem plots Stem plot A display for numerical data that involves splitting numbers into a stem (leading digits) and leaf (last digit). Sometimes called a stem-and-leaf plot.

21

163

10.5

stem leaf

stem leaf

stem leaf

Stem plots are best used for small to medium-sized numerical data sets where values are more than one digit.

Score of 10

Score of 56

Stem

Leaf

1

036

2

1 678

3

556

4

1 1 569

5

All of 0 3 6 8 these are

Features to note: • Stems are listed in consecutive, ascending order. • Leaves are placed next to their corresponding stems, also in ascending order. • Leaves are aligned in imaginary columns. • A key is provided to explain how the stems and leaves combine to form each data value.

in the 40s Key 2|1 = 21

The given stem plot represents the raw data set: 10, 13, 16, 21, 26, 27, 28, 35, 35, 36, 41, 41, 45, 46, 49, 50, 53, 56, 58

32

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Example 3 The given stem plot shows the age of people to enter through the gates of a concert in the first 5 seconds: Stem

Leaf

1

1 26899

2

02345778

3

2247

4 5

9 Key 1|2 = 12

a How many people passed through the gates in the first 5 seconds?

Create a strategy Count the number of leaves.

Apply the idea The number of people surveyed can be found by counting the number of leaves in the stem plot, since each data entry is represented by one leaf. There are 6 people aged 10 to 19, 8 people in their 20s, 4 people in their 30s, and 1 person in their 50s. So, the total number of people can be found by adding all these numbers. Total number of people = 6 + 8 + 4 + 1 = 19

Write the equation Evaluate

b What was the age of the youngest person?

Create a strategy Find the smallest number recorded.

Apply the idea The smallest number will be in the smallest stem, which is 1, and have the smallest leaf, which is also 1. This stem and leaf make the number 11. The youngest person is 11 years old.

1.03 Numerical data displays mathspace.co

33

c What was the age of the oldest person?

Create a strategy Find the largest number recorded.

Apply the idea The largest number will be in the largest stem which is 5, and have the largest leaf which is 9. This stem and leaf make the number 59. The oldest person is 59 years old.

d What proportion of the concert-goers were under 24 years old?

Create a strategy Divide the number of people whose ages are less than 24 by the total number of people.

Apply the idea There are 9 people aged less than 24 years old. Since the total number of people is 19, then the proportion of concert-goers is

9 . 19

Idea summary Stem

Leaf

1

036

2

1 678

3

556

4

1   1 5 6 9

5

0368

Key 2|1 = 21

34

A stem plot, sometimes called a stem-and-leaf plot, is used for organising and displaying numerical data. It is appropriate for small to moderately sized data sets. An advantage of a stem plot is the individual scores can be seen. The right-most digit in each data value is split from the other digits, to become the leaf. The remaining digits become the stem.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

1.03 Practice questions What do you remember? 1

2

3

Describe a data set that is suited to each data display: a

An ungrouped frequency table

b

A grouped frequency table

c

A stem plot

d

A dot plot

e

A histogram

Determine whether the data should be displayed in a frequency table with or without class intervals, and explain your reasoning: a

A researcher is studying the income distribution of a large population.

b

A survey collects data on individuals’ preferred hobbies.

c

A teacher surveys a class of 30 students to determine how many books each student read over the summer.

d

A teacher wants to analyse exam scores for a large class.

e

A community centre collects data on the number of pets owned by families in the neighbourhood.

If the range of a set of scores is 48, determine the appropriate group size to group the data into each number of groups: a

4

5

3 groups

b

8 groups

Determine whether each statement is true or false about an ordered stem plot: a

The scores are ordered.

b

A stem plot does not give an idea of outliers and clusters.

c

It is only appropriate for data where scores have high frequencies.

d

The individual scores can be read on a stem plot.

20 people were asked how many hours of sleep they had gotten the previous night. The results are shown: 1, 6, 9, 8, 7, 9, 7, 10, 2, 3, 8, 7, 7, 3, 7, 3, 3, 7, 10, 9 a

How many people got 7 hours sleep?

b

What is the maximum amount of sleep reported by the group?

1.03 Numerical data displays mathspace.co

35

6

The table shows the number of trains arriving either on time or late at a particular station: a

How many trains were late on Friday?

b

How many trains passed through the station on Wednesday?

c

How many trains were on time throughout the entire week?

d

What fraction of trains were on time over the whole week?

Day

Arriving on Arriving time late

Monday

23

38

Tuesday

12

25

Wednesday

22

29

Thursday

30

27

Friday

11

14

Saturday

15

11

Sunday

14

10

Practice 7

Ex 1

8

9

Construct a grouped frequency table for these data: a

46, 54, 35, 23, 24, 28, 26, 11, 19, 17, 32

b

83, 68, 39, 42, 86, 66, 64, 76, 63, 43, 65, 83, 63, 67, 49, 51, 32, 55, 38, 65, 41, 73, 35, 36, 74

A group of Year 11 students were surveyed on their height, and the results are shown in the frequency table: Height (cm)

Frequency

135 ≤ h < 145

4

145 ≤ h < 155

12

155 ≤ h < 165

13

165 ≤ h < 175

8

175 ≤ h < 185

2

185 ≤ h < 195

1

Total

40

a

What was the tallest a person in this survey could be?

b

How many people were at least 165 cm tall?

c

Construct a histogram to represent this data.

Consider this set of scores: 12, 59, 61, 27, 58, 18, 76, 27, 52, 19, 13, 56, 71, 31, 73, 60, 41, 17, 22, 68, 57, 15, 40, 19, 76, 44, 60, 55, 36

36

a

Determine the set of seven class intervals that should be used to organise this data.

b

Construct a grouped frequency table for the data.

c

Construct a histogram to represent the data.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

The histogram shows the number of hours that students in a particular class had slept for the night before:

Frequency

10

12 11 10 9 8 7 6 5 4 3 2 1 0

0

2

4

6 8 Time (hours)

a

Complete the frequency table for this set of data.

b

How many students are in the class?

c

How many students had at least 8 hours of sleep that night?

d

How many students have less than 6 hours of sleep?

e

What percentage of students had less than 6 hours of sleep?

10

14

12

Number of hours

Frequency

0≤t<2

⬚

2≤t<4

⬚

4≤t<6

⬚

6≤t<8

⬚

8 ≤ t < 10

⬚

10 ≤ t < 12

⬚ ⬚

Total 11

This frequency table shows the resting heart rate of people taking part in a study: a

How many people took part in the study?

b

How many people had a resting heart rate between 55 and 74?

c

12

How many people had a resting heart rate of below 65?

For the given frequency table: a

Complete the cumulative frequency column.

b

Calculate the total frequency.

c

Identify the class size.

d

Determine the value such that approximately half of the scores recorded are greater than.

Heart rate

Frequency

45 − 54

15

55 − 64

25

65 − 74

26

75 − 84

30

x

f

1→5

16

6 → 10

28

11 → 15

19

16 → 20

17

21 → 25

5

26 → 30

3

1.03 Numerical data displays mathspace.co

cf ⬚

⬚ ⬚ ⬚

⬚ ⬚ 37

13

Construct a dot plot that represents this data: 49, 49, 49, 50, 48, 47, 50, 49, 51, 50, 51, 49, 52, 47, 50, 50, 48, 51, 51, 50, 48, 49, 52, 49, 47, 47, 52, 52, 49, 47, 52, 48, 51, 50, 49, 52, 48, 48, 52, 50, 48, 47, 52, 52, 52, 48, 48, 49

14

The goals scored by a football team in their matches are represented in this dot plot. Construct the frequency distribution table for this data.

0 15

1

16

17

Ex 3

18

38

5

Consider this dot plot: a

Construct the frequency distribution table for this data.

b

How many students scored above 20?

c

How many students scored below 30? 10

Ex 2

2 3 4 Goals scored

20 30 40 50 Score

Christa is a casual nurse. She used a dot plot to keep track of the number of shifts she did each week for several weeks: a

Over how many weeks did Christa record her shifts?

b

For how many weeks did she work 5 shifts?

c

How many weeks did she work less than 6 shifts?

d

If Christa works at least 6 shifts a week, she buys a weekly train ticket. What fraction of the time did she buy a weekly train ticket?

3

4 5 6 7 Shifts per week

8

Construct a stem plot for these data sets: a

11, 12, 13, 13, 14, 14, 16, 16, 17, 20, 22, 24, 29, 29, 32, 35, 35, 35, 53

b

24, 26, 25, 37, 36, 37, 37, 38, 41, 46, 49, 56, 67, 63, 67, 68, 69, 75, 80, 80

c

9.2, 4.3, 9.8, 7.8, 8.9, 0.7, 1.8, 4.9, 6.7, 3.4, 2.0, 8.1, 2.2, 7.9, 6.7, 8.2, 9.0, 9.0, 6.8, 2.0

d

142, 173, 121, 157, 149, 126, 137, 162, 118, 152, 130, 115, 144, 176, 123, 138, 167, 119, 146, 175

The scores for a recent spelling test are shown in the stem plot. The maximum possible score on the test was 100: a

How many students took the test?

b

What was the highest score on the test?

c

What test score occurred the most frequently?

d

How many marks separate the highest score and the most frequent score?

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Stem     Leaf 6

233667

7

0 1 1   1 1   4 5 8

8

1 1 23456999

9

4579 Key 7|6 = 76

19

Consider the stem plot:

Stem Leaf

Construct a frequency table for the data.

1

1 237

2

347779

3

279

4

0 1 156

5

26

Key 1|2 = 12 20

Determine the best display for each data set. Justify your answer. a The heights of every student at a school with over 1000 students b The number of goals scored in 15 football (soccer) games c

The ages of 40 people on a tram

Extend your thinking 21

The dot plot shows the rounded maximum temperature (°C) in a town over several weeks: Temperature in a town

21

23 25 27 29 31

33 35

Explain the features of the dot plot that make it a better display than a stem plot. 22

A principal wants to investigate the performance of students at his school studying Engineering. To do this, he has the marks of each student collected into groups and put into a frequency table. Each group of marks is assigned a grade as shown:

a b c d

Grade

x

f

cf

E

0 ≤ x < 20

7

D

20 ≤ x < 40

16

C

40 ≤ x < 60

35

⬚

B

60 ≤ x < 80

102

A

80 ≤ x < 100

72

⬚ ⬚ ⬚ ⬚

Complete the cumulative frequency column. Determine the total frequency. Identify the class size. Determine the number that approximately three quarters of the scores are greater than.

1.03 Numerical data displays mathspace.co

39

23

Here are the luggage weights of 15 passengers, rounded up to the nearest kilogram: 20, 21, 18, 17, 19, 22, 20, 17, 22, 16, 21, 21, 18, 21, 20 a

Complete the frequency table.

b

How many times did someone check-in luggage that weighed more than 19 kilograms?

c

There is a baggage restriction of 20 kilograms and the airline charges each passenger $1.50 for every kilogram in their luggage above this restriction.

Weight (kg)

Frequency

15 < x ≤ 16

⬚

16 < x ≤ 17 17 < x ≤ 18 18 < x ≤ 19 19 < x ≤ 20

Calculate the total amount charged for excess baggage.

20 < x ≤ 21 21 < x ≤ 22

24

The speed of passing cars on a stretch of highway are recorded by a police officer. The results are shown in the first table:

40

Speed (km/h)

Frequency

80 < x ≤ 90

3

90 < x ≤ 100

55

100 < x < 110

10

110 ≤ x < 120

6

120 ≤ x < 130

0

130 ≤ x < 140

1

a

How many cars did the police officer observe?

b

The stretch of highway that the police officer was observing had a speed limit of 100 km/h. How many drivers were recorded to be speeding?

c

Use the information in the second table, to determine: Speed over the limit

Fine

Demerit points

x < 10

165

2

10 ≤ x < 20

350

3

20 ≤ x < 30

730

5

30 ≤ x < 45

870

7

i

The total value of fines to be issued

ii

The total number of demerit points to be dealt out

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

⬚ ⬚ ⬚ ⬚

⬚ ⬚

25

A car manufacturing company records the number of cars produced each day over a period of one month. The data collected is as follows: 10

Frequency

8 6 4 2 0

0

10 20 30 40 Number of cars

50

If the goal for the month was to produce at least 800 cars, did the company meet its target? Assume the company manufactured an average number of cars within each interval. 26

The level of mercury in 40 fishing waters was tested and recorded. The results are given in this stem plot: Stem

Leaf

9

1 3 5 5 6

10

1 2 4 4 5 5 6 8 9 9

11

1   1 2 3 3 3 5 6 7 7 7 8 8 8 8 9

12

222223469 Key 1|2 = 12 units

a

What was the second-highest reading?

b

In how many places was a reading of 117 or higher recorded?

c

What was the 11 th lowest score?

d

If the safe level of mercury is recommended to be 109 or lower, what percentage of the places have safe levels of mercury? Round your answer to one decimal place.

1.03 Numerical data displays mathspace.co

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1.04   Measures of centre After this lesson, you will be able to… • calculate measures of centre (mean and median) from data sets and data displays. • use technology to calculate and verify the mean and median of large data sets. • interpret measures of centre in context of the data. • compare measures of centre to analyse data distributions.

Mean Mean A measure of centre often referred to as the average. To calculate the mean, add all the scores in a data set, then divide this by the number of scores. The mean of a data set, denoted by x, is given by:

x

x n

Σx is the sum of the values n is the total number (count) of values in the data set Technology can be utilised to calculate the mean. The mean can also be found from data displays. For dot plots and stem plots, the raw data can be read off the display and used to calculate the mean. For histograms, the mean can only be estimated because the raw data cannot be determined. To find the mean from a histogram: 1. Write the information in a grouped frequency table. 2. Calculate the cumulative (total) frequency by adding up all of the frequencies. 3. Place two columns to the end of the table called ‘Class centre’ and ‘Class centre × frequency’. 4. Fill in the ‘Class centre’ column by finding the average of the two endpoints for that row. 5. Fill in the ‘Class centre × Frequency’ column by multiplying the class centre and frequency. 6. Add up the ‘Class centre × Frequency’ column. 7. Produce an estimated mean (average) by dividing the sum of the ‘Class centre × Frequency’ column by the total frequency.

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Class interval

Frequency

Class centre

Class centre × Frequency

45 ≤ time < 50

9

47.5

9 × 47.5 = 427.5

50 ≤ time < 55

7

52.5

7 × 52.5 = 367.5

55 ≤ time < 60

20

57.5

20 × 57.5 = 1150

60 ≤ time < 65

30

62.5

30 × 62.5 = 1875

65 ≤ time < 70

6

67.5

6 × 67.5 = 405

n = 72 Mean x

Σx = 4225 x n

4225 72

58.68

Did you know?

Measures of centre, like the mean, median, and mode, are crucial for understanding data! For instance, in sports statistics, the mean can show the average points scored by a player over a season, while the median can give a better sense of a player's consistent performance by eliminating outliers from extreme games. These measures are also widely used in education to understand student achievement across different assessments. No matter the field, measures of centre provide valuable insights that support smarter decision-making.

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Example 1 The histogram shows the speed of balls delivered by cricket bowlers: 16 14 Frequency

12 10 8 6 4 2 0

100

110

120

130 140 Speed (km/h)

150

160

170

a Complete the table: Speed (km/h)

Frequency

Class centre

Class centre × Frequency

110 ≤ x < 120

4

120 ≤ x < 130

⬚

⬚

⬚

⬚

⬚

130 ≤ x < 140 140 ≤ x < 150 150 ≤ x < 160 Total

Create a strategy

⬚

⬚

135

⬚ ⬚

⬚

⬚

⬚ ⬚ ⬚ ⬚

The frequencies can be read off from the height of the bars. The class centre can be found by calculating the average of the endpoints for that class interval.

Apply the idea

44

Speed (km/h)

Frequency

Class centre

Class centre × Frequency

110 ≤ x < 120

4

115

4 × 115 = 460

120 ≤ x < 130

12

125

12 × 125 = 1500

130 ≤ x < 140

8

135

8 × 135 = 1080

140 ≤ x < 150

5

145

5 × 145 = 580

150 ≤ x < 160

2

155

2 × 155 = 310

Total

n = 31

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Σx = 3930

b Determine the mean speed of the cricket balls, rounded to two decimal places.

Create a strategy The formula is x x , all of the needed information can be found in the table. n

Apply the idea

x x n

= 126.77

Write the formula

3930 31

Substitute the values for Σx and n Evaluate and round

The mean speed of balls delivered by these 31 cricket bowlers is 126.77 km/h.

Reflect and check This is reasonable because the tallest bar is 120 ≤ x < 130 and this falls in that range.

Example 2 The mean of 4 scores is 21. If three of the scores are 17, 3, and 8, find the 4th score, x.

Create a strategy Use the formula: x

x n

Apply the idea The given values are x = 21, Σx = 17 + 3 + 8 + x, and n = 4. x n

Write the formula

21

17 3 8 x 4

Substitute the values

21

28 x 4

Perform the addition

x

84 = 28 + x

Multiply both sides by 4

x = 56

Subtract 28 from both sides

1.04 Measures of centre mathspace.co

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Idea summary The mean, x, is a measure of centre often referred to as the average. It is calculated by adding all the scores in a data set and dividing by the number of scores. This is represented as:

x

x n

Σ x is the sum of the values n is the total number (count) of values in the data set Technology can be utilised to calculate the mean.

Median Median The middle score in a data set when the data is sorted in order. To find the median of a data set: 1. Sort the data points into order.

n 1 th score). 2 3. If there are an even number of scores, the median is calculated by finding the mean of the n n 1 th scores). two middle scores (the th and 2 2 2. If there are an odd number of scores, the median is the middle score (the

Sets ordered smallest to largest Median Odd number of scores:

Even number of scores:

Median:

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

2

Example 3 Determine the median of these nine numbers: 1, 1, 3, 5, 7, 9, 9, 10, 15

Create a strategy The median in an odd set of scores is the

n 1 th score, where n is the total number of scores. 2

Apply the idea There are 9 scores. Median position =

n 1 2

Write the formula

=

9 1 2

Substitute the values

10 2 = 5th score =

Evaluate the numerator Simplify

Therefore, the median will be the 5th score. Median = 7

Reflect and check This data set had an odd number of scores, so the median was just the middle score, as it was ordered from smallest to largest.

Example 4 Determine the median from the frequency distribution table: Score (x)

Frequency ( f )

23

2

24

26

25

37

26

24

27

25

Create a strategy Add a cumulative frequency column to the frequency table to determine the final cumulative frequency (total data points). Use the formula to find the median position. Find the median by locating the cumulative frequency that meets or exceeds this position.

1.04 Measures of centre mathspace.co

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Apply the idea Create a cumulative frequency column. Calculate the cumulative frequency by adding each frequency to the sum of all previous frequencies, row by row. Score

Frequency

Cumulative frequency

23

2

2

24

26

2 + 26 = 28

25

37

28 + 37 = 65

26

24

65 + 24 = 89

27

25

89 + 25 = 114

Since there are 114 scores, the median will be in the of the 57th and 58th values.

114 + 1 = 57.5th position, which is the average 2

Based on the table: • The first and second values had a score of 23. • The third to 28th values had a score of 24. • The 29th to 65th values had a score of 25. Therefore, the 57th and 58th values are both 25. Median =

25 25 25 2

Idea summary To find the median of a data set: 1. Sort the data points into order. 2. If there are an odd number of scores, the median is the middle score n 1 (the th score). 2 3. If there are an even number of scores, the median is calculated by finding n n 1 th scores). the mean of the two middle scores (the th and 2 2 To find the median using a frequency table: 1. Add a cumulative frequency column by summing frequencies row by row. 2. Find the total number of data points (final cumulative frequency, n). 3. Calculate the median position using the formula. 4. Locate the cumulative frequency that equals or exceeds this position. 5. Identify the corresponding data value or class interval as the median. Technology can be utilised to calculate the median of small or large data sets.

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Mode Mode The most frequently occurring score. A data set can have more than one mode, if two or more scores are equally tied as the most frequently occurring. For grouped data, it can be called a modal class and the class interval with the highest frequency is the mode.

Example 5 Determine the mode of these scores: 8, 18, 5, 2, 2, 10, 8, 5, 14, 14, 8, 8, 10, 18, 14, 5

Create a strategy Choose which number in the set appeared the most.

Apply the idea The number can be listed along with how many times it appears: • 8 appeared in the set 4 times. • 18 appeared in the set 2 times. • 5 appeared in the set 3 times. • 2 appeared in the set 2 times. • 10 appeared in the set 2 times. • 14 appeared in the set 3 times. Mode = 8

Idea summary The mode of a data set is the most frequently occurring score. The modal class is the class interval with the highest frequency for grouped data.

1.04 Measures of centre mathspace.co

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1.04 Practice questions What do you remember? 1

2

3

For these sets of scores: i

Determine the number of scores.

ii

Calculate the sum of the scores.

iii

Calculate the mean.

a

9, 5, 2, 3, 1

b

3, 2, 5, 1, 4

c

1, 3, 7, 5, 2

d

4, 5, 6, 7, 3

e

56, 89, 95, 71, 75, 84, 65, 83

f

−14, 0, −2, −18, −8, 0, −15, −1

ii

Determine the median.

For each set of scores: i

Sort the scores in ascending order.

a

17, 8, 6, 11, 9, 18

b

69.4, 66.7, 46.6, 76.6, 52.8, 80.6, 63.9

c

44.9, 45.6, –54.8, 74.7, –77.6, –42.6, 67.9, 40.6

A consumer group surveys the price of petrol per litre at six different petrol stations. The prices are as follows: $1.68, $1.64, $1.64, $1.71, $1.64, $1.57 Identify the mode petrol price.

Practice 4

Consider the given histogram: 6 5 Frequency

Ex 1

4 3 2 1 0

50

39

40

41 42 43 Scores

a

Determine the total number of scores.

b

Calculate the sum of the scores.

c

Calculate the mean, rounded to two decimal places.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

5

For each data set, use the midpoint of the interval to estimate the mean. Round your answer to one decimal place. a

6

Score Frequency

b

Score

Frequency

0 ≤ x < 20

4

20 − 24

5

20 ≤ x < 40

15

25 − 29

10

40 ≤ x < 60

23

30 − 34

16

60 ≤ x < 80

73

35 − 39

4

80 ≤ x < 100

45

40 − 44

3

45 − 49

2

50 − 54

1

A group of high school students wanted to convince their principal that the school needed air-conditioning. They measured the temperature in a classroom at 1 p.m. every day during February and summarised the results (in °C) in the grouped frequency table: a

Complete this frequency table: Class

Class centre, cc

Frequency, f

f × cc

22 − 24

23

5

25 − 27

26

6

28 − 30

29

8

⬚

31 − 33

32

5

34 − 36

35

4

⬚ ⬚ ⬚ ⬚ ⬚

Total:

7

b

Determine the modal class.

c

Calculate the average temperature, rounded to two decimal places.

Consider the frequency distribution table: a

Complete the table.

b

Determine the mean of the scores, rounded to two decimal places.

c

Identify the mode of the scores.

d

How many scores are less than the mode?

Frequency, f

4

11

5

⬚

35

43

365

⬚ 14

Total:

Ex 2

fx

Score, x

16

8

The mean of a set of scores is 35 and the sum of the scores is 560. Calculate the number of scores.

9

There are five numbers with a mean of 7. If four of the numbers are 10, 10, 8, 6 and x, determine the value of x.

10

The five numbers 16, 16, 17, 24 and 17 have a mean of 18. If a new number is added that is bigger than 24, determine whether the mean will become higher or lower.

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11

In each game of the season, a basketball team recorded the number of ‘three-point shots’ they scored. The results for the season are represented in the given dot plot: a

What was the total number of points scored from three-point shots during the season?

b

Considering the total number of points, what was the mean number of points scored each game? Round your answer to two decimal places.

c

0 1 2 3 4 5 6 Number of three-point shots

What was the mean number of three point shots per game this season? Round your answer to two decimal places.

12

A set of 69 scores is arranged in ascending order. In what position does the median score lie?

13

In a set of 152 scores between which two scores does the median lie?

14

Determine the median of each data set: 920, 709, 694, 667, 657, 630, 610, a 584, 567, 555

Ex 3, 4

Score

Frequency

11

24

12

9

13

21

14

9

15

6

For the given histogram: a

Determine the number of scores.

b

Identify the median.

10 8 Frequency

15

b

6 4 2 0

16

52

46

47

48 49 Scores

50

On Sunday, 20 planes were delayed at the airport. The dot plot shows the number of hours each departure was delayed: a

What was the median number of minutes a plane was delayed?

b

What percentage of planes were delayed for longer than the median time?

c

If a plane is delayed for more than 30 minutes, the airline must pay $5000. In total, how much were airlines fined that day?

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

0.5

1.0 1.5 2.0 Hours late

2.5

17

This stem plot shows the ages of 20 employees in a company: a

What is the mode age?

b

What is the median age of the employees?

c

What is the mean age of the employees?

d

How old is a typical employee at this company? Explain your answer.

Stem

Leaf

2

0356779

3

02227

4

44578

5

237

Key 1|2 = 12 years old

Ex 5

19

A diver measures how long she can hold her breath underwater over several dives. If the median time is 3.9 minutes, determine whether these statements are true: a

The longest she held her breath is 7.8 minutes.

b

Most of the time she held her breath for less than 3.9 minutes.

c

Half the dives she was able to hold her breath longer than 3.9 minutes.

d

The shortest time she held her breath is 1.95 minutes.

e

Most of the time she held her breath for longer than 3.9 minutes.

Determine the mode(s) of each data set: a

c

2, 2, 6, 8, 8, 8, 8, 12, 14, 14, 14, 14, 18, 18

Stem Leaf 2

25679

3

00568

4

0 1 389

b

Frequency

25

17

26

42

d

Key 2|4 = 24

20

Score

Frequency

18

This stem plot shows the number of hours students spent studying for a science exam: Calculate, rounded to two decimal places if necessary:

27

35

28

32

29

12

30

20

20 18 16 14 12 10 8 6 4 2 0

68 69 70 71 Scores

72

Stem

Leaf

6

0236778

7

3 69

a

Mean

8

0567

b

Median Mode

9

2

c

Key 1|2 = 12

1.04 Measures of centre mathspace.co

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21

The dot plot shows the number of goals scored across each of Eileen’s soccer games: a

How many games were played in total?

b

How many goals were scored in total?

c

Calculate the mean number of goals per game.

0

22

1 2 3 4 Number of goals scored

5

The dot plot shows the number of sit-ups achieved by students in a physical education exam: Calculate, rounded to two decimal places if necessary: a

Mean

b

Median

c

Mode 0 5 10 15 2025 30 35 40 45 5055 Numbers of sit-ups per student

The given histogram shows the number of hours that students in a particular class had slept for the night before: a

How many students are in the class?

b

Use the class centres to determine an estimate of:

c

24

54

i

Mean

ii

Median

Identify the modal class.

Frequency

23

12 11 10 9 8 7 6 5 4 3 2 1 0

0 2 4 6 8 10 12 14 Frequency

As part of a fuel watch initiative, the price of petrol at a service station was recorded each day for 14 days. The frequency table shows the findings: Price (cents per litre)

Class centre

Frequency

130.0 - 134.9

133.4

4

135.0 - 139.9

138.4

5

140.0 - 144.9

143.4

2

145.0 - 149.9

148.4

4

a

What is the modal class?

b

What percentage of prices were in the modal class? Round your answer to two decimal places.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Extend your thinking 25

The mean of a set of 41 scores is 18.6. If a score of 71.8 is added to the set, determine the new mean. Round your answer to two decimal places.

26

Han wants to try out as a batsman for a cricket team. In his last three matches, he scored 61, 75 and 66 runs. In his last match before trying out, he wants to lift his mean to 70. Determine the number of runs he needs to score to achieve this.

27

Identify 4 consecutive odd numbers whose median is 40.

28

Consider this set of data that represents the number of apps on six people’s phones: 10, 13, 14, 17, 20, 20 If each person downloads another 9 apps, which of the measures of centre would change?

29

30

Determine whether these statements are true or false: a

If two sets of data have the same median, then the data sets must be the same.

b

If two sets of data have different modes then, the highest values cannot be the same.

c

Two sets of data have the same highest and lowest values. This means they must have the same range.

d

Two sets of data that have the same highest and lowest values must have the same mean.

Consider the dot plot given: From which score can a dot be removed so that the mean, median and mode remain unchanged? 200

31

250

300

350

400

Carl has been recording his spelling test scores for the past semester: 14, 16, 2, 15, 15, 16, 15 a

Calculate the median of Carl’s scores.

b

Calculate the mean of Carl’s scores. Round your answer to two decimal places.

c

Which measure of centre most accurately describes the centre of this data set? Explain your answer.

Investigation: The problem with average

Investigate online

mathspace.co

1.04 Measures of centre mathspace.co

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1.05   Data distribution After this lesson, you will be able to… • identify the shape of different distributions. • interpret the shape and spread of data distributions represented with histograms, dot plots, and stem plots. • explain how summary statistics relate to the shape of distributions. • use statistical tools to analyse the effect of extreme values on the measures of centre.

Shape of data Exploration This histogram represents the points scored by a local football team over the past several years: 10 9 8 7 6 5 4 3 2 1 0 10 20 30 40 50 60 70 80 90 100 1. From this graph, can a statement be made about the score range that is typically achieved? Explain your answer. 2. What is the best measure of centre for this graph? Should the median or the mean be used? Provide an explanation for the choice. 3. 60 is in the middle of the range (20 to 100), would this be a good measure of centre? Explain why or why not?

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Data may be described as symmetrical or asymmetrical. There are many cases where the data tends to be around a central value with no bias left or right. In such a case, roughly 50% of scores will be above the mean and 50% of scores will be below the mean. In other words, the mean and median roughly coincide. Mean Median Mode The normal distribution is a common example of a symmetrical distribution of data. The normal distribution looks like this bell-shaped curve.

Percentage

30 25 20 15 10 5 75 80 85 90 95 100105 110

This shows how a data set that has an approximate normal distribution may appear in a histogram. The dark line shows the nice, symmetrical curve that can be drawn over the histogram that the data roughly follows.

Frequency

Weight

20 18 16 14 12 10 8 6 4 2 0

A special type of symmetrical distribution is called a uniform distribution. This is when all of the bars are approximately the same height on the histogram or the curve is actually a horizontal line.

0 10 20 30 40 50 60 70 Skipping rope jumps completed

If a data set is asymmetrical instead (i.e. it is not symmetrical), it may be described as skewed.

1.05 Data distribution mathspace.co

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Positive skew A data set that has positive skew (positively skewed) has a longer tail of values to the right of the peak of the graph, such that more than half of the scores are to the right of the peak. It is sometimes called right skew. As a result, the mean is greater than the median, which is greater than the mode:

Mode Median Mean

Frequency

Mode < Median < Mean

This graph shows how positively skewed data may appear in a histogram.

Score Negative skew A data set that has a negative skew (negatively skewed) has a longer tail of values to the left of the peak of the graph, such that more than half of the scores are to the left of the peak. It is sometimes called left skew. As a result, the mode is greater than the median, which is greater than the mean:

Median

Mode

Mean

Frequency

Mean < Median < Mode

This graph shows how negatively skewed data may appear in a histogram.

Score

58

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Example 1 Determine whether the scores in each histogram are positively skewed, negatively skewed or symmetrical (approximately): a

Frequency

10

5

8 9 10 11 12 13 14 15 16 17 Score

Create a strategy Check where the bulk of the data sits and look at the general shape of the distribution.

Apply the idea The scores are roughly even in both the high and low end, so the distribution is symmetrical.

b

Frequency

10

5

8 9 10 11 12 13 14 15 16 17 Score

Create a strategy Check where the bulk of the data sits and look at the general shape of the distribution.

Apply the idea Most of the scores on the histogram are relatively low, so the distribution is positively skewed.

1.05 Data distribution mathspace.co

59

c

Frequency

10

5

8 9 10 11 12 13 14 15 16 17 Score

Create a strategy

Apply the idea

Check where the bulk of the data sits and look at the general shape of the distribution.

Most of the scores on the histogram are relatively high, so the distribution is negatively skewed.

Idea summary The shape of a data distribution can be described as either symmetrical or asymmetrical. • In a symmetrical distribution, the left and right sides are mirror images. The mean, median, and mode are approximately equal. • The normal distribution is a common, bell-shaped symmetrical distribution. • A uniform distribution is a symmetrical distribution where all outcomes have roughly the same frequency. • An asymmetrical distribution is described as skewed. The direction of the “tail” determines the type of skew. • Positive skew (right-skewed): The tail is on the right, with most data concentrated on the left. Generally, Mean > Median. • Negative skew (left-skewed): The tail is on the left, with most data concentrated on the right. Generally, Mean < Median. Negatively skewed (Left skewed)

Frequency

Mode Median

Symmetrical distribution Mean Median Mode

Mean

Mean < Median < Mode Mean = Median = Mode

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Positively skewed (Right skewed) Mode Median Mean

Mean > Median > Mode

Cluster and modality In a set of data, a cluster occurs when a large number of the scores are grouped together within a very small range. Modality describes the prevalence of local peaks in a data set. The peaks do not necessarily need to be the mode of the whole data set, but rather a local cluster of data that is more frequent and stands out from the surrounding data. When looking at the modality of a data set, it is usually useful to examine a graph of the data. Modality is described by the number of peaks. Data with one peak is called unimodal. 20 15 10

A data set that has two distinct peaks, like in this histogram, is called bimodal.

5 0

5 10 15 20 25 30 35 40 Price

To determine the modality of a distribution, simply identify the number of modal peaks. For instance, the data shown in the dot plot has three modal peaks because there are local peaks at scores of 6, 12, and 20.

0 2 4 6 8 10 12 14 16 18 20 Score

Example 2 For the given stem plot: Stem 0 1 2 3 4 5 6 7 8 9

Leaf 5 78 08 133789 135888

2

Key 1|2 = 12

1.05 Data distribution mathspace.co

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a What is/are the modal class(es)?

Create a strategy Look at the class that occurs most often. Note that there may be more than one modal class.

Apply the idea

Reflect and check

Stems 3 and 4 have the same number of data points, and both have the highest frequency.

This means that the majority of the data lies in the interval 30 – 49. 12 out of 18 data points lie 2 in this range which is or 66.7%. 3

So the correct answers are the intervals 30 – 39 and 40 – 49.

b The distribution of the data is: A

Positively skewed

B

Symmetrical

C

Negatively skewed

D

Uniform

Create a strategy

Apply the idea

Ignore the extreme value and check where most of the scores lie.

Most of the scores are relatively high in the data set, which makes a negatively skewed distribution. So the correct answer is option C.

Example 3 How many peaks are there on the graph?

2 4 6 8 10 12 14 16 18 20 Score

Create a strategy

Apply the idea

Count how many high points are in the data distribution.

Scores on 6, 12, and 20 are observed to have high points or frequency in the data distribution, making them the peaks. So, 3 peaks are present on the graph.

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Idea summary A cluster occurs when a large number of data points are grouped together in a small range. Modality describes the number of peaks or prominent clusters in a data distribution. • A distribution with one peak is called unimodal. • A distribution with two distinct peaks is called bimodal.

Best choice of centre It is important to note that certain features in a data set can significantly affect one or more of the three measures of central tendency (the mean, median and mode). Remember the mode is the most frequently occurring score. It is most useful when analysing nominal data. The shape of the data is key in determining which measure of centre is the most appropriate of a numerical data set. If a data set is symmetrical, then the mean and median will be approximately equal. If the data is unimodal (has only one mode), then the mode will also be approximately equal. If the data has more than one mode (e.g. if it is bimodal) then the modes may be different to the mean and median. Recall that when data is positively skewed, the mean is the highest measure of central tendency and the mode is the lowest measure of central tendency. For positively skewed data: Mode < Median < Mean Similarly, when data is negatively skewed, the mode is the highest measure of central tendency and the mean is the lowest measure of central tendency. For negatively skewed data: Mean < Median < Mode Therefore, in skewed data, the most appropriate measure of central tendency will be the median. Here is a basic summary of selecting an appropriate measure of central tendency. Note that often it can be helpful to consider more than one measure. Data set ...

Mean

Has outliers

Median yes

Is nominal data

yes

Is skewed Is symmetrical

Mode

yes yes

yes

yes

Sometimes the context of the data being analysed lends itself to particular measures as well.

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Example 4 Determine the best choice of centre. a Which measure of centre would be best for this data set? 15, 14, 16, 16, 15, 15, 15, 12, 11, 13, 12, 15, 16, 14, 17, 12, 13, 11, 10, 9, 14 A

Mean

B

Mode

C

Median

Create a strategy Use the table to check the best measure of centre suited for the given data set. Data set ...

Mean

Has outliers

Median

Mode

yes

Is nominal data

yes

Is skewed

yes

Is symmetrical

yes

yes

yes

A display created with technology could also be used to help identify any skew.

Apply the idea 8

Frequency

6 4 2 0

64

By creating a histogram or dot plot, this data could be identified as negatively skewed. As the data is skewed, the median is the most effective measure of central tendency. The correct answer is option B. 9

11

13 15 Data

17

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b Every week for 45 weeks, a kayaking club runs social sessions that are open to the public. In each session, the number of people who attended is recorded. The results are displayed in the table provided: Number of people attending 12 13 14 15 16 17 18 19 20

Number of weeks 6 5 6 5 6 5 6 5 6

Considering the distribution of the responses, which of these statements is true? A The mean is a better indicator of the typical number of people who attended each session than the median. B The median is a better indicator of the typical number of people who attended each session than the mean. C The mean and median are equally accurate indicators of the typical number of people who attended each session.

Create a strategy

Apply the idea

Use the table to determine the distribution of the data set.

Notice that the distribution is uniform as the responses are fairly consistent. This distribution suggests that both mean and median can be equally accurate indicators of the data set. So the correct answer is option C.

Idea summary Here is a basic summary of selecting an appropriate measure of central tendency. Note that often it can be helpful to consider more than one measure. Data set ...

Mean

Has outliers

Median yes

Is nominal data

yes

Is skewed Is symmetrical

Mode

yes yes

yes

yes

Sometimes the context of the data being analysed lends itself to particular measures as well.

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1.05 Practice questions What do you remember? 1

Match each histogram with the best description: i

Symmetrical

ii

Uniform

iii

Negative skew

iv

Positive skew

v

Bimodal

vi

No clear shape, multimodal

a

12

b

8 6

8

Frequency

Frequency

10

6 4 2 0

c

5

d

Frequency

Frequency

10 15 20 25 30 Scores

5

10 15 20 25 30 Scores

5

10

15

6 4 2

e

5

20

8

0

2 0

10 15 20 25 30 Scores

10

4

10 5 0

5

10 15 20 25 30 Scores

12

f

6

8

Frequency

Frequency

10

6 4

4

2

2 0

66

5

10

15 20 25 30 Scores

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

0

15 20 25 30 Scores

2

3

a

For symmetrical data, about half of the data lies on either side of the mean.

b

For uniformly distributed data, the mean, median and mode are very similar.

c

For data with a negative skew, the mean is generally greater than the median.

d

For data with a positive skew, the mean is generally greater than the median.

e

Positive skew refers to a distribution where the tail is on the left side of the distribution.

f

Bimodal data has two modes or modal classes.

g

The shape can only be determined from histograms.

Determine whether the scores in each graph can be described as positively skewed, negatively skewed or symmetrical: a

Stem Leaf 1

677

2

2222333

3

33366677777

4

444444

5

77

b

6

Key 1 | 6 = 16 years old c

Histogram

7 8

15 Frequency

10

5

10

5 6 7 8 9 10 11 12 13 14 15 Score

4

9 10 11 12 13 14 15 Sample

d

15 Frequency

Ex 1

Is each statement true or false?

This stem plot shows the age of people to enter the library in the first five seconds: Is the data positively or negatively skewed?

6 7 8 9 10 11 12 13 Stem

Leaf

1

01 234566

2

0 1 1   2 2 5

3

479

4

4

5

1

Key 1|2 = 12 years old

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5

For each data set: i

Is the data set symmetrical or asymmetrical?

ii

Are there any extreme values in the data set?

iii

Would the mean be an appropriate measure of centre?

a

13, 14, 14, 15, 15, 15, 16, 16, 17

b

8, 10, 14, 18, 19, 91

c

18 19 21 22 24 27 32 48 Age

Practice 6

7

The table shows the number of crime novels in a bookshop for different price ranges: Price of crime novel, to the nearest $5

Frequency

$0 − < $5

3

$10 − < $15

9

$15 − < $20

19

$20 − < $25

6

$25 − < $30

9

$35 − < $40

3

a

Plot this data as a histogram.

b

Describe the shape of the data.

The given stem plot shows the age of people to enter through the gates of a concert in the first 5 seconds: a

What was the median age?

b

What was the mean age? Round your answer to two decimal places.

c

Is the data positively or negatively skewed?

Stem

Leaf

1

01 2233444888

2

1 7

3

455

4

0

5

4 Key 1|2 = 12 years old

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

8

For each of these graphs: i

Are there any clusters? If so where?

ii

What is the mode?

iii

Describe the distribution of the data.

a

Stem Leaf 0

b

2

1 2

0366

3

1 45667

4

04679

5

0

10 11 12 13 14 15 16 17 18 19 2021 2223 24 Sample

Key 5|2 = 52 hours 9

Describe the distribution of each graph: a

b 15 Frequency

Ex 2

10 5

5 10 15 20 25 30 35 40 45 50 Score 10

5 6 7 8 9 10 11 12

The number of peanuts in mixed nut packets were sampled and recorded in this stem plot: a

Complete the frequency distribution table: Score

Stem

Frequency

40 − 49 50 − 59 60 − 69 70 − 79 80 − 89 90 − 99 100 − 109 110 − 119

⬚ ⬚ ⬚ ⬚ ⬚ ⬚ ⬚

4

368

5

1 22

6

67788

7

0033459

8

1   1 1 4 6 8 8 9

9

024569

10

1   2 3 5 5 6 7 8

11

0457

⬚

Key 5|2 = 52

b

Describe the shape of the data.

c

Describe the modality of the data using one of these: i

Unimodal

ii

Leaf

Bimodal

iii

Multimodal but not bimodal

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Ex 3

11

How many peaks are there on this dot plot?

2 4 6

Ex 4a

12

8 10 12 14 16 18 20 Score

Oprah has been growing watermelons. The weight of the watermelons (in kilograms) are shown: 15, 6, 5, 2, 4, 4, 5

13

Calculate the median weight of Oprah’s watermelons.

b

Calculate the mean weight of Oprah’s watermelons. Round your answer to two decimal places if necessary.

c

Determine the most appropriate measure of centre of this data set.

Consider the histogram:

Frequency

Ex 4b

a

180 160 140 120 100 80 60 40 20 0

9

10

11

12

13

14

15 16 Scores

17

18

19

20

21

Determine the measure of centre that would be most appropriate to use to represent the data in this graph. Explain your answer. 14

A timed quiz consists of 6 puzzles. The data shows the times (in seconds) that it took Sophia to complete each question on her first and second attempts of the quiz: • Times in first attempt: 11, 15, 17, 20, 23, 34 • Times in second attempt: 20, 20, 20, 19, 21, 20

70

a

Calculate the mean time spent on each puzzle in the first attempt.

b

Calculate the mean time spent on each puzzle in the second attempt.

c

For which of Sophia’s attempts would the mean number of minutes spent per question be a better indicator of her performance than the median number of minutes spent per question? Explain your answer.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Extend your thinking 15

Four numbers have a range of 5, a median of 9 and a mode of 11. Write the four numbers of this data set.

16

A company records the number of sales made by its salespeople over a month. The data is heavily skewed to the right. What can be inferred about the sales performance of the team?

17

A teacher wants to compare the test scores of two different classes. One class has a symmetrical distribution, while the other has a positively skewed distribution. How might the teacher interpret and compare the performance of the two classes based on these distributions?

18

The price of petrol at a petrol station was recorded each day for two weeks. The results are presented in the given table: Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

Week 1

$1.70

$1.50

$1.62

$1.46

$1.49

$1.46

$1.55

Week 2

$1.25

$1.36

$1.25

$1.21

$1.21

$1.20

$3.30

The mean petrol price across the 14 days of records is $1.54 per litre. For which week is this mean a better indication of the price of petrol? Explain your answer.

Number of passengers

These histograms represent the luggage weight of each passenger on board an airline’s morning and afternoon flight:

Number of passengers

19

Morning flight check-in luggage weights 25 20 15 10 5 0

35 30 25 20 15 10 5 0

20

22

24

28 26 Weight (kg)

30

32

34

Afternoon flight check-in luggage weights

2

3

4

5 6 Weight (kg)

7

8

9

The airline wants to get an indication of how much luggage each passenger is checking in. For which flight’s data would the median be a more precise indicator of a typical passenger’s luggage weight than the mean? Explain your answer.

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1.06   Quartiles and spread After this lesson, you will be able to… • calculate quartiles and the interquartile range (IQR) for data sets, with and without technology. • interpret the five-number summary in context. • use the IQR to describe the spread of data distributions.

Range The simplest measure of spread for quantitative (numerical) data is the range. The range can be determined by calculating the difference between the maximum (highest value) and minimum (lowest value) in a data set. Range = Maximum – Minimum

Example 1 The range of a set of scores is 8, and the highest score is 19. What is the lowest score in the set?

Create a strategy Rearrange the formula for the range to solve for the minimum.

Apply the idea Range = Maximum – Minimum Minimum = Maximum – Range

Write the formula Rearrange

= 19 – 8

Substitute the given values

= 11

Evaluate

Reflect and check The difference between the minimum, 11, and the maximum, 19, is 19 – 11 = 8. This is what the question stated the range was equal to.

Idea summary The range is the simplest measure of spread. It is calculated using the formula: Range = Maximum – Minimum

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Quartiles The range tells us about the spread of the data based on the minimum and maximum, but does not provide information about the spread of data within these two values. The median is a measure of centre, and indicates the middle of the data set is. For a data set with n items: Median is the

n 1 th value 2

Interactive exploration

Discover this concept in action online

mathspace.co

To get a better picture of the internal spread in a data set, quartiles can be determined. Quartiles Values that split a data set into four equal sections. These equal sections can be used for a measure of spread called interquartile range (IQR). Quartiles are scores at particular locations in the data set, which is similar to the median, but instead of dividing a data set into halves, they divide a data set into quarters. • Minimum: Lowest value • Lower quartile, Q1: At most 25% of the data is below this value • Median, Q2: 50% of the data lies on either side of this value • Upper quartile, Q3: At most 25% of the data is above this value • Maximum: Highest value First quarter

Second quarter

2, 3, 5, 5, 7, 9, Minimum

Third quarter

10, 11,

13,

Fourth quarter

16,

18,

Median Q1 = 5

22,

39

Maximum Q3 = 17

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The position of the lower quartile (Q1), median (Q2), and upper quartile (Q3) depends on the number of scores in a data set. The following example demonstrates how to create quartiles in a data set with a even number of scores:

1

3

4

7

11

12

14

19

Ensure the data set is ordered before finding the quartiles or the median.

Median

1

3

4

7

11

12

14

19

Locate the median, between the 4th and 5th scores.

There are four scores in each half of the data set. Split each of the four scores in half to determine the quartiles. Q1

1

3

Q3

Median

4

7

11

12

14

19

The first quartile, Q1, is between the 2nd and 3rd scores. The third quartile, Q3, is between the 6th and 7th scores.

For an odd number of scores, the median is the middle score, and the lower and upper quartiles are determined from the remaining halves. For 9 scores, the 5th term is the median. There are four terms on either side of the median. So Q1 is still between the 2nd and 3rd scores and Q3 is between the 6th and 7th scores. Q1

8

8

Q3

Median

10

11

13

14

18

22 25

For 10 scores, the median is between the 5th and 6th scores. This time, however, there are 5 scores on either side of the median. So Q1 is the 3rd term and Q3 is the 8th term. Q1

12

13

14

Median

19

19

21

Q3

22 22 28 30

These quartiles are a special type of percentiles. A percentile is a percentage that indicates the value below which a given percentage of observations in a group of observations falls. For example, if a score is in the 75th percentile in a statistical test, it is higher than 75% of all other scores. The median represents the 50th percentile, or the halfway point in a data set.

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• Q1 is the first quartile (sometimes called the lower quartile). • It is the middle score in the bottom half of data and it represents the 25th percentile. • Q2 is known as the second quartile and is commonly referred to as the median, which has already been learned. • It represents the 50th percentile of the data set. • Q3 is the third quartile (sometimes called the upper quartile). • It is the middle score in the top half of the data set, and represents the 75th percentile. The interquartile range (IQR) is the difference between the third quartile and the first quartile. Fifty percent of scores lie within the IQR. Since it focuses on the middle 50% of the data set, the interquartile range often gives a better indication of the internal spread compared with the range, and it is less impacted by outliers.

IQR = Q3 – Q1 Q1 Lower quartile Q3 Upper quartile CAS technology statistic functions can automatically calculate the values of Q3 and Q1. The IQR cannot be automatically obtained through the same functions and will need to be calculated with the formula.

Example 2 Here are Ray’s scores from his last 13 rounds of golf played: 66, 66, 68, 68, 70, 78, 80, 84, 106, 116, 126, 130, 132 a What is his median score?

Create a strategy Use the formula, median.

n 1 th where n is the total number of scores, to determine the position of the 2

Apply the idea There are n = 13 scores in the list: Position of median = =

n 1 th 2

Write the formula

13 1 2

Substitute the vlaues

= 7th Median = 80

Evaluate Choose the 7th score

Reflect and check The data set was already ordered, otherwise it would have needed to be ordered first.

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b What is the lower quartile?

Create a strategy Determine the median of the lower half of the scores excluding the median.

Apply the idea The lower half of the scores are: 66, 66, 68, 68, 70, 78 Lower quartile =

68 68 2

Average the middle scores

= 68

Evaluate

c What is the upper quartile?

Create a strategy Determine the median of the upper half of the scores excluding the median.

Apply the idea The upper half of the scores are: 84, 106, 116, 126, 130, 132 Upper quartile =

116 126 2

= 121

Average the middle scores Evaluate

Reflect and check Check the answer visually: Q1

66

66

68

Q3

Median

68

70

78

80

84

106

From the diagram, the data seems to be separated into quarters.

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116

126

130

132

Example 3 Consider this data: 5, 14, 18, 18, 18, 24, 24, 32, 38, 38, 50, 50, 50, 50, 59 Determine the interquartile range of the set of scores.

Create a strategy Determine the median, lower and upper quartiles then use the formula: IQR = Q3 – Q1

Apply the idea There are 15 scores in the list: Position of median =

15 1 2

= 8th

Substitute n = 15 Evaluate

The 8th score is 32. Median = 32 There are 7 scores in the lower half excluding the median. The lower quartile will be the 4th score from the list. Lower quartile = 18 The upper quartile will be the 4th in the upper half, which is the 8 + 4 = 12th score overall. Upper quartile = 50 Solving for the interquartile range: IQR = Q3 – Q1

Write the formula

= 50 – 18

Substitute the quartiles

= 32

Evaluate

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Example 4 For this set of scores in the column graph: 5

Frequency

4 3 2 1 0

30

40

50 60 Scores

70

a Input the data in this distribution table: Score (x)

Frequency ( f )

Cumulative frequency (cf )

30

⬚

⬚

⬚

⬚

⬚

40

⬚

⬚

50 60

⬚

⬚

70

⬚

⬚

Total

Create a strategy Look for the frequencies of each score in the column graph. To find the cumulative frequency, add the frequency of each row to the cumulative frequency of the previous row.

Apply the idea Here is the complete distribution table:

78

Score (x)

Frequency ( f )

Cumulative frequency (cf )

30

5

5

40

5

5 + 5 = 10

50

5

5 + 10 = 15

60

1

1 + 15 = 16

70

3

3 + 16 = 19

Total

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

b Find the median score using the distribution table in part (a).

Create a strategy Use the cumulative frequency column in part (a) to determine the middle score.

Apply the idea Since there are 19 scores in total, the median will be the

19 1 th = 10th score. 2

Looking at the cumulative frequency table in part (a), the 10th score falls in 40. So, Median = 40

c Find the first quartile score.

Create a strategy Determine the middle value of the lower half of the scores, excluding the median, in the distribution table in part (a).

Apply the idea So, the first quartile is determined as the middle of the first 9 scores, which corresponds to the 9 1 th = 5th score. From the table, it can be observed that the 5th score is 30. 2

d Find the third quartile score.

Create a strategy Determine the middle value of the upper half of the scores, excluding the median, in the distribution table in part (a).

Apply the idea The median in part (b) is the 10th score. The third quartile is determined as the middle of the last 9 1 15th score. From the table, it can be observed that the 9 scores, which is computed as 10 2 15th score is 50. Q3 = 50

Reflect and check All of the quartiles can be verified using technology by inputting the raw data or frequency table into a spreadsheet or CAS.

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e Find the interquartile range.

Create a strategy

Apply the idea

Use the interquartile range formula: IQR = Q3 – Q1

Q1 = 30 from part (c) and Q3 = 50 from part (d) can be used. IQR = Q3 – Q1

Write the formula

= 50 − 30

Substitute the quartiles

= 20

Evaluate

Idea summary Quartiles are values that divide an ordered data set into four equal parts. They are specific types of percentiles. • The first quartile (Q1), or lower quartile, is the 25th percentile. It marks the middle of the lower half of the data. • The second quartile (Q2) is the median, or the 50th percentile, which splits the entire data set in half. • The third quartile (Q3), or upper quartile, is the 75th percentile. It marks the middle of the upper half of the data. The interquartile range (IQR) measures the spread of the middle 50% of the data and is less affected by outliers than the range. It is calculated using the formula:

IQR = Q3 – Q1 IQR is the interquartile range Q1 is the lower quartile Q3 is the upper quartile

1.06 Pratice questions What do you remember? 1

80

Is each statement true or false? a

Approximately 25% of data lies above Q1.

b

Approximately 50% of data lies above the median.

c

Approximately 75% of data lies below Q3.

d

The interquartile range is heavily influenced by outliers.

e

The formula for interquartile range is IQR = Q3 – Q1.

f

The interquartile range includes the middle 50% of the data set.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

2

Determine the range of these sets of scores: a 10, 7, 2, 14, 13, 15, 11, 4 b 15, –2, –8, 8, 15, 6, –16, 15 c

3

In a study, a group of people were shown 30 names, and after 1 minute they were asked to recite as many names by memory as possible. The results are presented in the dot plot: a What does one dot represent? b How many people took part in the study? c Identify the largest number of names someone remembered. d Identify the smallest number of names someone 10 11 12 13 14 15 16 remembered. e

4

–0.5537, 1.7444, –0.3381, 0.7200, –0.3381, 1.0435

Determine the range of the data.

Names remembered

What is the interquartile range for this data set? Q1 = 4

Q3 = 14.5

1, 2, 3, 5, 5, 7, 11, 14, 15, 16

Minimum 5

Median = 6

Maximum

Consider this set of scores: 13, 15, 5, 16, 7, 20, 12 a c e

Sort the scores in ascending order. Determine the median. Determine the first quartile.

g

Determine the interquartile range.

b d f

Determine the total number of scores. Determine the range. Determine the third quartile.

Practice Ex 1

6

The range of a set of scores is 14, and the highest score is 37. What is the lowest score in the set?

7

A group of students had a range in marks of 14 and the lowest score was 9. Determine the highest score in the group.

8

The stem plot shows the number of hours students spent studying for a science exam: From the data in the stem plot, determine the following: a

Mean

b

Median

c

Mode

d

Range

Stem Number of hours 6

27

7

1 22479

8

01 257

9

01 Key 1|2 = 12 hours

1.06 Quartiles and spread mathspace.co

81

9

For each data set, the median is shown, determine: i

The upper quartile

ii

The lower quartile

a

0, 1, 4, 6 8, 9, 11

b

22, 24, 25, 29, 32, 34

Median Median: 27 c

5, 11, 19, 20, 22 24, 26, 27, 27

d

14, 15, 16, 16, 17, 18, 19, 20, 24, 25, 30, 36

Median Median: 18.5 10

For each data set: i

Determine the number of scores.

ii

Determine the median.

iii

Determine the first quartile.

iv

Determine the third quartile.

v

Determine the interquartile range.

a

33, 38, 50, 12, 33, 48, 41

b

–3, –3, 1, 9, 9, 6, –9

c

Stem Leaf

d

Score

Frequency

2

256 79

5

3

3

00568

13

3

4

001 8 9

16

2

28

2

31

3

Key 1|2 = 12

Ex 3

Ex 4

11

12

82

38

4

48

2

Determine the interquartile range for each data set: a

31, 32, 34, 42, 49, 50, 51

b

21, 23, 26, 31, 34, 38, 42, 47, 48

c

2, 10, 28, 35, 50

d

41, 42, 43, 44, 49, 50, 51, 53, 58

e

8, 9, 11, 12, 13, 15

f

3, 4, 6, 7, 7, 8, 12, 14

g

5, 7, 12, 14, 15, 22, 24, 30

h

7, 8, 9, 12, 18, 21, 31, 33

Consider the set of scores in this column graph: a

Organise the data into a frequency table that includes an fx column and a cumulative frequency column.

b

Determine the median score.

c

Determine the first quartile.

d

Determine the third quartile.

e

Determine the interquartile range.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Frequency

Ex 2

26 24 22 20 18 16 14 12 10 8 6 4 2 0

7

8

9 Score

10

13

Dylan records the number of fish he catches for each fishing trip over a period of time:

Frequency

20 18 16 14 12 10 8 6 4 2 0

20

21

22

23

24

25

26

27

28

Number of fish caught

14

a

How many fishing trips did he go on?

b

Determine the range in the number of fish caught.

c

Determine and interpret the interquartile range.

Meteorologists predicted huge variations in temperatures throughout the month of April. The temperature each day for the first two weeks of April was recorded: 16, 18, 20.5, 21, 21, 21, 21.5, 22, 22, 24, 24, 25, 26, 27 a

Determine the range.

b

Determine the interquartile range. What does this number indicate?

c

Is the statement ‘Huge variation coming in April’ a valid statement based on the data? Explain.

Extend your thinking 15

A cyclist measured his heart rate immediately after finishing each event in which he competed. The results are recorded in this stem plot: a b

Determine the difference between his slowest and fastest heart rate. For each extra heartbeat per minute, it takes 10 seconds longer for him to recover. How much longer would it take to recover when he finishes with the fastest heart rate than when he finishes with the slowest heart rate?

Stem

Heart rate

16

0 1   4 4 6 7 8

17

1 2235556

18

4468

19

01 36 Key 13|2 = 132

16

In competition, a diver must complete 8 rounds of dives. Her scores for the first 7 rounds are: 7.3, 7.4, 7.7, 8.4, 8.7, 8.9, 9.4 Determine her score in the 8th round if the upper quartile of all of her 8 scores is 8.85

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17

In a competition, a contestant must complete 12 challenges earning as many points as possible. Her scores for the first 11 challenges are: 32, 38, 45, 55, 66, 86, 94, 98, 106, 108, 117 Determine her score in the 12th round if the lower quartile of all of her 12 scores is 47.5

18

Alex is analysing a set of 9 exam scores. He calculates the median and the interquartile range of the scores. Unfortunately, one of the scores is smudged and unreadable. Given that the median score is 70 and the interquartile range is 12, determine the missing score if the remaining scores are as follows: 58, 62, 67, 69, 72, 75, 78, 82

19

A data set has these known statistics: Median

Mode

Range

Interquartile range

12

8

18

7

Determine the new value for each of the statistics, given each value of the data set was: a 20

Increased by 6

b

Multiplied by 3

Sophia is conducting a biology experiment. She repeats the experiment 12 times and records the results. She calculates the first quartile (Q1) and the mode of her results and records these as well. Unfortunately, before she can write up the experiment, she spills water on her results sheet, and one of the data values is no longer readable. If Q1 = 3.0 and the mode is 3.0, and the 11 legible data values are as given: 2.9, 3.2, 3.0, 3.5, 3.1, 3.3, 3.0, 3.4, 3.1, 3.2, 3.5 Calculate the interquartile range of the completion times, including the missing value.

21

There is a test to measure the Emotional Quotient (EQ) of an individual. Here are the EQ results for two groups of 21 people listed in ascending order: Group 1: 90, 90, 91, 92, 93, 94, 95, 95, 95, 97, 99, 100, 108, 114, 116, 116, 117, 118, 118, 122, 129 Group 2: 85, 85, 85, 86, 86, 88, 90, 95, 97, 98, 100, 100, 100, 100, 102, 105, 105, 105, 110, 110, 115

84

a

Calculate the range, interquartile range, median and mean for both data sets.

b

Compare the centre of both data sets. What does this tell you about the EQ scores of the groups?

c

Compare the spread of both data sets. What does this tell you about the variability in EQ scores within the groups?

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

1.07   Box plots After this lesson, you will be able to… • construct box plots using the five-number summary. • interpret the shape, centre, and spread of box plots in context of the data. • compare histograms and box plots for the same data set.

Box plots The quartiles in combination with the two extremes of a data set simplify the data into a five-number summary. A five-number summary for data sets consists of: • Minimum value (Min) - the smallest value in the data set • Lower quartile (Q1) - 25% of the data set lies below this value • Median (Q2) - 50% of data lies on either side • Upper quartile (Q3) - 25% of the data set lies above this value • Maximum value (Max) - the largest value in the data set The five numbers from the five-number summary break up a set of values into four parts with 25% of the values in each quartile: Q1

Min 25%

Q3

Median 25%

25%

50%

Max 25%

50%

The four intervals have about the same number of data points (25%), but do not necessarily cover the same range of values. CAS technology or spreadsheets often have a statistics tool that will output the five-number summary given the data set. Box plots (also known as box-and-whisker plots) are a way of illustrating the five-number summary for a data set. Here is an example of a five-number summary for a set of data: Minimum

17

Lower quartile

52

Median

69

Upper quartile

87

Maximum

100

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This box plot visually shows a concise summary of this information: Minimum value

0

10

20

Median Lower quartile

30

40

50

60

70

Maximum value Upper quartile

80

90

100

The box plot consists of a number line, a rectangle with a line inside (the box) and two horizontal lines (the whiskers). The box represents the middle 50% of the scores and its width represents the interquartile range. Approximately 25% of the scores lie in each quartile. In this example, 25% of the scores would lie between 17 and 52, 52 and 69, 69 and 87, and 87 and 100.

Minimum and outlier

Lower quartile Q1

Median Q2

Upper quartile Q3 Maximum

∼25%

∼25%

∼25%

∼25%

Interquartile range

Example 1 The list shows the number of points scored by a basketball team in each game of their previous season: 52, 52, 55, 58, 59, 68, 69, 77, 80, 89, 96, 96, 96, 97, 99 a Find the maximum value.

Create a strategy Choose the last value when the data has been sorted in ascending order.

Apply the idea Maximum value = 99

b Find the minimum value.

Create a strategy Choose the first value when the data has been sorted in ascending order.

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Apply the idea Minimum value = 52

c Find the median value.

Create a strategy

Apply the idea

The median is the middle score, or the average of the two middle scores.

There are 15 scores, so the middle score will be the eighth score when the scores are in order. Median = 77

d Find the lower quartile.

Create a strategy Find the middle score of the scores below the median.

Apply the idea Scores below the median: 52, 52, 55, 58, 59, 68, 69 Q1 = 58

The middle score is 58

e Find the upper quartile.

Create a strategy Find the middle score of the scores above the median.

Apply the idea Scores above the median: 80, 89, 96, 96, 96, 97, 99 Q3 = 96

The middle score is 96

Example 2 Consider this data set: 20, 36, 52, 56, 24, 16, 40, 4, 28 a Complete the table for the given data: Minimum Lower quartile Median Upper quartile Maximum

⬚ ⬚ ⬚ ⬚ ⬚ 1.07 Box plots mathspace.co

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Create a strategy Order the numbers from smallest to largest to find the values of the five-number summary.

Apply the idea Ordered data: 4, 16, 20, 24, 28, 36, 40, 52, 56 Minimum = 4

The first score

Maximum = 56

The last score

Median = 28 Q1 =

16 20 2

The middle score Average the middle scores of 4, 16, 20, 24

= 18

Evaluate

40 52 Q3 = 2

Average the middle scores of 36, 40, 52, 56

= 46

Evaluate

The completed table is shown: Minimum

4

Lower quartile

18

Median

28

Upper quartile

46

Maximum

56

b Construct a box plot for the data.

Create a strategy Use the five-number summary in part (a) to construct the box plot.

Apply the idea Data

0

88

10

20

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50 60

Example 3 Using this box plot: Glass width

10.7 10.8 10.9 11.0 11.1 11.2 11.3 11.4 a Determine what percentage of scores lie between: • 10.9 and 11.2 • 11.1 and 11.3 • 10.8 and 10.9 • 10.9 and 11.3

• 10.8 and 11.2

Create a strategy Think about how many quartiles are in that range. One quartile represents 25% of the data set.

Apply the idea • 50% of scores lie between Q1 and Q3. So 50% of scores lie between 10.9 and 11.2 • 25% of the scores lie between the lowest score and Q1. So 25% of scores lie between 10.8 and 10.9 • 50% of scores lie between the median and the highest score. So 50% of scores lie between 11.1 and 11.3 • 75% of scores lie between Q2 and the highest score. So 75% of scores lie between 10.9 and 11.3 • 75% of scores lie between the lowest score and Q3. So 75% of scores lie between 10.8 and 11.2

b In which quartile (or quartiles) is the data the most spread out?

Create a strategy Which quartile takes up the longest space on the graph?

Apply the idea The second quartile is the most spread out.

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c Determine the interquartile range.

Create a strategy Read Q3 and Q1 off the box plot and then use the formula.

Apply the idea From the box plot, Q1 = 10.9 and Q3 = 11.2 IQR = Q3 – Q1

Write the formula

= 11.2 – 10.9

Substitute the values

= 0.3

Evaluate

Reflect and check This means that the spread of the middle 50% of glass widths was 0.3 mm, so it was fairly consistent.

Idea summary A five-number summary for data sets consists of: • • • • •

Minimum value (Min) - the smallest value in the data set Lower quartile (Q1) - 25% of the data set lies below this value Median (Q2) - 50% of data lies on either side Upper quartile (Q3) - 25% of the data set lies above this value Maximum value (Max) - the largest value in the data set

Technology, such as graphing calculators, can calculate the five-number summary. The key features of a box plot are:

Median Minimum value

0

10

20

Lower quartile

30

40

50

60

Maximum value Upper quartile

70

80

90

100

• The left-hand side of the box is Q1, and the right-hand side of the box is Q3. • The vertical line inside the box shows the median (the middle score) of the data. • The endpoint of the left line is at the minimum score, while the endpoint of the right line is at the maximum score. • Technology can be utilised to construct box plots.

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Shape of histogram and box plots Histograms and box plots can be used to represent the same set of data because they can both be used for numerical data sets.

Exploration A high school has scholarship programs for gymnastics and basketball. The histogram and box plot summarise the heights of 29 students in a class, in inches:

Number of students

Classmates’ heights 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 75 65 70 Height (in)

60

80

Classmates’ heights

60

62

64

66

68

70

72

74

76

78

80

1. What differences are noticeable between the various displays of the same data? 2. What information does the histogram show that the box plot does not? 3. What information does the box plot show that the histogram does not?

Histograms are useful representations as they can show the spread, modality, shape, and the frequencies of various scores or intervals. However, it can be difficult to read off summary statistics like the measure of centre. The five-number summary and spread of a data set can be clearly read off a box plot. Both types of data displays can assist in determining the shape of the data distribution.

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For box plots: Symmetric A symmetric data set is distributed around the centre with a similar amount of data on the left and right side of the median. Notice that the same number of data points are in each quartile. In this example, the data in the box is a little less spread out than in the whiskers, but is still symmetrical. 0

5

10

15 20 25 30 35 Uniform A uniform data set is evenly distributed across all values. In other words, all four quartiles appear to be the same size. In this box plot, notice that the whiskers and each part of the box is 7 units in length.

0

5

10

15 20 25 30 35

Negative (left) skew

0

5

10

15 20 25 30 35

A data set has a negative skew when the majority of the data points have higher values, with some data points at lower values. It is sometimes called a left skew because it looks stretched to the left. Remember, each quartile still contains the same number of data points. In this box plot, notice that the top 50% of the data is between the values 25 and 31. However, the bottom half of the data is much more spread out.

Positive (right) skew A data set has a positive skew when the majority of the data points have lower values, with some data points at higher values. It is sometimes called a right skew because it looks stretched to the right. In this box plot, notice that the bottom 50% of the data is between the values 5 and 12. However, the top half of the data is much more spread out.

0

5

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15 20 25 30 35

The shapes of box plots and histogram can be compared for the same data set:

Frequency

Symmetric 10 9 8 7 6 5 4 3 2 1 0 8 9 10 11 12 13 14 15 16 17 18 Score

Frequency

Positive (right) skewed 10 9 8 7 6 5 4 3 2 1 0

8 9 10 11 12 13 14 15 16 17 18 Score

Frequency

Negative (left) skewed 10 9 8 7 6 5 4 3 2 1 0 8 9 10 11 12 13 14 15 16 17 18 Score

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Example 4 Match the box plot shown to the correct histogram:

0 1 2 3 4 5 6 7 8 9 10 A

1

2 3 4

5 6 7 8

B

9 10

C

1

2 3 4

5 6 7 8

1

2 3 4

5 6 7 8

9 10

1

2 3 4

5 6 7 8

9 10

D

9 10

Create a strategy Look for characteristics of the data set like the value it is centred around, the range of the data, and the shape - skewed or symmetrical.

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Apply the idea

Reflect and check

The given box plot shows negative (left) skewed data, where the median is closer to the first quartile, and the left whisker is longer. Only option C shows this pattern with most data between 6 and 9 and a longer tail toward lower values.

If other histograms were negatively skewed, they could be matched by ensuring that the range of values is 1 to 10, and the data set is centred around 7, with most values between 6 and 8.5

The correct answer is option C.

Idea summary Histograms and box plots can both represent the same numerical data, but each display highlights different features. • Histograms are useful for showing the frequency of data within intervals, making it easy to see the data's spread, modality, and overall shape. • Box plots clearly display the five-number summary (minimum, first quartile, median, third quartile, and maximum), offering a concise look at the centre and spread. Both displays can be used to identify the shape of a data distribution, such as whether it is symmetric, uniform, has a positive skew (skewed right), or a negative skew (skewed left).

1.07 Practice questions What do you remember? 1

Fill in the missing numbers for the box plot: Minimum 26

2

Lower quartile

33

Median

41

Upper quartile

44

Maximum

45

Match each term with its definition: i

The largest data value

ii

50% of the data lies on either side of this value

iii

About 25% of the data is above this value

iv

The smallest data value

v

About 25% of the data is below this value

a

Minimum

b

Lower quartile, Q1

d

Upper quartile, Q3

e

Maximum

c

Median

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3

For each five-number summary, determine the: i

Range

ii

Interquartile range

a

Minimum 26

b

Minimum

16

Lower quartile

27

Lower quartile

4

5

33

Median

41

Median

35

Upper quartile

44

Upper quartile

42

Maximum

45

Maximum

50

Is each statement true or false for the data set shown in the box plot? a

The maximum is 16.

b

The minimum is 8.

c

The median is 11.

d

There is an error as the median is not exactly in the middle of Q1 and Q3.

e

The interquartile range is 3.

6

8

10

For the box plot shown, determine the: a

Minimum value

b

Maximum value

c

Range

d

Median

e

Interquartile range

12

14

16

18

20

Score

0 2 4 6 8 10 12 14 16 18 20 6

Match each box plot with the correct description: i

Negative (left) skew

ii

Positive (right) skew

iii

Uniform

iv

Symmetric

a

b

0 c

10

15 20 25 30 35

0

96

5

5

10

15 20 25 30 35

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

0

5

10

15 20 25 30 35

0

5

10

15 20 25 30 35

d

Practice 7

Construct a box plot to represent the data in these tables: a

Ex 1, 2

8

Minimum 5

b

Minimum

8

Lower quartile

25

Lower quartile

10

Median

35

Median

16

Upper quartile

60

Upper quartile

24

Maximum

75

Maximum

28

Consider this data: 20, 36, 52, 56, 24, 16, 40, 4, 28 a

9

10

Determine the five-number summary.

b

For each data set, use technology to: i

Complete the table for the given data.

ii

Create a box plot.

a

15, 8, 13, 15, 7, 15, 2, 7, 16

b

20, 36, 52, 56, 24, 16, 40, 4, 28

c

15, 17, 10, 18, 19, 11, 12, 13

d

71, 73, 79, 85, 89, 80, 79, 76

Construct a box plot for the data. Minimum Lower quartile Median Upper quartile Maximum Interquartile range

For each box plot, determine the:

⬚ ⬚ ⬚ ⬚ ⬚ ⬚

i

Minimum

ii

Lower quartile

iii

Median

iv

Upper quartile

v

Maximum

vi

Interquartile range

a

0 c

10

20

b

30

40

0 1

2 3 4 5 6 7 8 9 10

0 2 4 6 8 10 12 14 16 18 20 d

40 42 44 46 48 50 52 54 56 58 60

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Ex 3

11

Consider the box plot shown: Score

0 a

b 12

13

10

15

20

Determine what percentage of scores lie between: i

7 and 15

ii

1 and 7

iv

7 and 19

v

1 and 15

iii

19 and 9

In which quartile is the data the least spread out?

The box plot shows the age at which a group of people got their driving licenses:

Age

a

Determine the oldest age at which someone got their license.

b

Determine the youngest age at which someone got their license.

c

What percentage of people were aged from 18 to 22?

d

The middle 50% of responders were within how many years of one another?

e

In which quartile are the ages the least spread out?

f

The bottom 50% of responders were within how many years of one another?

15

20

The box plot shows the number of hours a group of students studied per week for their final exams: a

Formulate two questions that can be answered by the box plot.

b

Select the option that describes the shape of the box plot:

c

98

5

A

Symmetric

B

Uniform

C

Negatively skewed

D

Positively skewed

25

30

35

Hours per week

0

5

10

15

20

25

Use centre and spread to describe the typical number of hours students in this group studied per week.

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Ex 4

14

Match the histograms on the left to the corresponding box plots on the right: Box plot 1

Histogram A

10 20 30 40 50 60 70 80 90 1 2 3 4 5 6 7 8 9 Box plot 2

Histogram B

0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 Box plot 3

Histogram C 10

5 0 1 2 3 4 5 6 7 8 9 10

10 20 30 40 50 60 70 80 90 Box plot 4

1

2

3

4

5

6

7

8

Histogram D

9 1 2 3 4 5 6 7 8 9

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Box plot 5

Histogram E 10

5 0 10 20 30 40 50 60 70 80 90100

10 20 30 40 50 60 70 80 90 Box plot 6

Histogram F 10

5 0 10 20 30 40 50 60 70 80 90100

10 20 30 40 50 60 70 80 90

Extend your thinking 15

100

The glass windows for an airplane are cut to a certain thickness, but machine production means there is some variation. The thickness of each pane of glass produced is measured (in millimetres), and the dot plot shows the results: a

Determine the median thickness.

b

Determine the interquartile range.

c

Plot a box plot that represents the data.

d

What percentage of thicknesses were between 10.8 mm and 11.2 mm inclusive? Round your answer to two decimal places.

10.8 10.9 11.0 11.1 11.2 11.3 Glass thickness

e

According to the box plot, in which quartile are the results the most spread out?

f

Determine whether these can be determined from the box plot: i

The mode thickness

ii

The frequency of each thickness

iii

The median thickness

iv

The spread of thicknesses

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

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Two groups of people, athletes and non-athletes, had their resting heart rate measured. The results were displayed in this pair of box plots: a

Identify the median heart rate of athletes.

b

Identify the median heart rate of the non-athletes.

c

Which group has the lower heart rates?

d

Determine the interquartile range of the athletes’ heart rates.

e

Determine the interquartile range of the non-athletes’ heart

Athletes

40

50 60 70

80

90

Non-athletes

rates. f

Which group has more consistent heart rate measures?

40 17

In training, a fighter pilot measures the number of seconds he blacks out over several flights. He constructs this box plot.

50 60 70

80

90

Seconds

As long as the pilot is not unconscious for more than 7 seconds, he will be safe to fly. The pilot concludes that he is safe to fly all the time. Explain why his conclusion is incorrect. 0 1 2 3 4 5 6 7 8 9 10 18

VO2 Max is a measure of how efficiently the body uses oxygen during exercise. The more physically fit you are, the higher your VO2 Max. A group of people had their VO2 Max measured, the results are: 21, 21, 23, 25, 26, 27, 28, 29, 29, 29, 30, 30, 32, 38, 38, 42, 43, 44, 48, 50, 76

19

a

Determine the median VO2 Max.

b

Determine the upper quartile value.

c

Determine the lower quartile value.

d

Construct a box plot for this data.

Explain why these pairs of histograms and box plots do not match: a

b

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1.08   Outliers After this lesson, you will be able to… • identify outliers using the upper and lower fences: Q1 – 1.5 × IQR and Q3 + 1.5 × IQR. • construct box plots with outliers to visually represent data distributions. • explain the significance of outliers in a data set. • evaluate the impact of outliers on summary statistics.

Quantify outliers Outlier A data point that varies significantly from the body of the data. An outlier will be a value that is either significantly larger or smaller than other observations. Outliers are important to identify as they often indicate that further investigation may be in order or that there was a measurement or data entry error. Previously, outliers were visually identified, but there was no quantitative rule that defined an outlier.

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To determine if a data value is an outlier, there is a rule that involves the interquartile range (IQR). This rule calculates the upper and lower fences of a box plot. A fence refers to the upper and lower boundaries, and any score that lies outside of the fences are classified as outliers. A data point is classified as an outlier if it lies above the upper fence or below the lower fence. These are calculated as follows: Lower fence = Lower quartile (Q1) – 1.5 × (Interquartile range) Upper fence = Upper quartile (Q3) + 1.5 × (Interquartile range) A box plot can be constructed, and any outliers can be identified using the five-number summary and the upper and lower fences. Smallest value that’s not an outlier

Lower fence equation: Q1 − 1.5 (IQR)

Largest value that’s not an outlier Outliers

Upper fence equation: Q3 + 1.5 (IQR)

Q1 Q3 First 25% of data First 75% of data

The diagram shows the construction of the box plots and how the upper and lower fences are constructed. Any data points outside of these are outliers.

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Example 1 Consider the data set: 1, 1, 3, 21, 7, 9, 10, 6, 11 a Construct a five-number summary.

Create a strategy Determine the minimum and maximum scores, the lower and upper quartiles, and the median.

Apply the idea First, the data set should be sorted in order. The ordered data set is: 1, 1, 3, 6, 7, 9, 10, 11, 21 It can be clearly observed that the minimum value is 1 and the maximum value is 21. Since there are 9 scores, the middle score, which is the fifth score in order, represents the median. The median is 7. To determine the quartiles, the data set is divided into two parts on either side of the median. The lower part consists of 1, 1, 3, 6, and the upper part consists of 9, 10, 11, 21. Since both parts contain an even number of scores, the quartiles are calculated as the averages of their middle scores: Q1

1 3 2

2   Q3

10 11 2

Minimum

1

Q1

2

Median

7

Q3

10.5

Maximum

21

10.5

So the five-number summary is:

b Determine the values of the lower and upper fences.

Create a strategy Calculate the lower fence and upper fence using the formulas: Lower fence = Lower quartile - 1. 5 × IQR Upper fence = Upper quartile + 1.5 × IQR

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Apply the idea The IQR is solved first. As found in part (a), Q1 = 2 and Q3 = 10.5 IQR = Q3 – Q1

Write the formula

= 10.5 – 2

Substitute the quartiles

= 8.5

Evaluate

Lower fence = 2 – 1.5 × 8.5 = – 10.75 Upper fence = 10.5 + 1.5 × 8.5 = 23.25

Substitute the values Evaluate Substitute the values Evaluate

c Identify any outliers.

Create a strategy Compare the largest and smallest values to the fences found in part (b).

Apply the idea

Reflect and check

1 > –10.75, so there is no lower outlier.

Notice that although 21 is quite a bit larger than the other scores in the set, it is not far enough away to be considered an outlier. The upper and lower fence values should always be checked to determine whether a score is an outlier or not.

21 < 23.25, so there is no upper outlier. So there are no outliers in this data set.

d Construct a box plot to represent this data.

Create a strategy

Apply the idea

Use the five-number summary from part (a).

0 2 4 6 8 10 12 14 16 18 2022

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Example 2 The five-number summary for the given dot plot is shown:

Minimum

1

Q1

2

Median

3

Q3

4

Maximum

9

1

2

3

4

5 6 Score

7

8

9

a Calculate the interquartile range.

Create a strategy Use the interquartile range formula: IQR = Q3 – Q1

Apply the idea IQR = 4 – 2 =2

Substitute the quartiles Evaluate

b Determine if there is an outlier. Justify your answer.

Create a strategy Calculate the lower fence and upper fence using the formulas: Lower fence = Lower quartile − 1.5 × IQR Upper fence = Upper quartile + 1.5 × IQR

Apply the idea It was given that Q1 = 2 and Q3 = 4. In part (a), it was found that IQR = 2. Lower fence = Q1 – 1.5 × IQR

Write the formula

= 2 – 1.5 × 2

Substitute the values

= –1

Evaluate

Upper fence = Q3 + 1.5 × IQR

Write the formula

= 4 + 1.5 × 2

Substitute the values

=7

Evaluate

Since 9 is above the upper fence of 7, then the outlier is 9.

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c Construct a box plot of this data.

Create a strategy Use the five-number summary, but the upper whisker will end at the highest non-outlier.

Apply the idea

0 1

2 3 4 5 6 7 8 9 10

Reflect and check When verifying with technology, the outlier may not be shown. However, statistically, it should be shown.

Idea summary A data point is classified as an outlier if it lies above the upper fence or below the lower fence. These are calculated as follows: Lower fence = Lower quartile - 1.5 × Interquartile range Upper fence = Upper quartile + 1.5 × Interquartile range Technology can be used to automatically obtain the quartile values using statistics functions. Fence calculations, however, will typically still need to be completed manually with technology support. When graphing a box plot using technology it may or may not show outliers.

Effects of outliers Once an outlier is identified, the underlying cause of the outlier should be investigated. If the outlier is simply a mistake, then it should be removed from the data. If the data is not a mistake, it should not be removed from the data set as while it is unusual, it is representative of possible outcomes. When data contains an outlier, its impact on any calculations should be considered. Observe the effect that outliers have on the three measures of centre - mean, median and mode.

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Outliers can skew or change the shape of a data set. This can be problematic (especially for small data sets) because the mean, median and range might not be a true representation of the data. This can be counteracted by removing outliers. Removing outliers will have these impacts: Removing a low outlier

Removing a high outlier

The range will decrease.

The range will decrease.

The median might increase.

The median might decrease.

The mean will increase.

The mean will decrease.

The mode will not change.

The mode will not change.

Data will become less skewed.

Data will become less skewed.

Did you know?

Skewed data distributions can reveal surprising insights! For example, in real estate, the distribution of house prices is often right-skewed—meaning most properties are moderately priced, but a few luxury homes dramatically raise the average. By identifying this skew, analysts avoid misleading conclusions and use the median (rather than the mean) to better represent typical house prices. Understanding skewness helps ensure data tells the real story, especially when extremes are involved!

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Example 3 Consider this set of data: 37, 46, 35, 56, 56, 35, 125, 36, 48, 56 a Fill in this table of summary statistics. ⬚

Mean

⬚

Median

⬚

Mode

Create a strategy To find the mean, use the formula: Mean To find the median, find the middle score.

Sum of scores Number of scores

To find the mode, find the most frequent score.

Apply the idea 37 46 35 56 56 35 125 36 48 56 10

Use the formula

530 10

Evaluate the addition

= 53

Evaluate the division

Mean

To find the median, order the scores: 35, 35, 36, 37, 46, 48, 56, 56, 56, 125 The middle scores are: 46, 48 46 48 Median 2

Find the average

= 47

Evaluate

To find the mode, choose the score which occurs most often. Mode = 56 By filling in the table:

108

Mean

53

Median

47

Mode

56

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b Which data value is an outlier?

Create a strategy Choose the value that is much greater or much smaller than the rest of the data set.

Apply the idea Outlier = 125

c Fill in this table of summary statistics after removing the outlier. ⬚

Mean

⬚

Median

⬚

Mode

Apply the idea

37 46 35 56 56 35 36 48 56 9

Mean

Use the formula

405 9

Evaluate the addition

= 45

Evaluate the division

To find the median, order the scores: 35, 35, 36, 37, 46, 48, 56, 56, 56 The middle score is 46. Median = 46 To find the mode, choose the score which occurs most often. Mode = 56 By filling in the table: Mean

45

Median

46

Mode

56

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d Let A be the original data set and B be the data set without the outlier. Complete the table using the symbols >, < and = to compare the statistics before and after removing the outlier. With outlier

>, <, =

Without outlier

⬚

B

Mean

A

Median

A

Mode

A

B

⬚

B

⬚

Create a strategy Compare the statistics in part (a) and in part (c).

Apply the idea Statistics from parts (a) and (c): With outlier

Without outlier

Mean

53

45

Median

47

46

Mode

56

56

Comparison table: With outlier

>, <, =

Without outlier

Mean

A

>

B

Median

A

>

B

Mode

A

=

B

Idea summary The mean will be significantly affected by the inclusion of an outlier: • Including a high outlier will increase the mean. • Including a low outlier will decrease the mean. The median is the middle value of a data set, the inclusion of an outlier will not generally have a significant impact on the median. The mode is the most frequent value, as an outlier is an unusual value it will not be the mode. The inclusion of an outlier will have no impact on the mode.

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Suitability of measures of centre The mean, median, or mode can be used to describe the centre of a data set. At times, one measure may better represent the data than another, and in certain cases, a single statistic may be preferred for an article or report instead of detailing multiple measures. When choosing which measure to use, it is important to consider which one best represents the type of data. Some main considerations are: • Is there a repeated value? If there are no repeated values or only a couple of randomly repeated values, then the mode will not be representative of the data. If there are one or two highly frequent data points, these may be a fair representation of the centre of the data. • Is there an outlier? As observed, an outlier can significantly affect the mean — potentially distorting the perceived centre of the data. For example, if a list of houses sold in an area includes a historic mansion sold at a price far above the others, using the median would be a better representation of average house prices in the area than the mean. • Do you need all the data values to be taken into account? Only the mean uses all the values in its calculation.

Example 4 The salaries of part-time employees at a company are given in the dot plot. Which measure of centre best reflects the typical wage of a part-time employee? Salaries (in $1000’s)

18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Create a strategy Choose the measure that is appropriate for data with extreme values.

Apply the idea The data set has extreme values that are away from the rest of the set at 37 and 38. So the median is the best measure of centre, since it will not be affected by the extreme values.

Idea summary Main considerations when choosing a suitable measure of centre: • If there are no repeated values or only a couple of randomly repeated values, then the mode will not be representative of the data. • If there are outliers, the median will be more suitable than the mean. • Only the mean uses all the values in its calculation.

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1.08 Practice questions What do you remember? 1

For each data set, calculate: i

The interquartile range

ii

The value of the lower fence (Q1 – 1.5 × IQR)

iii

The value of the upper fence (Q3 + 1.5 × IQR)

a

Minimum 5 Q1

6

Median

12

Q3

17

Maximum

28

b

2 2

4

6

Mode

b

Range

c

Interquartile range

For each of these scenarios, determine whether the outlier that was removed must have had a value smaller or larger than the values that remain: a

A set of data has an outlier removed and the mean lowers.

b

A set of data has an outlier removed and the mean rises.

c

A set of data has an outlier removed and the median lowers.

d

A set of data has an outlier removed and the median rises.

For the given dot plot, the current range is 11. If the outlier is removed, determine the new range.

0 1 2 3 4 5 6 7 8 9 10 11 Three-pointers

112

8 10 12 14 16 18

When an outlier is removed from a data set, describe the effect on the: a

3

4

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Practice 5

For the data set: 3, 10, 9, 2, 7, 5, 6

6

7

Ex 1

Ex 2

8

9

a

Construct the five-number summary.

b

Calculate the interquartile range.

c

Calculate the value of the lower fence.

d

Calculate the value of the upper fence.

e

Would the value –5 be considered an outlier?

f

Would the value 16 be considered an outlier?

Determine the outlier(s) in these data sets: a

73, 77, 81, 86, 131

b

69, 79, 86, 72, 86, 77, 73, 82, 81, 76, 83, 47, 87, 70, 80, 85

c

58, 63, 58, 59, 64, 68, 68, 30, 73, 25, 72, 61, 65, 69, 75, 72

The table shows the average temperature (°C) in a particular city over several years. Identify the year in which the temperature is an outlier. Year

2002

2003

2004

2005

2006

2007

2008

Temperature (°C)

34.2

26.5

22.6

22.5

24.2

23.0

24.1

For each data set: i

Construct the five-number summary.

ii

Calculate the value of the lower fence.

iii

Calculate the value of the upper fence.

iv

Identify any outliers.

v

Construct a box plot of the data with the outlier(s) displayed separately.

a

6.8, 4.0, 3.5, 5.1, 2.4, 1.6, 3.9, 3.5, 3.1, 3.6, 7.6, 3.7, 4.0, 5.1, 3.6, 3.8, 3.6, 6.7

b

10, 15, 12, 26, 18, 15, 11, 38, 25, 12, 19, 17, 16, 17, 11, 36, 9, 2, 21, 18, 16

For the given data set and five-number summary: 82, 87, 92, 76, 80, 85, 71, 84, 61, 79, 81, 81, 86, 97, 101, 80, 71, 76, 78, 86, 84

Minimum

61

Q1

77

Median

81

Q3

86

Maximum

101

a

Calculate the interquartile range.

b

Determine if there is an outlier. Justify your answer.

c

Construct a box plot of this data

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10

Redraw the box plots showing the outlier(s) as points: a

104, 115, 215, 127, 109, 118, 121, 130, 116, 134, 131, 129, 125

b

0

100 120 140 160 180 200 220 c

11

d

61, 80, 108, 68, 70, 72, 63, 65, 75, 78

60

70

80

90

100

4.0, 3.5, 5.1, 4.4, 0.3, 3.9, 3.5, 3.1, 4.9, 4.3, 4.7, 5.0, 3.7, 3.8

110

1

2

3

4

5

6

22, 37, 39, 27, 29, 30, 60, 31, 32, 34, 28, 29, 35, 42, 25, 45

20 26 32 38 44 50 56 62

Consider the given stem plot: Stem Leaf 3

449

4

6689

5

14

6 7 8

4

Key 2|3 = 23 If the outlier is removed, what is the new mean? Round your answer to two decimal places. 12

Consider the given stem plot:

Stem Leaf 2

5

3 4

99

5

00457

6

26

Key 1|2 = 12 If the outlier is removed, determine the new range.

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Ex 3

13

14

For each set of data: i

Determine the mean, median, mode and range. Round your answers to two decimal places where necessary.

ii

Which data value is an outlier?

iii

Remove the outlier from the set and recalculate the values found in part (i).

iv

Describe how each of the four statistics changed after removing the outlier.

a

27, 50, 24, 37, 47, 41, 27, 126, 44, 27

b

4.7, 2.8, 1.9, 0.9, 0.9, 2.2, 2.2, 1.2, 1.5, 0.9

c

4700, 4700, 4700, 4500, 5300, 4900, 5200, 4800, 1500, 5100

The selling price of recently sold houses is given: $760 000, $650 000, $810 000, $780 000, $760 000, $590 000, $1 360 000

Ex 4

15

a

Determine the mean selling price. Round your answer to the nearest thousand dollars.

b

Identify the median selling price.

c

Recalculate the mean selling price excluding the outlier.

d

Recalculate the median selling price excluding the outlier.

e

Which measure of centre best identifies the typical selling price of recently sold houses? Explain your answer.

The number of goals scored by different soccer teams in a tournament is given in the dot plot. Which measure of centre best reflects the typical number of goals scored? Explain your answer.

0 1 2 3 4 5 6 7 8 9 10 Goals scored 16

A group in a study take a test to assess their reaction time. The participants clicked a button as soon as they heard a sound which was played at random intervals. The reaction time, in milliseconds, of each participant is shown: 220, 280, 210, 220, 215, 180, 185, 190, 190, 195, 150, 190, 195, 195 a

Construct the five-number summary.

b

Identify any outliers and use statistical calculations to justify your answer.

c

Create a box plot of the data with the outlier displayed separately.

d

Give a possible explanation for the outliers present.

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17

VO2 Max is a measure of how efficiently the body uses oxygen during exercise. The more physically fit you are, the higher your VO2 Max. Here are some people’s results when their VO2 Max was measured: 46, 27, 32, 46, 30, 25, 41, 24, 26, 29, 21, 21, 26, 47, 21, 30, 41, 26, 28, 26, 76 a

Arrange the values into ascending order.

b

Determine the median VO2 Max.

c

Determine the upper quartile value.

d

Determine the lower quartile value.

e

Calculate 1.5 × IQR, where IQR is the interquartile range.

f

Identify any outliers using upper and lower fences.

g

Construct a box plot of the data with the outlier displayed separately.

h

An average untrained healthy person has a VO2 Max between 30 and 40. Using the box plot, what level of exercise is likely to describe the majority of people in this group?

18

When an outlier is removed from a data set, describe the effect on the: a

Mean

b

Median

Extend your thinking 19

The data point 5 is below the lower fence and is considered an outlier. The interquartile range is 12. Determine the smallest integer value the lower quartile can be.

20

The data point 37 is above the upper fence and is considered an outlier. The interquartile range is 10. Determine the largest integer value the upper quartile can be.

21

Consider this frequency table: Weight in kilograms

Frequency

12

2

13

5

14

1

15

2

16

0

17

0

18

1

If the outlier is removed what is the new mean? Round your answer to two decimal places if needed.

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22

The glass windows for an aeroplane are rolled to a certain thickness, but machine production means there is some variation. The thickness of each pane of glass produced is measured (in millimetres), and the dot plot shows the results: Glass thickness (mm)

10.5

23

10.7

10.9

11.1

11.3

a

The current median is 11.15. If the outlier is removed what is the new median?

b

The current mean is 11.1. If the outlier is removed what is the new mean? Round your answer to two decimal places.

A journalist wanted to report on road speed cameras being used as revenue raisers. She obtained data that showed the number of times 20 speed cameras issued a fine to motorists in one month. The results were: 101, 102, 115, 115, 121, 124, 127, 128, 130, 130, 143, 143, 146, 162, 162, 163, 178, 183, 194, 977 The journalist wants to give the impression that speed cameras are just being used to raise revenue. Which measure of centre should she use in her article? Explain your answer.

24

The selling prices of artworks sold at an auction are given: $18 000, $11 000, $17 000, $20 000, $18 000, $16 000, $15 000, $218 000 Which measure of centre best identifies the typical selling price of recently sold artwork? Explain your answer.

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1.09   Standard deviation After this lesson, you will be able to… • calculate the standard deviation using technology. • interpret sample standard deviation as a measure of spread. • analyse symmetrical and asymmetrical data distributions using appropriate summary statistics. • calculate and interpret the percentage of data lying within several standard deviations of the mean.

Standard deviation Standard deviation A measure of spread, which helps give a meaningful estimate of the variability in a data set when the range and interquartile range do not tell the full story. Consider these data sets and their summary statistics: Set A

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Set B

1

1

1

4

8

8

8

8

8

8

8

12

15

15

15

Summary statistic

Set A

Set B

Mean

8

8

Median

8

8

Range

14

14

Q1

4

4

Q3

12

12

Interquartile range

8

8

The data sets are clearly different, but the existing summary statistics do not show that. While the quartiles describe the spread about the median, the standard deviation describes the spread from the mean. It is a weighted average of the distance of each data point from the mean. • A small standard deviation indicates that most scores are close to the mean. • A large standard deviation indicates that the scores are more spread out away from the mean value. The standard deviation can be calculated for a population or a sample. The symbols used are: Population standard deviation = σ (lowercase sigma) Sample standard deviation = s

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In statistics mode on a calculator, these symbols might be used: Population standard deviation = σn Sample standard deviation = σn–1 Note: It is only required to calculate standard deviation using the automatic function in the statistics mode on CAS calculators, so use of the formal equation is not required. The standard deviation is found by calculating the square root of the sample or population variance. Variance is the average of the squared differences from the mean. Here is the population variance formula:

2

( x1

)2

( x2

)2 n

( xn

)2

σ 2 is the population variance n is the population size x is a particular data value μ is the population mean The population standard deviation is then: 1 ( xi )2 n In this formula, • The numbers xi are the values in the data set. There is one value for each subscript i. • There are n numbers xi in the data set. So, i goes from 1 to n in the summation. • The symbol μ (Greek letter ‘mu’) is the population mean. • The Greek letter σ (sigma) is used for the population standard deviation. • The symbol Σ (upper case sigma) is the summation symbol. The sample standard deviation is:

( x x )2 n 1

s s x x

n

is the standard deviation is a data value is the sample mean is the sample size

Standard deviation describes the spread of data by comparing the distance of each score to the mean. It is a tedious process to calculate, but it gives a lot of information about the spread of data because it takes into account every data point in the set. Standard deviation is also a very powerful way of comparing different data sets, particularly if there are different means and population numbers.

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Here are some examples of what a data set can look like with the same measures of centre, but different measures of dispersion: • Mean = 5, Median = 5, Standard deviation = 2.84 7 6 5 4 3 2 1 0 1

2

3

4 5

6

7

8

9 10

6

7

8

9 10

6

7

8

9 10

6

7

8

9 10

• Mean = 5, Median = 5, Standard deviation = 2 7 6 5 4 3 2 1 0

1

2

3

4

5

• Mean = 5, Median = 5, Standard deviation = 0.93 8 7 6 5 4 3 2 1 0

1

2

3

4

5

• Mean = 5, Median = 5, Standard deviation = 0 7 6 5 4 3 2 1 0 1

120

2

3

4

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Example 1 Determine the population standard deviation of this set of scores using technology with statistics functions: 8, 20, 16, 9, 9, 15, 5, 17, 19, 6 Round your answer to two decimal places.

Create a strategy Enter all the scores into a spreadsheet or list, then use technology to find the population standard deviation.

Apply the idea σ = 5.30

Reflect and check The sample standard deviation would be s = 5.58, because it is the same total, but divided by a smaller number, the sample standard deviation is larger.

Example 2 Data was collected and organised in a grouped frequency table. Class

Frequency

1−9

8

10 − 18

6

19 − 27

4

28 − 36

6

37 − 45

8

Total a Complete the table: Class

Class centre

Frequency

fx

1−9

⬚

8

⬚

10 − 18 19 − 27 28 − 36 37 − 45 Total

⬚

6

⬚

8

⬚ ⬚

4 6 ⬚

⬚ ⬚ ⬚ ⬚ ⬚

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Create a strategy To find the class centre, determine the average of the upper and lower bounds of each class. To find the relative frequency, f x, multiply the class centre with its corresponding frequency.

Apply the idea Here is the complete table: Class

Class centre

Frequency

fx

1−9

1 9 2

5

8

5 × 8 = 40

10 − 18

10 18 2

14

6

14 × 6 = 84

19 − 27

19 27 23 2

4

23 × 4 = 92

28 − 36

28 36 2

32

6

32 × 6 = 192

37 − 45

37 45 2

41

8

41 × 8 = 328

32

736

Total

b Use the class centres to estimate the mean of the data set, rounded to two decimal places.

Create a strategy Use the formula: Mean

fx f

Apply the idea Looking at the table in part (a), the sum of the relative frequencies is found at the bottom of the fourth column, which is Σ fx = 736. The total number of scores or the sum of the frequencies can also be seen at the bottom of the third column, which is Σ f = 32. Mean =

736 32

= 23

Substitute the values Evaluate using a calculator

Reflect and check It is possible to enter the data as a frequency table on most CAS calculators, but there is often an additional step required to convert to a list or define it as a frequency table before calculating the summary statistics.

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c Use the class centres to estimate the population standard deviation, rounded to two decimal places.

Create a strategy Use the Statistics mode on your calculator to find the population standard deviation.

Apply the idea Enter the class centres and their corresponding frequencies.

σn = 13.87

d If the original ungrouped data were used to calculate the standard deviation, would the ungrouped data result in a higher or lower standard deviation?

Create a strategy Recall the definition of the standard deviation.

Apply the idea A small standard deviation indicates that most scores are close to the mean, while a large standard deviation indicates that the scores are more spread out away from the mean value. So, having the original ungrouped data would have a higher standard deviation because there would be more variables and a wider spread in the data set.

Idea summary Standard deviation is a weighted average of how far each piece of data varies from the mean. The standard deviation can be calculated for a population (σ) or a sample (s). The standard deviation is a more complex calculation but takes every data point into account. It is significantly impacted by outliers. For each measure of spread: • A larger value indicates a wider spread (more variable) data set. • A smaller value indicates a more tightly packed (less variable) data set.

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Standard deviations from the mean Exploration Consider the data sets represented by the histograms:

7

11

15

19 23 27 31

8

Histogram 1

12

16 20 Histogram 3

10 12

14 16 18 20 22

Histogram 2

24

5

10 15 20 25 30 35 40 Histogram 4

1. Match each of the histograms to the correct mean and standard deviation: • μ = 16, σ = 3 • μ = 15, σ = 2 • μ = 19, σ = 4 • μ = 18, σ = 2 2. Justify your choices. 3. These histograms are all symmetrical and bell-shaped (normally distributed). How many standard deviations away from the mean are the maximum and minimum for each histogram?

Depending on the shape of a data set, the percentage of the data that lies within 1, 2, or 3 standard deviations of the mean varies. For most data sets, it is a manual process of determining the boundaries of ±1, ±2, and ±3 standard deviations from the mean and finding the percentage of the data values that lie in those intervals.

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Normal distribution

y

A statistical distribution represented with symmetrical bell-shaped graph.

1 x −1

1

For symmetrical, bell-shaped data, there is a special property called the 68 – 95 − 99.7% rule, also known as the empirical rule. This rule estimates the spread only for data that is approximately normally distributed. According to this rule, the majority of scores (68%) fall within one standard deviation of the mean, most scores (95%) fall within two standard deviations of the mean, and nearly all scores (99.7%) fall within three standard deviations of the mean.

34%

34% 13.5%

13.5%

2.35%

2.35%

µ −σ

µ −2σ

µ − 3σ

µ +σ

µ

µ + 2σ µ + 3σ

68% 95% 99.7% Empirical rule

Empirical rule (68 – 95 – 99.7% rule) A statistical rule that provides an estimate for the distribution of approximately normal data.

Example 3 Consider the normally distributed data sets shown:

10

11

12

13

Distribution A

14

12

13

14

15

16

Distribution B

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a Which data set has a higher mean?

Create a strategy A normal curve is symmetric about the mean. To compare the means of the distributions, the value on the line of symmetry for each curve can be identified, then compare those values.

Apply the idea

The mean of Distribution A is 12 since the curve is symmetric about that value. 10

12

11

13

12

14

13

15

14

16

The mean of Distribution B is 14 since the curve is symmetric about that value.

Since 14 > 12, Distribution B has a higher mean value.

Reflect and check The shape of the curve does not affect the mean. The mean determines the curve’s line of symmetry in a symmetric distribution.

b Which data set has a smaller standard deviation?

Create a strategy The standard deviation affects the spread of a normal curve. To compare the standard deviations of the distributions, the spread of each curve can be analysed and compared.

Apply the idea In Distribution A, approximately 100% of the data lies between 10 and 14. In Distribution B, the curve stretches beyond 12 and 16. The curve is wider and less peaked, showing it has a larger spread. Distribution A has a smaller deviation.

Reflect and check Because the normal curve has an area of 1, curves with a smaller standard deviation will be tall and thin. As the variation in the data increases (as the data values spread further from the centre), the curve will become shorter and wider.

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Example 4 For each display, the mean and standard deviation are given. Determine the percentage of the data that lies in the given range. a What percentage of the data lies in the range 72°F to 88°F, given that the mean is 78°F and the standard deviation is 6°F? Daily temperatures 12

Frequency

10 8 6 4 2 0

60.00 64.00 68.00 72.00 76.00 80.00 84.00 88.00 92.00 96.00 Temperature

Create a strategy This data set is fairly symmetrical, but is a small sample. The empirical rule might be a poor estimate.

Apply the idea 29 There are 57 data values and 8 + 11 + 10 = 29 of them lie in this range. So 51% of the data 57 values lie in this range.

Reflect and check Since the mean is 78°F and the standard deviation is 6°F, the range 72°F = 78°F – 6°F to 84°F = 78°F + 6°F is within one standard deviation of the mean. However, there is significantly less than 68% of data values in this range because it is a small sample and not quite normally distributed.

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b What percentage of data values lie between 12 and 56, given that the mean is 34 and the standard deviation is 11? Stem   Leaf 1

588

2

1 233444

3

23477 788888

4

233557

5

28 Key 2|3 = 23

Apply the idea

28 96.6% of the data values lie in There are 29 data values and 28 of them lie in this range. So 29 this range.

Reflect and check This is fairly close to what the empirical rule predicts.

c What percentage of data values lie between 1.5 and 17.1, given that the mean is 9.3 and the standard deviation is 2.6?

6 7 8 9 10 11 12 13 14 15 Sample

Create a strategy This data set is not symmetrical, so the empirical rule does not apply.

Apply the idea

Reflect and check

This range includes all of the data, so 100% of data values lie in this range.

In general, 3 standard deviations should include most of the data for any set, unless it is very skewed.

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Example 5 The grades on a recent exam are approximately normally distributed with a mean score of 72 and a standard deviation of 4. a Construct a normal curve and label the boundaries for the empirical rule.

Create a strategy A normal curve will have a symmetric bell-like appearance with the mean as the central value and divisions for: • Mean ± 1 standard deviation • Mean ± 2 standard deviations • Mean ± 3 standard deviations

Apply the idea Subtract the standard deviations from the mean to find the values on the left side of the curve: • 1 standard deviation below the mean: 72 – 4 = 68 • 2 standard deviations below the mean: 72 – 2 (4) = 64 • 3 standard deviations below the mean: 72 – 3 (4) = 60 Add the standard deviations to the mean to find the values on the right side of the curve: • 1 standard deviation above the mean: 72 + 4 = 76 • 2 standard deviations above the mean: 72 + 2 (4) = 80 • 3 standard deviations above the mean: 72 + 3 (4) = 84

60

64

68

72

76

80

84

b Determine the percentage of students who scored between 64 and 68 on the exam.

Create a strategy To use the empirical rule, first determine how many standard deviations 64 and 68 are away from the mean score of 72.

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Apply the idea 64 is two standard deviations below the mean and 68 is one standard deviation below the mean.

60

64

68

72

76

80

84

According to the empirical rule, the percentage of data between 1 and 2 standard deviations below the mean is 13.5%.

c If 32 students took the exam, determine the number of students expected to score 80 or more on the exam.

Create a strategy First, determine the number of standard deviations 80 is away from the mean score of 72. Then, multiply the percentage found from the empirical rule by 32 students to determine the number of students who may have scored more than 80.

Apply the idea 80 is two standard deviations above the mean. According to the empirical rule, 95% of the data is within 2 standard deviations of the mean. Since all of the data lies below the curve, then 1 − 0.95 = 0.05 or 5% of the data lies above and below 2 standard deviations of the mean. Divide it in half to find the percentage of data that is only 2 standard deviations above the mean: 0.05 0.025 or 2.5%. 2 2.5% of the 32 students are expected to score 80 or more. This gives 32 (0.025) = 0.8. If the data is approximately normal, not even one student will score above an 80 in a class of 32.

Reflect and check Use the empirical rule to check the reasonableness of this solution. According to the rule, 95% of the students will receive scores between 64 and 80 because these are 2 standard deviations from the mean. 32 × 0.95 = 30.4 Because of the rounding error, there are still 2 students that scored below 64 or above 80 on the test. A conclusion that 1 student scored 80 or above on the test would still be valid.

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Idea summary A data set that is symmetric and bell-shaped is said to have an approximately normal distribution. The mean, median, and mode are approximately equal in a normal distribution. The centre of the normal distribution is at the arithmetic mean, μ. The standard deviation σ, describes the spread of the data. The normal curve represents a probability distribution, and the area under the entire curve is equal to 100%, or 1. The percentage of data between 1, 2, and 3 standard deviations can be accurately summarised using the empirical or 68 – 95 − 99.7% rule. However, for non-normal distributions or for small data sets, the percentages need to be calculated manually.

1.09 Practice questions What do you remember? 1

2

Ex 1

3

Match each term with its description: i

The whole group from which the information will be gathered

ii

The spread of the middle 50% of the data set

iii

The average distance from the mean calculated as the square root of the variance

iv

A portion of the group chosen for gathering information

v

The average of the square difference from the mean

a

Standard deviation

b

Variance

c

Sample

d

Population

e

Interquartile range

Are the statements always, sometimes, or never true? a

If all values in a data set were increased by the same amount, the standard deviation would also increase by that amount.

b

If the range of a data set is large, the standard deviation will be large.

c

If an extreme value is added to a data set, it will not affect the standard deviation.

d

The standard deviation of a data set will always be positive.

e

The variance is larger than the standard deviation.

Determine the standard deviation of the sets of scores using the statistics mode on the calculator. Round the answer to two decimal places. a

12, 18, 25, 7, 10, 15, 20, 13, 17

b

− 17, 2, − 6, 9, − 17, − 9, 3, 8, 5

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4

The shoe size of 16-year-olds are normally distributed. Which is closest to the percentage of these shoe sizes that is within 3 standard deviations of the mean? D

99.7%

For the given data set, the mean is 55 and the standard deviation is 30.

Stem

A 5

0.3%

B

95%

C

5%

Describe each interval in relation to the mean: a

25 to 85

b

– 15 to 115

c

– 45 to 145

Leaf

0

47

1

01 79

2

4 88

3

5

4

47

5

3 466

6

1 1 57

7

05

8

3 457

9

3 4688

Key 1|2 = 12

Practice 6

i

Determine the sample variance of each set of scores, rounded to four decimal places.

ii

Determine the sample standard deviation of each set of scores, rounded to two decimal places.

a

20, 44, 27, 25, 21, 28, 41, 24, 27, 39, 35, 43, 30, 17, 40

b

3, 14, 11, 17, 3, 18, 15, 6, 17, 15

For each data set, determine: i

Population standard deviation

ii

a

20 18 16 14 12 10 8 6 4 2 0 18 19 20 21 22 23 24

b

Sample standard deviation

Frequency

7

For each data set, use technology to:

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25

26

27

28

29

30

Ex 2

8

a

Complete the table: Class

Class centre

Frequency

fx

1−9

⬚

8

⬚

10 − 18 19 − 27 28 − 36 37 − 45 Totals

9

⬚

6

⬚

8

⬚

⬚ ⬚

4

⬚

⬚

6

⬚

⬚

⬚

b

Use the class centres to estimate the mean, rounded to two decimal places.

c

Use the class centres to estimate the standard deviation, rounded to two decimal places.

d

If the original ungrouped data were used to calculate the standard deviation, would the ungrouped data have a higher or lower standard deviation? Explain your answer.

Consider the normally distributed data sets shown:

Ex 3

Ex 4

10

62 66 70 74 78 82 86

75 76 77 78 79 80 81

Distribution A

Distribution B

a

Which data set has a higher mean?

b

Which data set has a smaller standard deviation?

Consider the dot plot with mean of 25.14 and standard deviation of 1.36: a

Determine the actual percentage of data that lie within 1 standard deviation of the mean.

b

Determine the percentage of data that lie within 2 standard deviations of the mean. 23

11

The dot plot shows the number of sit-ups completed by students in a Physical Education exam. The mean number of sit-ups was 27.73 and the standard deviation was 11.59 a

Determine the actual percentage of data that lie within 1 standard deviation of the mean. Round your answer to one decimal place.

b

Determine the actual percentage of data that lie within 2 standard deviations of the mean.

24

25 26 Sample

27

Number of sit-ups per student

0 5 10 15 20 25 30 35 40 45 50 55

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12

For the given stem plot, determine the values, rounded to two decimal places:

Stem Leaf

a

Mean

b

Standard deviation

c

Percentage of data that lie within 1 standard deviation of the mean.

d

Percentage of data that lie within 2 standard deviations of the mean.

1

339

2

38

3

4

4

479

5

58

6

8

7

5

8

3

9

7

Key 1|2 = 12 13

The table shows the number of goals scored by a football team in each game of the year: a

In how many games were 0 goals scored?

b

Determine the median number of goals scored.

Score

Frequency

c

Calculate the mean number of goals scored, rounded to two decimal places.

0

3

d

Calculate the standard deviation, rounded to two decimal places.

1

1

2

5

3

1

4

5

5

5

e f

14

Determine the percentage of data that lie within 1 standard deviation of the mean. Determine the percentage of data that lie within 2 standard deviations of the mean.

The histogram and statistics for a data set are shown: n

25

Mean

22.5

σ

10.4019

s

10.6164

Σx 2

134

562.5

Σx

15361.25

Min

2

Q1

14.5

Median

22.5

Q3

30.5

Max

43

y 5 4 3 2 1 x 0 0

5 10 15 20 25 30 35 40 45 50

a

What is the range of scores that fall within 1 standard deviation of the mean?

b

Approximately what percentage of scores lie within 1 standard deviation of the mean?

c

What is the range of scores that fall within 2 standard deviations of the mean?

d

Approximately what percentage of scores lie within 2 standard deviations of the mean?

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A normally distributed data set has a mean of 0 and a standard deviation of 0.75. Which is the closest to the percent of values between –1.5 and 1.5? A

16

34%

B

95%

D

68%

The number of offspring in litters of puppies is normally distributed, as shown:

3

17

50%

C

5

7

9

11

a

Determine the percentage of litters that have no more than 9 puppies.

b

Determine the percentage of dogs that have at least 7 puppies.

c

Determine the percentage of puppies that are within 2 standard deviations of the mean.

d

99.7% of the litters will be between what numbers of puppies?

The mean income of people in Country A is $19 069. This is the same as the mean income of people in Country B. The standard deviation of Country A is greater than the standard deviation of Country B. In which country is there likely to be the greatest difference between the incomes of the high-income and low-income?

Ex 5

18

The operating times of phone batteries are approximately normally distributed with a mean of 34 hours and a standard deviation of 4 hours. a

Construct a normal distribution curve that models the operating times of phone batteries and label the boundaries for the empirical rule.

b

Determine the percentage of batteries that last between 22 and 42 hours.

c

Any battery that lasts less than 22 hours is deemed faulty. The company CEO asked one of the managers. “If we manufactured 51 000 batteries, approximately what number of batteries would we be able to sell?” Answer the CEO’s statistical question.

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Extend your thinking 19

Consider the test results for four Geography classes, labelled as Class W, Class X, Class Y, and Class Z:

Class W

10 9 8 7 6 5 4 3 2 1 0

10 9 8 7 6 5 4 3 2 1 0 0

Class Y

2

4

6

0

8 10

10 9 8 7 6 5 4 3 2 1 0

Class X

2

4

6

8 10

10 9 8 7 6 5 4 3 2 1 0 0

2

4

6

8 10

Class Z

0

2

4

6

8 10

Rank the classes from largest standard deviation to smallest standard deviation. Explain your reasoning. 20

21

Seven millionaires with an average net wealth of $41 million with a standard deviation of $7 million are having a party. Suddenly Carlos Slim, who has a net wealth estimated to be $31 billion, walks into the room. a

Determine the new average net wealth in the room. Round your answer to the nearest million.

b

Will the new standard deviation be higher, lower or unchanged from before?

c

Will the new mode be higher, lower or unchanged from before if at least two of the original seven millionaires have the same net wealth?

d

Will the range be higher, lower or unchanged from before?

The scores obtained by two classes are given: • Red Class: 55, 57, 49, 58, 68, 57, 60, 53, 56, 51 • Blue Class: 53, 57, 62, 51, 56, 62, 58, 55, 58, 51

136

a

Which class performed better on average? Use statistical calculations to justify your answer.

b

Which class produced more consistent results? Use statistical calculations to justify your answer.

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23

Han, a cricketer, has achieved scores of 52, 20, 68, 70 and 150 in his first five innings this season. In his sixth innings, he scored 0. a

Describe how his season batting average changed from before to after the sixth inning.

b

Describe how his standard deviation changed from before to after the sixth inning.

c

Describe how his median score changed from before to after the sixth inning.

d

Describe how his range changed from before to after the sixth inning.

The table shows the heart rate data of a group of people after exercise on the Stair Champion 10 000:

a

Height of step

Stepping rate

Heart rate

Short step

Slow

89

Short step

Slow

91

Short step

Medium

106

Short step

Medium

105

Short step

Fast

124

Short step

Fast

128

Tall step

Slow

100

Tall step

Slow

96

Tall step

Medium

125

Tall step

Medium

129

Tall step

Fast

132

Tall step

Fast

127

Complete this table. Round all values to one decimal place. Height of step Short step Tall step

Data

Slow

Medium

Fast

Average heart rate

90.0

Standard deviation of heart rate

1.0

Average heart rate

⬚

⬚

⬚

⬚

⬚

Standard deviation of heart rate

⬚

⬚ ⬚

⬚ ⬚

b

Which of the combinations of step height and stepping rate generated the higher heart rate? Quote relevant statistics in your answer.

c

Which combination of step height and stepping rate showed the least variability? Quote relevant statistics in your answer.

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1.10   Compare sets of data After this lesson, you will be able to… • construct back-to-back stem plots to compare data distributions across two groups. • construct parallel box plots to compare data distributions across two or more groups. • compare distributions across groups using centre (mean, median) and spread (range, IQR, standard deviation, outliers). • interpret differences between data sets in context using appropriate statistical language.

Compare data sets The measures of centre, measures of spread and shape of the data are all used to analyse data sets. Measures of centre summarise a data set and indicate where the majority of the data lies. Advantages

Drawbacks

Mean

Includes all of the data in the calculation, widely used

Heavily impacted by outliers

Median

Indicates the middle value, not impacted by outliers

Does not include all data values

Mode

Quick to identify, indicates the most frequent value(s)

Does not include all data value, is not necessarily representative of the data

Measures of spread tell us about the variability and consistency. Advantages

Drawbacks

Easy to calculate, utilises the extremes of the data

Heavily impacted by outliers

Interquartile range

Indicated the middle 50% of the data, not impacted by outliers

Does not include all data values

Standard deviation

Indicates how far values are from the mean, widely used in other areas of statistics

Impacted by outliers, difficult to calculate without technology

Range

The shape of the data provides insight into the distribution of values within a data set, which differs from measures of centre or spread. While measures of centre summarise the central tendency of the data, and measures of spread describe how dispersed the data points are, the shape of the data indicates the overall pattern of distribution.

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Advantages

Drawbacks

Identifies whether the data is evenly distributed or if it has a tail on one side.

Does not provide insight into the spread or exact values of the data.

Modality

Reveals whether the data has one or multiple peaks, which can suggest distinct groups or patterns.

Modality focuses on peaks, which may overlook important information about the tails of the distribution.

Outliers

Outliers are critical in detecting Focusing too much on outliers can anomalies or unusual observations detract from understanding the that could influence results. overall pattern of the data.

Skew/symmetry

Describing and comparing data sets using multiple measures of centre, spread and shape, provides an accurate and comprehensive understanding of the data’s characteristics. By incorporating various measures such as the mean, median, standard deviation, range and skewness, the strengths of one measure can offset the weaknesses of another. This multi-faceted approach minimises the limitations of relying on a single measure, allowing for a more balanced assessment of the data. As a result, it leads to better-informed conclusions about data sets. Using a variety of well-chosen summary statistics makes it easier to compare data sets.

Example 1 The given data sets show the time to get to school for 20 students at two different schools, rounded to the nearest minute. Some summary statistics are given: School A 3

4

5

8

9

11

11

11

12

13

14

15

15

18

19

27

33

33

42

45

School B 15

15

17

17

17

18

18

18

18

19

19

22

22

22

23

23

23

24

24

24

School A

School B

Mean

17.4

19.9

Median

13.5

19

Variance

142.14

9.09

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a Interpret and compare the means for both schools.

Create a strategy

Apply the idea

The mean is also called the average and describes a typical value that balances all of the data points. Check the difference between the means provided in the summary.

From the given summary, it shows that School A has a lower mean of 17.4 compared to School B that has a mean of 19.9. This shows that the typical travel time for students going to School A is shorter by 2.5 minutes compared to students going to School B.

b Interpret and compare the medians for both schools.

Create a strategy Median is the middle value of a data set, which is not impacted by extreme values. Check the difference between the medians provided in the summary.

Apply the idea

Reflect and check

It can be seen from the given summary that School A has a lower median of 13.5 compared to School B that has a median of 19. This shows that the typical travel time for students going to School A is shorter by 5.5 minutes compared to students going to School B.

The two measures of centre tell the same story that School B students have a longer journey to school than School A students. However, the median makes this conclusion more obvious because the difference is larger.

c Calculate the standard deviations for both schools.

Create a strategy Use the given variance to find the standard deviation which is the square root of the variance.

Apply the idea

Variance

School A

School B

142.14

9.09

Get the square root: School A

School B

Variance

142.14

9.09

Standard deviation

142.14 11.92

9.09 3.01

Reflect and check The units for the standard deviation are the same as the original context, so the standard deviation for School A is 11.92 minutes and the standard deviation for School B is 3.01 minutes.

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d Interpret and compare the standard deviations for both Schools.

Create a strategy Observe the difference between the standard deviations for School A and School B. Remember that the standard deviation quantifies how spread out the values in the data set are relative to the mean value.

Apply the idea The standard deviation for School A is 11.92 while it is 3.01 for School B. This suggests that the time students take to get to School A have greater variability or the values are more spread out, compared to School B. On average, there is 11.92 minutes of difference from the mean for School A, indicating higher variability. On the other hand, there is less variability or the values are closer to each other for School B, with an average difference from the mean of 3.01 minutes.

Reflect and check The higher standard deviation or variability for students travelling to School A suggests that they may have encountered heavier traffic conditions, frequent road closures, or other factors that contribute in increased travel time.

Idea summary The measures of centre, measures of spread and shape of the data are all used to analyse data sets. By incorporating multiple measures, the strengths of one measure can offset the weaknesses of another, leading to better informed conclusions about data sets. When comparing data sets: • To determine which data set scored higher, compare the measures of centre. • To determine which data set is more consistent, compare the measures of spread.

Compare data displays Data displays are helpful because they allow for a visualisation of a data set which is often less cumbersome than looking at a long list of data values. For numerical data, the most common displays are: • Histograms - used for continuous data and clearly show shape, range, clustering/modality, and mode (modal class) • Box plots - used for numerical data and clearly show shape, spread, outliers, median, and the five-number summary • Stem plots - used for discrete data and show the shape, range, and the raw data values • Dot plots - used for small discrete data and show the shape, range, mode, and the raw data

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All displays also have disadvantages. Based on the advantages and disadvantages of each display, it may be easier or more difficult to compare two data sets using the displays. Histograms Advantages

Disadvantages

Good for large quantities of data

Individual data values are lost

Easy to read off the spread, clusters, and trends

Bin size can affect the conclusion

We do not need to round when collecting data Box plots Advantages

Disadvantages

Good for large quantities of data

Individual data values are lost (other than maximum and minimum)

Easy to read off shape, interquartile range, and outliers

Cannot easily see gaps or trends

Median is clearly shown Stem plots Advantages

Disadvantages

Data is sorted

Difficult to show large quantities of data

Keeps the original data

Not helpful if the data only varies in the last digit

Can see some clusters in the data Dot plots (line plots) Advantages

Disadvantages

Simple to create

Difficult to show for large quantities of data

Can read off the most and least common response

Difficult to show a large number of categories or a large spread

Can see some clusters in the data To compare two or more sets of data with data displays, it is important to be able to identify key features and compare and contrast them. If outliers are identified in the data displays, then it is important that these are mentioned in the comparison, with an explanation of how they are handled. In particular, a statement should be made if outliers are included or excluded from comparisons, and the effect that this has on the analysis.

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Example 2 Data on player height was collected from the Year 8 and the Year 12 basketball teams: Year 8

10

Number of students

Number of students

Year 12

8 6 4 2 0

155 160 165 170 175 180 185 190 195 200 Height (cm)

10 8 6 4 2 0

135 140 145 150 155 160 165 Height (cm)

a Compare and contrast the shape, modality, and presence of outliers for the two teams.

Create a strategy Identify the overall shape of the distributions. Then determine whether the data is single modal or multimodal and identify any outliers that stand out from the main distribution.

Apply the idea Both distributions are approximately symmetrical and single modal. The modal class for Year 12 (170 – 175 cm) is higher than the modal class for Year 8 (150 – 155 cm). There is a possible outlier in Year 12 around 195 – 200 cm, but Year 8 has no significant outliers.

b Compare the overall heights of the two teams.

Create a strategy Compare the mean, median, or modal class of both distributions. Consider whether the data is clustered around a certain height range.

Apply the idea Year 12 has a higher overall height distribution, with most students clustered around 170 – 180 cm. Year 8’s height distribution is lower, mostly between 145 – 160 cm. The modal class for Year 12 (170 – 175 cm) is much higher than for Year 8 (150 – 155 cm), confirming that Year 12 consists of older students.

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c Compare the variability within each team.

Create a strategy Assess the range and observe how spread out the data is from the centre.

Apply the idea If the outlier values are ignored in Year 12 (195 – 200 cm), both distributions have similar ranges (35 cm for Year 12 and 30 cm for Year 8). The heights for Year 12 appear to be more spread out overall compared to Year 8. Year 12’s data is slightly more dispersed, suggesting greater height variability among older students.

Idea summary Data sets can be compared using data displays. The data displays can be the same or different. The numeric or descriptive information that can be determined from the display varies by display. Strengths Histograms

Box plots

Stem plots

Dot plots

144

Drawbacks

Easily display large or spread out Cannot see individual data values. data sets. The shape, mode, and Depending on the number or size of spread are visible. the bins or intervals, the shape can look different. Easily display large or spread out Cannot see individual data values or gaps data sets. The shape, median, in data spread. interquartile range, and outliers are visible. Useful for individual data values. Do not work well for large data sets. Can The highest and lowest values be slower to make by hand. are easy to see. Useful for individual data Do not work well for large data sets. Can values. The highest and lowest be slower to make by hand. categories are easy to see.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Parallel box plots Parallel box plots Used to compare two or more sets of data visually. These box plots are presented parallel to each other along the same number line using the same scale. When comparing box plots, the five-number summary is important to consider. Other statistics can be derived such as the range and interquartile range, and visual observations can be made of symmetry, skew, and outliers that should be considered in any comparisons. The term parallel is used because the box plots are presented parallel to each other along the same number line for comparison. They must therefore be on the same scale, so a visual comparison is fairly straightforward. Key comparisons for box plots: When comparing two sets of data, compare the five-number summary as shown. There are key questions that should be asked: • How do the spreads of data compare? • How do the skews compare? Is one set of data more symmetric? • Is there a big difference in the medians? • Can we see regions on one box plot that extend past the comparable region on the other? Always consider what factors are more important for the given situation. In some cases, it might be necessary to make judgements by simply comparing the median value; sometimes the minimum or the maximum value is the critical measurement. In other situations, the consistency will be more important than extreme values so it is also worth considering measures of spread to make judgements. If outliers are identified in the box plots, then it is important that these are mentioned in the comparison, with an explanation of how they are handled. That is, there should be a statement if outliers are included or excluded from comparisons, and the effect that this has.

Example 3 Data was collected on the time it took for people to complete an online task. The data was grouped by age using parallel box plots: 25

18 20

12

5 Under 30s

16

6

22

27

30

Over 30s

0

2

4

6

8

10 12 14 16 18 20 22 24 26 28 30 Seconds

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a Compare the five-number summary for the two groups.

Create a strategy Determine the minimum and maximum scores, the lower and upper quartiles, and the median of the data set for each age group.

Apply the idea Under 30s

Over 30s

Minimum

5

6

Q1​

12

16

Median

18

22

Q3

20

27

Maximum

25

30

It can be seen that each statistic for the ‘Under 30s’ group is lower than that of the ‘Over 30s’ group, indicating faster completion times on average for the younger group. b Which group was more consistent? Explain.

Create a strategy Determine and compare the ranges (difference between maximum and minimum) and interquartile ranges (difference between Q3 and Q1) of the two groups.

Apply the idea Under 30s

Over 30s

Range

25 – 5 = 20

30 – 6 = 24

IQR

20 – 12 = 8

27 – 16 = 11

The ‘Under 30s’ group was more consistent because it has both a smaller range and IQR, which indicate greater consistency in completion times compared to the ‘Over 30s’ group.

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c Which statement is incorrect? A

100% of the under 30s completed the task before the slowest 25% of the over 30s.

B

t least 75% of the under 30s completed the task in under 22 seconds, which is the median A time for the over 30s.

C

They both have a positive skew.

D

The under 30s had more consistent times than the over 30s

Create a strategy Analyse each statement based on the data provided in the box plots and the properties of distributions such as skewness.

Apply the idea All ‘Under 30s’ completed the task before the slowest 25% of the ‘Over 30s’. So Option A is true. 75% of ‘Under 30s’ finished in 20 seconds or less, while the median for ‘Over 30s’ is 22 seconds. So Option B is true. The box plots show that the ‘Over 30s’ have a distribution that extends further right relative to the median compared to the ‘Under 30s’, indicating a positive skew for the ‘Over 30s’ and not necessarily for the ‘Under 30s’. So Option C is false. Option D is true based on the IQR and range discussed. So, the incorrect statement is option C.

d Overall, which group performed better?

Create a strategy

Apply the idea

Evaluate overall performance by comparing all aspects of the box plots, focusing on measures of centre and data distribution.

Every one of these measures is in favour of the under 30s so, overall, it can be concluded that the under 30s performed better.

Idea summary Parallel box plots are used to compare two or more sets of data visually. These box plots are presented parallel to each other along the same number line using the same scale. When comparing two box plots, compare the five-number summary and overall distribution. There are key questions that could be asked: • • • •

How do the spreads of data compare? Are there any outliers? How do the skews compare? Is one set of data more symmetrical? Is there a big difference in the medians? Can we see regions on one box plot that extend past the comparable region on the other?

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Back-to-back stem plots Back-to-back stem plots Allow for the display of two data sets at the same time using a common stem. These types of plots are a great way to make comparisons between data sets.

Group A

Group B

73

1

036

50

2

1 678

655

3

556

1 1

4

1 1 569

84 3

5

0368

Key 3|1 = 13

One set of data is shown with its leaves on the left and the other on the right. The leaves are written in ascending order, moving outward from the stem.

Key 1|0 = 10

Example 4 The back-to-back stem plots show the number of pieces of paper used over several days by Maximillian’s and Charlie’s students: Maximillian’s students

Charlie’s students

7

0

7

3

1

1 23

8

2

8

43

3

234

765

4

9

32

5

2

Key 2|1 = 21

Key 6|1 = 61

Which of these statements are true? I. Maximillian’s students did not use 7 pieces of paper on any day. II. Charlie’s median is higher than Maximillian’s median. III. The median is greater than the mean in both groups.

148

A

I and II

B

II and III

C

None of the statements are correct

D

III only

E

II only

F

I only

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Create a strategy Examine both stem plots and assess the validity of each statement.

Apply the idea Statement I: Based on the stem plot of Maximillian’s students, they used 7 pieces of paper on any day. This means that statement I is incorrect. Statement II: Both groups have 10 data points. For Charlie’s students, the median lies between 28 and 32 as these are the 5th and 6th data points, respectively. For Maximillian’s students, the median lies between 34 and 45 as these are the 5th and 6th data points, respectively. Calculating the median for each group: Maximillian’s median =

34 45 2

= 39.5 Charlie’s median =

28 32 2

= 30 This means that Maximillian’s median is higher than the median of Charlie’s, so statement II is incorrect. Statement III: Calculating the mean for each group, Maximillian’s mean =

7 13 28 33 34 45 46 47 52 53 10

= 35.8 Charlie’s mean =

7 11 12 13 28 32 33 34 49 52 10

= 27.1 Comparing the calculated mean of both groups with their respective median, the median is greater than the mean in both groups. This means that the statement III is correct. So, the correct answer is option D, III only.

Idea summary A back-to-back stem plot are very similar to a regular stem plots, in that the “stem” is used to group the scores and each “leaf” indicates the individual scores within each group. When a stem plot is being created, it is easier for the scores to be written in ascending order before they are placed into a back-to-back stem plots.

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1.10 Practice questions What do you remember? 1

Marge grows two different types of bean plants. She records the number of beans that she picks from each plant for 10 days: • Plant A: 10, 4, 4, 5, 7, 10, 3, 3, 9, 10 • Plant B: 8, 7, 5, 5, 9, 7, 8, 7, 5, 6

2

a

Determine the mean number of beans picked per day for Plant A.

b

Determine the mean number of beans picked per day for Plant B.

c

Determine the range for Plant A.

d

Determine the range for Plant B.

e

Which plant produces more beans on average? Explain your answer.

f

Which plant has a more consistent yield of beans? Explain your answer.

The parallel box plots show the data collected by the manufacturers on the lifespan of light bulbs, measured in thousands of hours: Manufacturer A

Manufacturer B

0

1

2

3

4

5

6

7

8

Thousands of hours a

Complete the table. Express each answer in terms of hours. Manufacturer A

Manufacturer B

⬚

⬚

⬚

⬚

Median Lower quartile Upper quartile Range Interquartile range b

150

⬚ ⬚ ⬚

⬚ ⬚ ⬚

Which manufacturer produces light bulbs with the best lifespan? Explain your answer.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

3

A mathematics test is given to two classes. The score out of 20 received by students in each class are represented in the box plots: Class 8P

Class 8Q

0

4

2

4

6

8

10

12

14

16

18 20 22

a

Which class tended to score better?

b

Which class(es) had an outlier?

c

Which class(es) had symmetrical shape?

d

Would the range or interquartile range be better for describing the spread of Class 8Q?

e

Which class had more consistent results?

The back-to-back stem plots show the batting scores of two cricket teams, A and B. Determine: Team A Team B a The median score of Team A. 762 6 268 b The median score of Team B. 8665 2 7 1 57 c The range of Team A’s scores. 84 8 1 4 79 d The range of Team B’s scores. 9 47 e The interquartile range of Team A’s scores. f

The interquartile range of Team B’s scores.

Key 6|1 = 16

Key 1|2 = 12

Practice 5

The residents of two blocks of townhouses were asked about the number of pets they owned. The frequencies of various responses are presented in the dot plots:

0

1

2 3 Block A

4

0

1

2 3 4 5 6 Block B

a

In which block is pet ownership lower?

b

How many pets do most households have in Block A?

c

How many pets do most households have in Block B?

d

Describe the shape of the data for Block A.

e

Determine the range for Block A.

f

Which block has more variability in the number of pets per household?

g

Does either block have an outlier?

1.10 Compare sets of data mathspace.co

151

6 Ex 1, 2

A group of students recorded the resting heart rates for 30 students and 30 teachers. The results are represented with the histograms: Resting heart rate of teachers 12

10

10

8

8

6

6

4

4

2

2

0

40 45 50 55 60 65 70 75

Teachers

7

Resting heart rate of students

12

0

40 45 50 55 60 65 70 75 Students

Mean

55.7

Mean

58.4

Median

56

Median

58

Range

21

Range

20

Interquartile range

6

Interquartile range

7

Standard deviation

4.3

Standard deviation

5.0

a

Compare the spread (dispersion) of the student and teacher data sets. Explain what this means in the context.

b

Compare the centres of the student and teacher data sets. Explain what this means in the context.

The pulse rates of two groups are given: • Group 1: 82, 85, 88, 65, 73, 89, 79, 90, 76, 68, 88, 65, 63, 62, 88, 82 • Group 2: 75, 88, 74, 73, 80, 76, 67, 81, 71, 83, 89, 62, 63, 80, 71, 78

152

a

Determine the mean pulse rate of Group 1, rounded to two decimal places.

b

Determine the mean pulse rate of Group 2, rounded to two decimal places.

c

Determine the standard deviation of Group 1, rounded to two decimal places.

d

Determine the standard deviation of Group 2, rounded to two decimal places.

e

What is the range for Group 1?

f

What is the range for Group 2?

g

Which group has the greater spread?

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

8

The median house price in the suburb of Humbleton is $950 000 with a mean price of $1 000 000 and the median house price in the suburb of Brockway is $950 000 with a mean price of $880 000. Which suburb is more likely to have very expensive houses? Explain your answer.

9

The hours of sleep per night for two people over a two-week period are shown:

a

Person A

8

5

10

7

9

7

6

10

6

9

7

7

10

5

Person B

8

8

8

7

7.5

8

7.5

7

7

7

7.5

7

7

7.5

Calculate the following (rounded to one decimal place where necessary) for Person A: i

b

ii

The median

iii

The mode

The range

iv

Calculate the following (rounded to one decimal place where necessary) for Person B: i

10

The mean

The mean

ii

The median

iii

The mode

The range

iv

c

Which person is the least consistent in their sleep habits? Explain your answer.

d

Which person has the most sleep over the 14 nights? Explain your answer.

The test scores of 11 students in Drama and German are listed: • Drama: 75, 85, 62, 65, 52, 76, 89, 83, 55, 91, 77 • German: 82, 86, 76, 84, 64, 73, 89, 62, 54, 69, 78 Construct parallel box plots to represent both data sets.

Ex 3

11

Two bookstores recorded the selling price of all their books. The results are presented in the parallel box plots: Bookstore A Bookstore B

50

60

70

80

90

100

110

120

130

140

150

Prices ($) a

Which bookstore had the more consistent prices?

b

Comparing the most expensive books in each store, how much more expensive is the one in Bookstore B?

c

True or False: 25% of the books in Bookstore B are at least as expensive as the most expensive book in Bookstore A.

d

True or False: 25% of the books in Bookstore B are cheaper than the cheapest book in Bookstore A.

1.10 Compare sets of data mathspace.co

153

12

The parallel box plots show the distances, in centimetres, jumped by two high jumpers: Matt

Paul

150

140

160

180

170

Distance (cm)

13

a

Who had a higher median jump?

b

Who made the highest jump?

c

Who made the lowest jump?

d

Who has the larger range?

The parallel box plots show the prices, in dollars, of the items on the menu of an upmarket restaurant and the menu of a fast food restaurant: Upmarket restaurant

Fast food

0

10

20

30

40

50

60

Price

154

a

Which restaurant has the higher median price for the items they sell?

b

What is the difference between the median prices of the items sold by each restaurant?

c

Which restaurant has a greater price range for the items on their menu?

d

What is the price difference between the most expensive items sold by each restaurant?

e

What amount of the cheapest item at the fast food restaurant could be bought for the same price of the most expensive item at the upmarket restaurant?

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Ex 4

14

The back-to-back stem plots show the number of books read in a year by a random sample of university and high school students. Which of these statements are true? University

High school 7

0

663

1

0035

432 1

2

1 2 446

9886

3

1 89

82

4

01

5 6 3

7

Key 2|1 = 12 books

15

a

Compare the medians of both groups.

b

Compare the range of both groups.

c

Which group of students read more books? Explain your answer.

The back-to-back stem plots show the length (in minutes) of a random sample of phone calls made by Sharon and Tricia: a

16

Key 1|2 = 12 books

Determine Sharon’s mean, rounded to one decimal place.

Sharon

Tricia 3

1

34

76432

2

678

98

3

24

b

Determine Sharon’s median.

43

4

1 2

c

Determine Tricia’s mean, rounded to one decimal place.

76

5

678

d

Determine Tricia’s median.

e

Who generally makes slightly longer phone calls?

The back-to-back stem plots show the number of desserts ordered at Hotel A and Hotel B over several randomly chosen days:

Key 2|1 = 12 minutes

Key 1|2 = 12 minutes

Hotel A

Hotel B

3

0

432

1

34

a

Interpret the lowest score for Hotel A.

b

Which hotel’s median is higher?

76

2

7

Is the mean greater than the median in both groups?

43

3

34

6

4

67

2

5

234

c

Key 2|1 = 12

Key 1|3 = 13

1.10 Compare sets of data mathspace.co

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17

Two friends compete in hammer throw competitions and train together over a season. They compete in 15 competitions and their final throw for each competition is shown: • Tim: 29.8, 37.4, 33.9, 38.8, 34.3, 36.5, 34.5, 30.0, 35.2, 38.4, 33.0, 33.2, 39.6, 35.0, 36.9 • Odi: 32.2, 35.4, 34.8, 33.0, 38.4, 26.0, 40.0, 37.2, 39.5, 42.4, 38.6, 42.3, 38.4, 42.8, 37.2 a

Complete this table of statistics: Tim

Odi

Minimum

29.8

Q1

33.2

⬚

Median

35.0

Q3

37.4

Maximum

39.6

Mean (1 d.p.)

37.2

Sample standard deviation (2 d.p.)

⬚

2.93

Range

⬚

⬚

⬚

Interquartile range

⬚ ⬚ ⬚ ⬚ ⬚ ⬚

b

Which competitor throws more consistently? Explain your answer.

c

Identify any outliers and use statistical calculations to justify your answer.

d

Create parallel box plots of the two sets of data with the outlier(s) displayed separately.

e

Who is the better hammer thrower? Explain your answer.

f

When considering Odi’s average throw is it reasonable to remove the outlier before calculating the mean? Explain your answer.

Extend your thinking 18

156

The ages of employees at two competing fast food restaurants on a Saturday night are recorded. Some statistics are given in the table: Mean

Median

Range

Berger’s Burgers

18

17

6

Fry’s Fries

18

19

2

a

If the data for Berger’s Burgers was represented using a histogram, would it be positively or negatively skewed?

b

Which restaurant has the oldest employee on the night the data is recorded?

c

Which restaurant has the most consistent ages among employees? Explain your answer.

d

Which restaurant has an older workforce? Explain your answer.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

19

The weight (in kilograms) of a group of men and women were recorded and presented in a back-to-back stem plots as shown: Men

Women 5

01 234445557

988766 653

6

0223477 8

64322 1 0000

7

0

0

8

Key 2|4 = 42 kg

20

Key 4|2 = 42 kg

a

Find the mean weight of the group of men.

b

Find the mean weight of the group of women.

c

Which group is heavier overall? Explain your answer.

A cinema is showing three films, labelled A, B and C. The ages of people watching each of the films are illustrated in the parallel box plots: A B C 0

21

10

20

30

40

50

60

a

Which film do you think has an adults only rating, restricting it to viewers 18 years of age and older? Explain your answer.

b

Which film would be recommended for a group of 15-year-olds to watch? Explain your answer.

c

Which film would you recommend to a family of two parents in their 40’s and two teenagers? Explain your answer.

Cooper and Marion are racing go-karts. The times (in seconds) for the 12 laps of their qualifying race are shown: • Cooper: 58.9, 46.5, 52.6, 66.6, 58.4, 53.1, 45.0, 52.1, 52.4, 52.7, 44.8, 51.7 • Marion: 47.8, 54.6, 68.5, 68.0, 62.8, 57.2, 54.8, 63.4, 58.1, 64.3, 66.2, 47.1 a

Construct the five-number summary for each set.

b

Identify any outliers and use statistical calculations to justify your answer.

c

Create a parallel box plots of the two sets of times with the outlier(s) displayed separately.

d

Which racer will be in pole position for the final race, if it is given to the racer with the fastest qualifying lap time?

e

Does spinning out on a lap, causing a high outlier, impact the selection for pole position? Explain your answer.

1.10 Compare sets of data mathspace.co

157

22

23

These box plots summarise results from a medical study. The treatment group received an experimental drug to relieve cold symptoms, and the control group received a placebo. The box plots show the number of days each group continued to report symptoms: Control group

Treatment group

0 2 4 6 8 10 12 14 16 18 20

0 2 4 6 8 10 12 14 16 18 20

a

Describe the shape of the data from the control group.

b

Describe the shape of the data from the treatment group.

c

Does the drug have a positive effect on patient recovery? Explain your answer.

The box plots drawn shows the number of repetitions of a 70 kg bar that Weightlifter A and Weightlifter B can lift. They both record their repetitions over 30 days:

A

B 0

2

4

6

8 10 12 Repetitions

14

16

a

Which weightlifter has the more consistent results? Explain your answer.

b

Which weightlifter can do the most repetitions of the 70 kg bar? Explain your answer.

Chapter 1 review

Online review

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

mathspace.co

“Mathematics is a place where you can do things which you can’t do in the real world.” – Marcus du Sautoy

Big ideas • Linear relationships are characterised by a constant rate of change and can be represented algebraically as equations or visually as straight-line graphs. • The key features of a linear graph, specifically its slope and intercepts, provide critical information for interpreting, sketching, and defining the unique equation of the line.

2 Linear relationships Chapter outline 2.01 2.02 2.03 2.04 2.05 2.06

Solve linear equations Linear equation applications Graphs from tables Slopes and intercepts Sketch linear functions Equations of straight lines Investigation: Filling a pool Chapter 2 review

162 168 175 187 200 208

The slope of a line isn’t just math; it represents change! A steeper slope means a faster change – just like climbing a hill!

2.01

Solve linear equations After this lesson, you will be able to… • solve multi-step linear equations with rational coefficients and constants. • verify solutions using appropriate technology. • apply inverse operations to solve equations systematically.

Solve linear equations Solving linear equations involves applying two fundamental principles: • Using inverse (or opposite) operations • Keeping both sides of the equation balanced Inverse operation An operation that is used to reverse the mathematical operations applied to the variable. Inverse operations allow the equation to be simplified step by step until the variable is isolated. Operation

Inverse operation

Example

Addition

Subtraction

x+4−4=x

Subtraction

Addition

x–3+3=x

Multiplication

Division

y×4÷4=y

Division

Multiplication

y÷2×2=y

Applying the same mathematical operation to both sides of the equation keeps it balanced and ensures both sides remain equal, much like keeping a scale balanced. The objective is to determine the value of the variable that makes both sides of the equation equal, ensuring the solution is valid. This method is effective for equations requiring either single or multiple steps to isolate the variable. To solve multi-step equations, consider how the equation is built, then apply inverse operations in reverse order to both sides of the equation until the variable is isolated. This process is also known as making a variable the subject of an equation.

Interactive exploration

Discover this concept in action online

Equations can be solved and solutions can be verified using technology.

162

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

mathspace.co

Example 1 Solve each equation. a 24 + p = 59

Create a strategy

Apply the idea 24 + p = 59

The inverse of addition is subtraction, therefore, subtract 24 from both sides to isolate the variable.

b

Write the equation

p = 35 Subtract 24 from both sides

u 7 5 2

Create a strategy Apply the inverse operations in the reverse order of how the equation was built.

Apply the idea The equation was built by: 7 u

2 u 7

u 7 2

The order of inverse operations will be: 1. Multiply by 2. 2. Subtract 7. u 7 5 2

Write the equation

u + 7 = 10

Multiply both sides by 2

u=3

Subtract 7 from both sides

Reflect and check Substitute u = 3 into the original equation to check. u 7 5 2

Write the equation

3 7 5 2

Substitute u = 3

10 5 2

Evaluate the addition

5=5

Evaluate and check that the two sides are equal

2.01 Solve linear equations mathspace.co

163

c – 3 (–2h – 5) = 33

Apply the idea The equation was built by: × –2 h

× –3

–5 – 2h

– 2h – 5

– 3 (– 2h – 5)

So, the order of inverse operations will be: 1. Divide by – 3. 2. Add 5. 3. Divide by – 2. – 3 (– 2h – 5) = 33

Write the equation

– 2h – 5 = – 11

Divide both sides by –3

– 2h = – 6

Add 5 to both sides

h=3

Divide both sides by – 2

Reflect and check This equation can also be solved using technology by entering the equation into a solve function or command.

d

5x 9 4 8

Apply the idea The equation was built by: ×5

x

÷8 5x

–9 5x 8

5x –9 8

So, the order of inverse operations will be: 1. Add 9. 2. Multiply by 8. 3. Divide by 5. 5x 9 4 8 5x 5 8 5x = 40 x=8

164

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Write the equation Add 9 to both sides Multiply both sides by 8 Divide both sides by 5

Idea summary A linear equation is a type of equation where the highest power of the variable is one. Solving linear equations involves applying two fundamental principles: • Using inverse (or opposite) operations • Keeping both sides of the equation balanced To solve multi-step equations, consider how the equation is built, then apply inverse operations in reverse order to both sides of the equation until the variable is isolated. Equations can also be solved using technology.

Equations with variables on both sides Inverse operations can be used to combine terms for equations with variables on both sides. Once all the terms containing the variable are on one side, that side can be simplified by collecting like terms. Then the equation can be solved like a multi-step equation applying the inverse operations in reverse order. Best practice is to eliminate the variable term with the smaller coefficient to avoid unnecessary introduction of negative coefficients. To keep the equation balanced when moving terms, remember to do the same operations to both sides of the equation.

Example 2 Digital only example

Explore in the online lesson

mathspace.co

Example 3 Solve the equation 4 (x + 3) − 2x = 3 (x − 1).

Create a strategy First, expand the brackets on both sides then collect like terms.

Apply the idea 4 (x + 3) − 2x = 3 (x − 1)

Write the equation

4x + 12 − 2x = 3x − 3

Expand the brackets on both sides

2x + 12 = 3x − 3

Combine like terms on the left-hand side

12 = x − 3

Subtract 2x from both sides

15 = x

Add 3 to both sides

x = 15

Rewrite

2.01 Solve linear equations mathspace.co

165

Idea summary To solve an equation with variables on both sides, begin by using inverse operations to move all variable terms to one side. Then, continue applying inverse operations in reverse order to isolate the variable. Expanding any brackets may be necessary before starting to solve the equation.

2.01 Practice questions What do you remember? 1

For each of the equations, identify the inverse operation that must be applied to both sides to solve the equation: 1 3 x a x–4=6 b 4x = 92 c 7 d x 5 5 5

2

For each pair of operations:

3

i

Determine the expression if one starts with x, and applies the two given operations.

ii

Determine the inverse operations in reverse order of application.

a

Add 6 then multiply by 4

b

Multiply by – 3 then subtract 7

c

Subtract 9 then divide by 3

d

Divide by 11 then add 8

For each equation: i

Determine the inverse operation needed to make the variable the subject of the equation.

ii

Solve the equation.

a

y+5=9

n 4 9

b

9p = 36

c

t – 3 = – 45

d

n + 27 = 11 t 1.6 5

d

6 = – 16 + n v 5 5

Practice 4

Ex 1a

5

Solve: a

x – 4 = 10

b

3x = 18

c

e

t – 15 = 0

f

9t = 3

g

b

4x − 6x = 4

c

Solve: a

166

h

3x + 11 = 5

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

− 10 + 3k = 5

d

8m + 9 = 65

Ex 1b

Ex 1c

6

7

Solve: 5x 11 a 6

8

9

10

Ex 3

11

12

m 6 4.25 4

d

n 36 11 2

b

3 (q + 3) = – 18

c

2 (p – 5) = – 12

d

7 (11 – n) = 210

4

b

3x 8

12 15

c

d

8 p 13 8

b

6x = 5x + 37

c

10x = 9x + 50

d

39x + 125 = 40x

b +5 3

1 3

7 4

4

7x = 6x + 2

Solve and verify with technology: a

2x – 9 = 5x

b

7x = x + 30

c

3x = x – 18

d

8x = 6x + 10

e

7x = 3x – 12

f

3x = − 8x − 44

g

4x = − x + 5

h

6x = 16 − 2x

Solve: a

6y − 3 (y + 4) = − 12

b

4x + 3 − 4 = 8 1 x 1 2

c

5 (x − 2) − 3x = 2 (x + 5)

d

13 = 4m − (2m − 7)

b

3y 3 3

d

3.9 k 28 5 4

Solve each equation and justify each step:

c

14

c

Solve:

a

13

2 (q + 11) = 20

Solve: y 2 a 6

a

Ex 2

x 15 5 2

Solve: a

Ex 1d

b

14 k 76 2 2k 2 5x 2 2x 8 3

2y 4 4

2 y 10 0.6k

Identify the step where Angus made an error and explain what he should have done instead: 1

4x – 8 = 12

2

4x = 4

Subtract 8 from both sides

3

x = 1

Divide both sides by 4

Write the equation

Ryan attempted to solve the equation 9 (4x – 6) = 18, and his working is shown: 9 (4x – 6) = 18 4x – 6 = 91 4x = 15 15 x 4 a

Identify the error made.

b

Solve the equation correctly.

2.01 Solve linear equations mathspace.co

167

15

Jamie had x pencils in her pencil case. She finds one more, and then shares all her pencils equally with her friend, Teddy. Her sister then gives her 5 more pencils. Jamie now has 7 pencils left in her pencil case: a

Explain why the three-step equation that can represent this situation is given by: x 1 5 7 2

b

Solve the equation to determine the number of pencils Jamie started with.

Extend your thinking 16

A square with side length 4x − 11 m has the same perimeter as a square with side length 2x + 9 m. Determine the value of x.

17

A car rental company charges a flat fee of $50, plus $0.20 per kilometre driven. If a customer returns a car with a total rental fee of $74, after driving it, how many kilometres did they drive?

18

In a parallelogram, the base is 3 times the height. If the area of the parallelogram is 48 cm2, determine the base and height of the parallelogram.

2.02   Linear equation applications After this lesson, you will be able to… • develop linear equations from contextual information. • interpret the solutions of linear equations in real-world contexts.

Equations from worded descriptions To develop a linear equation from a worded problem: • Define the unknown quantity and represent it with a variable. • Identify the relationships between the quantities described. • Form an equation using the variable to express these relationships mathematically, applying the four operations (addition, subtraction, multiplication, division) as appropriate.

168

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Keywords in the problem often indicate which operation to use. This table highlights common keywords and their corresponding operations: Addition (+)

Subtraction (−)

Multiplication (×)

Division (÷)

Sum

Difference

Product

Quotient

Total

Decrease

Times

Shared

Increase

Minus

Of

Per

Plus

Less than

Double

Out of

More than

Reduce

Twice

Ratio

Combined

Fewer

Triple

Half

For every

Split

Per

Average

Together

Groups of Notice that “per” appears in both multiplication and division because it can signal different operations depending on the context: • Multiplication: When “per” indicates a rate (how much of one quantity for each unit of another), it typically signals multiplication to find a total amount over multiple units. For example, “60 kilometres per hour” means 60 km are travelled for each hour. To find the total distance covered after t hours, multiply the rate by the number of hours: Distance = (60 × t) km. • Division: “Per” can also signal division when the context is about distributing some quantity. For example, “100 dollars per 5 people” suggests dividing the total amount by the number of people: 100 ÷ 5.

Example 1 Write each situation as an algebraic equation. a Four less than twice a number equals the same number increased by six

Create a strategy

Apply the idea

Represent the ‘number’ with a variable. ‘Less than’ indicates subtraction. ‘Twice’ means multiply by 2, and ‘increased’ indicates addition.

Let the number be x. The term ‘equals’ means separating the expressions. So, the equation is: 2x – 4 = x + 6

b The product of six and the sum of a number and three is the difference of the number and twelve

Create a strategy

Apply the idea

Use a variable to represent the ‘number’. ‘Product’ indicates multiplication, ‘sum’ means addition, and ‘difference’ means subtraction.

Let the number be x. Parentheses are needed to group the sum (x + 3) before multiplying. The equation can be written as: 6 (x + 3) = x – 12

2.02 Linear equation applications mathspace.co

169

c The quotient of four more than triple a number and two is nine

Create a strategy Use a variable to represent the ‘number’. ‘Quotient’ indicates division, ‘triple’ means multiplied by 3, and ‘more than’ indicates addition.

Apply the idea The word and acts as the separator between the numerator and denominator for the fraction representing the quotient. This means ‘four more than triple a number’ is the numerator and ‘two’ is the denominator. Using the word is to separate the expressions, the equation can be written as: 3x 4 2

9

Idea summary To develop a linear equation from a worded problem: • Define the unknown quantity and represent it with a variable. • Identify the relationships between the quantities described. • Form an equation using the variable to express these relationships mathematically, applying the four operations as needed.

Solve practical problems Once a linear equation has been developed from a worded description, solving it involves applying inverse operations in reverse order to isolate the variable. For practical problems, it is important to interpret the solution in the context of the problem, checking it is valid and reasonable.

Example 2 Carrie is going shopping to buy school supplies. She plans to buy 6 notebooks, 2 packages of pens and 1 three-ring binder. The total cost will be $30. She remembers that the binder costs twice as much as a notebook, and a pack of pens costs the same as six notebooks. a Write an algebraic equation to represent the situation.

Create a strategy Assign a variable to represent the cost of one notebook. Since the binder costs twice as much as a notebook and a pack of pens costs the same as six notebooks, the costs of the binder and pens can be expressed in terms of the notebook’s cost. Finally, use the total amount Carrie can spend to set up the equation.

170

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Apply the idea The sum of everything Carrie buys is $30, so: 6 notebooks + 2 pens + 1 binder = $30 Let n represent the cost of one notebook in dollars. So the cost of 6 notebooks is 6n. The cost of one pack of pens is the same as the cost of 6 notebooks or 6n, so the cost of 2 packs of pens is 2 × 6n. The cost of one binder is twice (or two times) the cost of a notebook, n, so the cost of a binder is 2n. Adding these up and setting them equal to $30, the equation is: 6n + 2 × 6n + 2n = 30

b Carrie solves the equation using the information she knows. Her value for n is 1.5. Verify and interpret the solution.

Create a strategy To verify Carrie’s solution, substitute her value for n into the equation, 6n + 2 × 6n + 2n = 30, and check if the result matches the total cost of $30.

Apply the idea Substituting n = 1.5 into the equation: 6 (1.5) + 2 × 6 (1.5) + 2 (1.5) = 30 9 + 18 + 3 = 30 30 = 30 This shows that the value of 1.5 for n makes the equation true, so it is the solution. Since n represents the cost of one notebook, this shows that, the cost of one notebook is $1.50, a pack of pen is $9.00 and a binder is $3.00.

Example 3 A rectangle has a length that is 4 units more than its width. If the perimeter of the rectangle is 36 units: a Determine an equation to represent this situation using x for the width.

Create a strategy Observe the relationships between the quantities by identifying keywords and construct a linear equation.

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Apply the idea The unknown variable, defined as x, represents the width of the rectangle. x = the width of the rectangle Keywords such as “more than” indicate addition. x + 4 = the length of the rectangle To calculate the perimeter of a rectangle, twice the width needs to be added to twice the length. Perimeter = 2 × Width + 2 × Length

Write the formula for the perimeter of a rectangle

36 = 2 × x + 2 × (x + 4)

Substitute the variable and expression

36 = 2x + 2x + 8

Simplify and expand brackets

36 = 4x + 8

Collect like terms

b Determine the length and width of the rectangle.

Create a strategy Solve for the unknown by applying the inverse operations in reverse order.

Apply the idea 36 = 4x + 8

Write the equation

28 = 4x

Subtract 8 from both sides

7=x

Divide both sides by 4

The width of the rectangle is 7 units. Since the length of the rectangle is 4 more units than the width, the length is 11 units.

Reflect and check Calculate the perimeter of the rectangle with a width of 7 and length of 11 to verify the solution is correct. Perimeter = 2 × Width + 2 × Length

Write the formula for the perimeter of a rectangle

= 2 × 7 + 2 × 11

Substitute the values

= 14 + 22

Evaluate the multiplication

= 36

Evaluate the addition

Idea summary For practical problems, develop the equation, solve for the unknown by applying inverse operations in reverse order to isolate the variable. Interpret the solution in the context of the problem, checking it is valid and reasonable.

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2.02 Practice questions What do you remember? 1

2

Express these statements as equations and solve them. Let x represent the unknown number: a

If 11 is subtracted from a number, the result is – 7.

b

The quotient of a number and – 3 is – 20.

Determine an expression for the perimeter of the rectangle in terms of n. 2 cm

8n cm 3

Determine an expression for the perimeter of the triangle in terms of b. 5b m

4

Solve these linear equations for x: a

x + 5 = 12

b

3x – 4 = 11

c

2x + 7 = 5x – 8

d

4(x – 3) + 2 = 3x – 7

Practice Ex 1

5

Write each situation as an algebraic equation, where x represents the number: a

Five less than 2 times a number equals the number increased by 7.

b

Twice a number increased by 15 totals the difference of the number and 1.

c

8 times a number increased by the product of 6 and the sum of the number and 11 is 25. 25 Seven times a number is equal to more than twice the number. 3

d e f 6

The quotient of 9 and the difference of 2 and a number is the number. 4 The quotient of a number and – 3 is – . 3

The perimeter of this triangle is 189 cm. a

Determine the perimeter of the triangle in terms of x.

b

Solve for the value of x.

9x cm

5x + 21 cm

7x + 84 cm

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173

7

Given that the perimeter of this triangle is 98 cm, determine an equation in terms of x and solve for the unknown.

2x + 17 cm

2x + 13 cm

8

Ex 2

9

10

Ex 3

11

Express the statements as equations and solve them for x: a

The sum of 8 and 12x is equal to 92.

b

The product of 5 and the sum of x and 7 equals 50.

Kate and Isabelle do some fundraising for their sporting team. Together they raised $600. If Kate raised $272 more than Isabelle, and Isabelle raised $p: a

Write an equation in terms of p that represents the relationship between the different amounts, and then solve for p.

b

How much did Kate raise?

John and Uther do some fundraising for their sporting team. Together they raised $403. If John raised $m, and Uther raised $71: a

Determine an equation that represents the relationship between the amounts each contributed.

b

Determine the value of m.

Vanessa is cutting out a rectangular board to construct a sign. The board is to have a perimeter of 48 centimetres, and its length is to be 3 centimetres shorter than double the width. Let x be the width of the board. a

12

5x + 14 cm

Solve for x, the width of the board.

b

Determine the length of the board.

Marge is looking at accommodation prices in Paris. One particular hotel charges $184.70 for the first night, and then $153.97 for every additional night. Marge has a budget of $1108.52. Use an equation to determine how many nights she can afford to stay.

Extend your thinking 13

The cost of a cricket ball is 136 cents more than the cost of a rugby ball. Let $x be the cost of a cricket ball and $y be the cost of a rugby ball.

174

a

Express x in terms of y

b

8 cricket balls and 9 rugby balls cost $119. Express this statement as an equation in terms of x and y.

c

Determine the value of y, the cost of a rugby ball.

d

Determine the value of x, the cost of a cricket ball.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

14

15

Beth is 7 times as old as James. Let x be the present age of Beth in years and y be the present age of James in years. a

Express x in terms of y.

b

In 2 years time, Beth will be 5 times as old as James. Express this sentence as an equation in terms of x and y.

c

Determine the value of y, James’ present age.

d

Determine the value of x, Beth’s present age.

A website advertises properties for sale. If the property is sold within the month, the website charges $130 for the advertisement, but if the property is not sold within the month, the website pays the advertiser $20. In a particular month, 58 advertisements were placed, and a total revenue of $3040 was earned from all advertisements. Use an equation to determine the number of advertisements that resulted in a sale of the property.

16

Three consecutive integers are such that the sum of the first and twice the second is 15 more than twice the third. a

Let x be the smallest of the integers. Solve for x.

b

Determine the three consecutive integers.

c

Are there any other sets of three consecutive integers that could fulfill the requirements? Explain how you know.

2.03

Graphs from tables

After this lesson, you will be able to… • determine if a table, graph, or equation represents a linear relation. • create tables of values from linear equations using substitution. • plot points to construct linear graphs. • identify constant rate of change in tables of values.

Linear relationships Linear relationship A relationship between two quantities, typically denoted by x and y, where there is a constant rate of change.

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Linear functions have graphs that are straight lines. The following are examples of linear graphs and their corresponding equations. 8 6 4 2

y

4 3 2 1

x

−8 −6 −4 −2 −2 −4 −6 −8

y

x

−4 −3 −2 −1 −1

2 4 6 8

1

2 3 4

−2 −3 −4

y=x

y

1–

1 x 2

Not all relationships between two quantities form a linear graph on the Cartesian plane. The following are examples of non-linear graphs and their corresponding equations. 4 3 2 1 −4 −3 −2 −1

−1 −2 −3 −4

y

x 1

2 3 4

y = x2 + 1

4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

y

x 1

2 3 4

y = 2x3 − 1

The equation of a linear relationship can be arranged into the forms: • y = a + bx, where y, is the response (dependent) variable and the subject of the equation in terms of x, the explanatory (independent) variable. • Ax + By = C, where A, B, C are constants The x-term and y-term in a linear relationship will always have a power of 1. Non-linear functions are represented by equations that often have higher powers, and cannot be represented by y = a + bx or Ax + By = C.

176

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A table of values can also be used to determine if a relationship is linear or non-linear. If there is a common difference between y-values as x changes by a constant amount, then, the table represents a linear relation. A constant rate of change is a key feature of linear relationships. +2

+2

+2

+2

Weeks (x)

0

2

4

6

8

Growth in cm (y)

3

4

5

6

7

+1

+1

+1

+1

A key feature of linear relationships is a constant rate of change. In the table, for every increase of 2 in the x-values, the y-values consistently increase by 1, demonstrating this linear relationship.

Example 1 Determine whether these equations are linear: a y=

11 – 9x 2

Create a strategy Consider that if x and y have linear relationship, then their equation is of the form y = a + bx or Ax + By = C.

Apply the idea The equation is in the form y = a + bx, therefore, it is linear. b y = x10 + 5x + 26

Create a strategy The x term in a linear relationship must always have a power of 1.

Apply the idea The first term is raised to the power of 10, therefore, it is not linear.

Example 2 Digital only example

Explore in the online lesson

mathspace.co

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177

Example 3 Determine if each table of values represents a linear relationship. a

x

3

6

9

12

15

y

−7

−14

−21

−28

−35

Create a strategy Check if for constant increases in x, y changes by a constant amount.

Apply the idea +3

+3

+3

+3

x

3

6

9

12

15

y

–7

–14

–21

–28

–35

–7

–7

–7

–7

From the table, notice that the x-values go up by 3 each time. For each 3 unit increase in x, y decreases by 7. So, the given table represents a linear graph as there are constant changes in the x-values and y-values. The relationship is linear.

b

x

7

21

35

49

63

y

15 2

15

45 2

45

135 2 ​

Apply the idea +14

+14

+14

+14

x

7

21

35

49

63

y

15 2

15

45 2

45

135 2

+15 2

+15 2

+45 2

+45 2

From the table, notice that the x-values go up by 14 each time. For each 14 unit increase in x, the y-values do not increase or decrease by a constant amount. So, the given table does not represent a linear relationship.

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Reflect and check To further confirm that the relationship is not linear, plot the points from the table on a Cartesian plane and observe the resulting shape. y 70 60 Connecting these points does not form a straight line, as the rate of change in y varies. This graph confirms that the relationship is not linear as there are no constant differences.

50 40 30 20 10

x 0

10 20 30 40 50 60

Idea summary Linear relations have graphs that are straight lines. Linear relations have equations that are written or can be arranged in the form of y = a + bx or Ax + By = C. When given a table of values relating two quantities, a linear relationship exists if there is a common difference between y-values as x changes by a constant amount.

Graph lines with tables of values Table of values A table that represents ordered pairs which satisfy a given equation (or rule defining a relationship). A table of values for a linear relationship can be created by substituting several x-values into the equation and evaluating to get the corresponding y-values. Each pair, (x, y), is called an ordered pair and represents a point on the line. For example, for y = –6 + 3x: x

0

1

2

3

y = −6 + 3x

−6 + 3(0) = −6

−6 + 3(1) = −3

0

3

Once the table of values is complete, plot the points on the Cartesian plane and connect with a line.

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179

Using a spreadsheet tool on a computer or graphing calculator, creating a table of values can be very efficient. These steps illustrate this process: A

B

x

y

1

C

2

Start by labelling the header row with x and y.

3 4 5

A

B

1

x

2

0

3

= A2 + 1

4

⫶

C

y

Next, fill in x-values by starting with the initial value and using a formula to fill down.

⫶

5

A

B

1

x

2

0

= − 6 + 3 × A2

3

1

4

2

5

3

⋮

C

y

⋮

With the x-values filled in using a formula or manually, use a formula for the equation and fill down.

⋮

A

B

1

x

y

2

0

–6

3

1

–3

4

2

0

5

3

3

C This gives a completed table of values. Optionally, these points can be plotted using built-in functions on the calculator.

Alternatively, linear relationships can be graphed from the equation using the graphing page on a graphing calculator.

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Example 4 3x 1. 2 a Complete the table of values. Consider the equation y

x

−2

0

2

4

y

⬚

⬚

⬚

⬚

Create a strategy

3x To complete the table, each given x-value will be substituted into the equation y - 1 to 2 calculate the corresponding y-value.

Apply the idea • For x = –2: y=

3x 1 2

y=

3( 2) 2

=

6 2

Write the equation Substitute x = –2

1 1

Evaluate the multiplication

= –3 – 1

Evaluate the division

= –4

Evaluate

• For x = 0: y= y= =

3x 1 2

Write the equation

3(0) 0 2

Substitute x = 0

1

2 1

Evaluate the multiplication

=0–1

Evaluate the division

= –1

Evaluate

• For x = 2, substituting and evaluating gives y = 2. • For x = 4, substituting and evaluating gives y = 5. Therefore, the completed table of values with the calculated y-values: x

−2

0

2

4

y

−4

−1

2

5

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181

b Plot the points from the completed table of values to draw the graph of y =

3x 1. 2

Create a strategy To plot the points, each (x, y) pair from the completed table will be located and marked on a Cartesian plane. Then join the points with a line.

Apply the idea 1. Identify the coordinate pairs from the completed table of values. The coordinate pairs are (– 2, – 4), (0, – 1), (2, 2), and (4, 5). 2. Plot these points on a Cartesian plane. The x-axis should range from at least –2 to 4, and the y-axis from at least –4 to 5. 5 4 3 2 1

y

(4, 5)

(2, 2) x

−3 −2 −1 1 2 3 4 5 −1 (0, −1) −2 (−2, −4) −3 −4

3. Connect the points with a straight line. 5 4 3 2 1

y

(4, 5)

(2, 2) x

−3 −2 −1 1 2 3 4 5 −1 (0, −1) −2 −3 (−2, −4) −4

Reflect and check Since points from the table of values form a straight line, y =

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3x 1 is a linear relationship. 2

Idea summary Each pair in a table of values can be written as coordinates in the form (x, y). This pairing of numbers is known as an ordered pair. A table of values can be created by substituting each x-value into the given equation by hand or with technology. Using a table of values, a linear function can be graphed.

2.03 Practice questions What do you remember? 1

2

Determine if each statement is true or false: a

Linear equations cannot have powers higher than 1 on variables.

b

Linear relationships involve a constant rate of change.

c

Linear functions have graphs that are curved.

d

The x-values need to go up by 1s in a table of values for it to represent a linear relationship.

For each table of values: i

Plot the points on the Cartesian plane.

ii

Determine whether the graph of the equation is linear.

a

y = –3x

x 3

y

b

x −3 −2 −1 0 x y 3

9

6

3

y

0

−3

0

3

6

1

0

−1

−2

Given the linear equation y = 2x – 3, complete the table of values by substituting the given x-values: x

−1

0

2

4

y

⬚

⬚

⬚

⬚

Practice Ex 1

4

Determine whether these equations are linear: 12 11 a y = x5 + 26 b y c y 26 x 4x 3 2 1

e

y

2x 2

2

f

y

16

20 x

g 3–3=0

10

20 x+ 16 16 = d y +y x6x6+ 20x

21 2

h 9x2 + 3 = 0

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5 Ex 2

Determine if each graph represents a linear or non-linear relationship: a

y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

c

6

d

4 3 2 1

x

2 3 4

f

4 3 2 1

x

−4 −3 −2 −1 −1 −2 −3 −4

2 3 4

2 3 4

1

2 3 4

1

2 3 4

y

x

−4 −3 −2 −1 −1 −2 −3 −4

x 21 42 63 84 y

184

x 1

y

x

Determine whether these tables of values represent linear relationships: a

7

1

y

−4 −3 −2 −1 −1 −2 −3 −4

2 3 4

y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

Ex 3

1

4 3 2 1

x

y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

e

1

b

11 2

55 2

77 2

55

105

b

66

x

5

10

25

30

50

y

11

22

55

66

110

d

y = 1 – 2x

For each equation: i

Construct a table of values for x = − 1, 0, 1, 2.

ii

Plot the points from the table of values on the Cartesian plane.

iii

Sketch the graph of the equation.

a

y = 4x

b

y = – 4x

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

c

y = 1 + 3x

Ex 4

8

a

9

13

14

−3

0

3

y

⬚

⬚

⬚

⬚

x . 5 Complete the table of values: x

−10

−5

0

5

y

⬚

⬚

⬚

⬚

x

−4

−2

0

2

y

⬚

⬚

⬚

⬚

b

Plot the points from the table on the Cartesian plane.

c

Is the relationship linear?

b

Plot the points from the table on the Cartesian plane. x . Sketch the graph of y 5

c

x . 2 Complete the table of values: 5

b c

x . 4 Complete the table of values:

Consider the equation y a

12

−6

Consider the equation y a

11

x

Consider the equation y a

10

2x 4. 3 Complete the table of values:

Consider the equation y

Plot the points from the table on the Cartesian plane. x Sketch the graph of y 5 . 2

5

x

−8

−4

0

4

y

⬚

⬚

⬚

⬚

Consider these equations:

b c

Plot the points from the table on the Cartesian plane. x . Sketch the graph of y 5 4

i

Construct a table of values for x = –1, 0, 1.

ii

Sketch the graph of the equation.

a

y = 8x

b

y = 1 + 9x

c

d

y = –7x

e

y = –3 – 7x

f

4 5x 3 9x – y = –6 y

–

a

3 5x : 4 Create a table of values using reasonable values of x. Explain your choice.

b

Sketch the graph of y

For the linear relationship y

3 5x . 4

Sketch each linear relationship: x x a y b y 7 4

c

y

5

x 6

d

y

6

x 3

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185

Extend your thinking 15

Let the height of a candle be y cm. If the candle is lit, the height decreases according to the equation y = 8 – 2t, where t is the elapsed time in minutes. a

16

186

t

0

1

2

3

y

⬚

⬚

⬚

⬚

b

Sketch the graph of y = 8 – 2t.

c

The height of the candle and the time elapsed must be non-negative. For which values of t does the equation y = 8 – 2t accurately model the height of the candle?

Let the volume of water in a tank be V L. If the tank is initially empty, then the amount of water in the tank is given by V = 5t, where t is the elapsed time in minutes and 5 is the rate that the tank fills in litres per minute. a

17

Complete the table of values:

Complete the table of values: Time (t)

0

1

2

3

Volume of water (V)

⬚

⬚

⬚

⬚

b

Sketch the graph of V = 5t.

c

The volume of the tank is 15 L and the time elapsed is non-negative. For which values of t does the equation V = 5t accurately model the volume of the tank?

A photography drone starts on a cliff 20 m above sea level. Its flight path consists of two distinct legs. For the first leg, its height above sea level is modelled by h1 (t) = 20 – 4t, where h1 is in metres and t is the time in minutes from the start of its flight. For the second leg, which begins immediately after the first, its height above sea level is modelled by h2 (t) = 8 + 2t, where h2 is in metres and t is the total time in minutes from the initial start of its flight. a

Determine the time t and the height h (in metres) at which the drone transitions.

b

Construct a graph showing the height of the drone above sea level for the first 6 minutes of its flight.

c

Assess whether the drone’s path for these two legs could be modelled by a single linear equation instead. Justify your answer using the rates of change for each leg.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

2.04

Slopes and intercepts

After this lesson, you will be able to… • calculate slope given two points. • identify slope and x- and y-intercepts from a graph or equation. • use slopes and intercepts in practical situations.

Slope (gradient) Slope (gradient) Describes the steepness of a straight line. Rise . The rise (vertical length) and Run run (horizontal length) are calculated by forming a right-angled triangle using any two points on the line. The slope can be calculated by using the formula: Slope =

y

This line has a slope of

2 1 x

4 −3

−2

1

−1 −1

Rise 4 . Run 3

In this case, the slope is positive, because over the 3 unit increase in the x-values, the y-values have also increased. Increasing lines have positive slopes, and decreasing lines have negative slopes (as x-values increase, the y-values decrease).

−2 3

Interactive exploration

Discover this concept in action online

mathspace.co

2.04 Slopes and intercepts mathspace.co

187

• If the line goes up to the right, then it is increasing, and has positive slope.

• If the line goes down to the right, then it is decreasing, and has negative slope.

y

y x

x

• If the size (absolute value) of the slope is larger, the line will be steeper. For example, if |b| = 5:

• If the size (absolute value) of the slope is smaller, the line will be less steep. For example, if |b| = 0.3:

y

y x

x

Sometimes, a graph may not be available, in which case, coordinates may be used to determine the slope. The slope can be found by recognising that the rise and run values are related to the x- and y-values in the coordinates. 7 6 5 4 3 2 1 −1 −1

y

Slope = =

Difference in y-values Difference in x-values

=

2 6 7 3

(x2, y2)

(x1, y1) 1

Rise Run

8 4 = –2 =

x

2 3 4 5 6 7

Therefore, for any two points (x1, y1) and (x2, y2), the slope can be found using the formula: Slope =

188

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

y2 y1 x2 x1

Example 1 Create a right-angled triangle and determine the slope of the line shown using points A and B. y

9 8 7 6 5

B

4 3 2 1

A

0

x 1 2 3 4 5 6 7 8 9

Create a strategy Draw a right-angled triangle to find the rise and run, and then use the slope formula: Rise Slope Run

Apply the idea y

9 8 7 6 5

Using the right-angled triangle, this line has a run of 3 and a rise of 2. B 2

4 3 2 1

A

0

3

Rise Run 2 = 3

Slope =

x 1 2 3 4 5 6 7 8 9

Reflect and check The line is increasing so it should have a positive slope.

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189

Example 2 Determine the slope of a line that passes through the points A (3, 5) and B (– 2, 10).

Create a strategy Use the slope formula: Slope

y2 x2

y1 x1

Apply the idea Let A be coordinate 1, x1 = 3 and y1 = 5 and B be coordinate 2, x2 = –2 and y2 = 10. Slope =

y2 x2

y1 x1

10 5 2 3 5 = 5 =

Write the slope formula Substitute the coordinates Evaluate the subtraction

= –1

Simplify

Idea summary The slope (gradient), denoted as b, describes the steepness of a straight line. When given a graph, it can be calculated using the formula:

Slope =

Rise Run

Rise is the vertical length Run is the horizontal length When given any two coordinates (x1, y1) and (x2, y2), the slope can be found using the formula:

Slope =

y2 y1 x2 x1

(x1, y1) are the coordinates of the first point (x2, y2) are the coordinates of the second point

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Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Intercepts Most straight line graphs extend across the Cartesian plane and either cross the x-axis, y-axis or both, passing through the origin, (0, 0). Intercept The point where the line crosses or intersects an axis • The x-intercept is the point where the line crosses the x-axis. The coordinates of the x-intercept will always have a y-coordinate of zero. • The y-intercept, often denoted as a, is the point where the line crosses the y-axis. The coordinates of the y-intercept will always have an x-coordinate of zero. Here are some examples: y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

4 3 2 1

x 1

2 3 4

–4 –3 –2 –1 –1 –2 –3 –4

x-intercept: (−3, 0); y-intercept: (0, 2)

y

x 1

2 3 4

A line passing through the origin has an intercept at (0, 0)

To calculate intercepts algebraically, use substitution. • To calculate the x-intercept, substitute y = 0 into the equation. • To calculate the y-intercept, substitute x = 0 into the equation.

Example 3 Consider this graph: 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

y

x 1

2 3 4

2.04 Slopes and intercepts mathspace.co

191

a Determine the x-value of the x-intercept.

Create a strategy

Apply the idea

The x-intercept is the point where the line intersects the x-axis.

The line intersects the x-axis at (3, 0). So, the x-value of the x-intercept is x = 3.

b Determine the y-value of the y-intercept.

Create a strategy

Apply the idea

The y-intercept is the point where the line intersects the y-axis.

The line intersects the y-axis at (0, 4). So, the y-value of the y-intercept is y = 4.

Example 4 A line has the equation y = – 3 + 3x: a Determine the y-value of the y-intercept.

Create a strategy Substitute x = 0 into the given equation.

Apply the idea y = –3 + 3x

Write the equation

= – 3 + (3 × 0)

Substitute x = 0

= –3

Evaluate

b Determine the x-value of the x-intercept.

Create a strategy Substitute y = 0 into the given equation.

Apply the idea y = –3 + 3x

Write the equation

0 = –3 + 3x

Substitute y = 0

3x = 3

Add 3 to both sides

x=1

Divide both sides by 3

Idea summary An intercept is the point where the line crosses or intersects an axis. • The x-intercept is the point where the line crosses the x-axis. The coordinates of the x-intercept will always have a y-coordinate of zero. • The y-intercept, denoted as a, is the point where the line crosses the y-axis. The coordinates of the y-intercept will always have an x-coordinate of zero.

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Interpret the slope and y-intercept In real-world problems, the slope represents a constant rate of change, while the y-intercept represents an initial or starting value. Response variable 9 8 Rate of change 7 6 5 4 3 2 Initial value 1 Explanatory variable 1 2 3 4 5 6 7 8 9 In many contextual situations, specific words or phrases will indicate what part of slope-intercept form they will represent. Often indicates the y-intercept: • Starting value • One-time fee or cost • Initial position

Often indicates slope: • Each • Per • For every • Changes over time

Example 5 A diver starts at the surface of the water and begins to descend below the surface at a constant rate. The table shows the depth of the diver, in metres, over several minutes: Time passed (min)

0

1

2

3

4

Depth (m)

0

−1.4

−2.8

−4.2

−5.6

a Identify the explanatory and response variables.

Create a strategy The value of the response variable changes based on the value of the explanatory variable.

Apply the idea

Reflect and check

How far the diver has descended depends on the amount of time passed.

The explanatory variable and response variable are often associated with consistent variables across different representations if specific variables are not listed.

The explanatory variable is time passed in minutes, and the response variable is the depth in metres.

The explanatory variable is an input, so it is associated with x. The response variable is an output, so it is associated with the variable y.

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b Determine and interpret the slope in the situation.

Create a strategy The slope, or rate of change, is represented by the ratio

Change in y . Change in x

Apply the idea To find the slope, observe that the x-values change by 1, so to find the change in y-values, subtract the previous value from the current value, checking all pairs to make sure the rate is constant. Slope = =

Change in y Change in x y2 y1 x2 x1

Write the formula for slope as rate of change Write the slope formula

=

1.4 0 1 0

Substitute x1 = 0, y1 = 0, x2 = 1, and y2 = –1.4

=

1.4 1

Evaluate the subtraction

= – 1.4

Simplify

The slope is – 1.4, which means that the diver is descending at 1.4 metres per minute.

c Determine and interpret the y-intercept of the situation.

Create a strategy

Apply the idea

The y-intercept is represented by the pair (0, y) in a table.

The pair (0, 0) is in the table, so this is the y-intercept. This means that initially, the diver is at the surface of the water.

Idea summary Linear relationships (functions) as equations, tables, and graphs can be used to solve a variety of real-world problems. For y = a + bx: • The slope, b, will represent a rate of change. • The y-intercept, a, represents the initial or starting value (the value of y when x = 0).

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2.04 Practice questions What do you remember? 1

Match each term with its definition and representation on the graph. Slope

i

The point where the line crosses or intersects with the x-axis, where y = 0. Rise The rate of change of the line, defined as the . Run

ii iii

2

b

x-intercept

c

The point where the line crosses or intersects with the y-axis, where x = 0.

2 1

y B 1 2 3 4 5 6

–2 –1 –1 –2 –3 –4 –5 –6 A

x

C

Identify what each variable represents for the equation of a line: y = a + bx: a

3

y-intercept

a

b

b

a

y-intercept

For each line segment, determine the: i

Slope

ii

a

y 4 3 2 1

b

–4 –3 –2 –1 1 –1 –2 A –3 –4

y 5 4 A 3 2 1

B x

3 4 2 3

–4 –3 –2 –1 –1 –2 –3

4

y Given the slope formula b = 2 x2 which point is (x2, y2)?

5

A line is drawn on the Cartesian plane:

1

B 2 3 4

x

y1 and two points, does it matter which point is (x1, y1) and x1

Which of these describe the relationship between the slopes of the segments? A

Slope of AB is equal to slope of AC.

B

Slope of AB is twice slope of AC.

C

Slope of AC is twice slope of AB.

y 4 C

3 2 B

1 –4 –3 –2 –1 –1

1

x

2 3 4 A

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Practice Ex 1

6

y 16 B 12 8 4

Create a right-angled triangle and determine the slope of the line shown using points A (4, 0) and B (0, 16).

–5–4–3–2–1 –4 –8 –12 –16 7

Determine the slope, b, of these lines: a

y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

c

1

b

x

5 4 3 2 1 –6–5–4–3–2–1 –1 –2 –3 –4 –5

2 3 4

y 5 4 3 2 1 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5

196

A 1 2 3 4 5

d

y

x 1 2 3 4 5 6

y 4 3

x

1 2 3 4 5

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

2 1 –6–5–4–3–2–1 –1 –2

x 1 2 3 4 5 6

x

8

Is the slope of each line positive or negative? a

y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

c

Ex 2

10

11

x

y

x

–4 –3 –2 –1 –1 –2 –3 –4

d 4 3 2 1

x

1

2 3 4

1

2 3 4

d

(– 6, – 6) and (6, 2)

y

x

–4 –3 –2 –1 –1 –2

2 3 4

–3 –4

i

Sketch the line that passes through the given points.

ii

Determine the slope.

a

(– 1, 3) and (0, 6)

b

(– 1, 0) and (0, – 5)

c

(2, 1) and (8, – 2)

Determine the slope of the line that passes through: a

Point A (3, – 9) and the origin

b

Point A (– 4, 0) and Point B (0, 4)

c

Point A (– 2, 4) and Point B (5, 1)

d

Point A (– 3, – 1) and Point B (– 5, 1)

e

Point A (1, – 1) and Point B (– 1, – 2)

Complete the coordinates using the points and slope provided: a c

12

1

4 3 2 1

2 3 4

y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

9

1

b

(4, – 3) and (1, ⬚), slope = – 2 (5, 3) and (⬚, 63), slope = 4

b d

(5, 3) and (2, ⬚), slope = – 4

(11, ⬚) and (– 20, 16), slope =

4 7

The points A (– 11, – 9), B (– 5, 1) and C (– 2, 6) are plotted on the Cartesian plane: a

Determine the slope of AB.

b

Determine the slope of BC.

c

Do the points A, B and C lie in a straight line? Explain your answer.

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Ex 3

13

For each graph: i

Determine the x-value of the x-intercept.

ii

Determine the y-value of the y-intercept.

a

y 3

b

2 1

x

–3 –2 –1

1

2

3

–1

–3

y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

Ex 4

14

15

198

x

–2 –1 –1 –2

–2

c

y

6 5 4 3 2 1

1

1

d

2 3 4 5 6

y 1 x –2 –1 –1

x

2 3 4

1

2

3

4

5

–2 –3 –4

For each equation: i

Determine the value of the x-intercept.

ii

Determine the value of the y-intercept.

a

y = 2x – 2

b

3x + y = –3

c

y = 4x – 8

d

e

y = 9x

f

2y + 2x = 4

g

3x – 5y = 1

h

2y + x = –3 3 y 10 x= 5

A cellphone salesman earned $600 in a particular week during which he sold 26 phones and $540 in another week during which he sold 20 phones. a

What are the explanatory and response variables?

b

Determine the slope of this linear relationship.

c

Interpret the slope in the situation.

d

Determine the y-intercept.

e

Interpret the y-intercept in the situation.

Mathspace VCE General Mathematics Units 1 & 2 mathspace.co

Ex 5

16

The number of fish in a river is approximated over a five year period. The results are shown in the table: Time in years (t)

0

1

2

3

4

5

Number of fish (F)

4800

4600

4400

4200

4000

3800

a

Identify the explanatory and response variables.

b

Determine and interpret the slope in the situation.

c

Determine and interpret the y-intercept in the situation.

Extend your thinking 17

A line passing through the points (– 1, 4) and (– 4, t) has a slope of –3. Determine the value of t.

18

19

4 A line passes through the points (11, c) and (– 20, 16) and has a slope of . 7 Determine the value of c. Consider the ramp: a

b

20

Determine the slope of this skateboard ramp if it rises 0.9 metres above the ground and runs 1 metre horizontally at the base.

Rise

The ramp can only be classified as a ‘beginner’s ramp’ if for every 1 metre horizontal run, it has a rise of at most 0.5 metres. Can it be classified as a ‘beginner’s ramp’?

Run

A certain ski resort has two ski runs as shown in the diagram: a

Determine the slope of Run A. Round your answer to two decimal places.

b

Determine the slope of ski run B. Round your answer to two decimal places.

c

Run A

21 m 7m

Which run is steeper?

Run B

20 m 29 m 21

Gas costs a certain amount per litre. The table shows the cost of various amounts of gas in dollars: Number of litres (x)

0

10

20

30

40

Cost of gas (y)

0

15

30

45

60

Calculate, if there is any, the change someone will receive if they use a $100 bill to pay for 47 L of gas. 22

A racing car starts the race with 150 L of fuel. From there, it uses fuel at a rate of 5 L per minute. How many minutes will it take for the car to run out of fuel?

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2.05   Sketch linear functions After this lesson, you will be able to… • sketch linear functions in slope-intercept and general forms using key features. • sketch horizontal and vertical lines. • use technology to graph and verify linear functions.

Sketch from slope-intercept form Slope-intercept is a helpful form because two key features, the slope and the y-intercept can be read off without any calculations required.

y = a + bx a is the y-intercept of the line b is the slope (gradient) of the line

Interactive exploration

Discover this concept in action online

mathspace.co

To graph a straight line, only two points are needed. These points can be determined from a linear equation in the form y = a + bx by following these steps: 1. Identify the y-intercept, a, and plot the point (0, a) on the Cartesian plane. Rise 2. Express the slope (gradient), b, as a fraction in the form: . Move from the y-intercept Run according to the rise and run values to determine the second point. 3. Draw a straight line through the two points, extending it across the graph. Alternatively, a second point can be found by substituting any non-zero value for x into the equation. To graph using technology, equations in the form y = a + bx, can be input as functions. While equations in other forms may need to be input as relations.

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Example 1 Sketch the graph of the line y = – 1 + 4x using its slope and y-intercept.

Apply the idea 1. Identify the y-intercept a, and plot the point (0, a) on the Cartesian plane. The y-intercept is at (0, – 1). Rise . Run Move from the y-intercept according to the rise and run 4 values to determine the second point. Since b 4 , 1 move 1 unit right then 4 units up from the y-intercept

2. Express the slope b as a fraction in the form:

leading to the point (1, 3). 3. Draw a straight line through the two points, extending it across the graph. Connect the points to form the line y = – 1 + 4x.

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

y y = 4x − 1 x 1 2 3 4 5

Reflect and check Technology can be used to verify the line’s accuracy by inputting the equation into the graphing page on a calculator and ensuring the sketched line matches.

Idea summary To graph a straight-line from a linear equation in the form y = a + bx, first plot the y-intercept (0, a), then use the slope (gradient), b, to determine a second point by moving according to the rise and run. Then, join the two points to construct a line.

Sketch using intercepts Alternatively, a linear function can be graphed by finding the x-intercept and y-intercept. The steps to graph using intercepts are: 1. Substitute x = 0, and solve for y to get the coordinates of the y-intercept: (0, y). 2. Substitute y = 0, and solve for x to get the coordinates of the x-intercept: (x, 0). 3. Plot the two points on the Cartesian plane. 4. Connect the points with a straight line. Note that for functions where the x- and y-intercepts are equal, this will result in only one point - the origin (0, 0). Another point is needed, so choose another point, say x = 1, and substitute x = 1 into the equation and solve for y to find a second point that lies on that line.

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In all forms, the strategy of setting x = 0 and then y = 0 to reveal the two intercepts can be used. • The general form is usually written as Ax + By = C. • The slope-intercept form y = a + bx where the slope of the line is given by the coefficient b and the y-intercept given by a.

Example 2 Consider the linear equation –6x + 2y – 12 = 0. a Determine the y-value at the point where x = 0.

Create a strategy Substitute x = 0 into the given equation.

Apply the idea – 6x + 2y – 12 = 0

Write the equation

– 6 (0) + 2y – 12 = 0

Substitute x = 0

2y – 12 = 0

Evaluate the multiplication

2y = 12

Add 12 to both sides

y=6

Divide both sides by 2

The coordinates of the y-intercept at (0, 6).

b Determine the x-value at the point where y = 0.

Create a strategy Substitute y = 0 into the given equation.

Apply the idea – 6x + 2y – 12 = 0

Write the equation

– 6x + 2(0) – 12 = 0

Substitute y = 0

– 6x – 12 = 0 – 6x = 12 x = –2 The coordinates of the x-intercept at (– 2, 0).

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Evaluate the multiplication Add 12 to both sides Divide both sides by –6

c Use these points to graph the line.

Create a strategy Plot the points found and draw a line through them.

Apply the idea Plot the x-intercept at (– 2, 0) and the y-intercept at (0, 6). y 9 8 7 6 5 4 3 2 –6x + 2y – 12 = 0 1 x –4 –3 –2 –1 –1

1

2

3

4

Idea summary The x-intercept occurs at the point where y = 0. The y-intercept occurs at the point where x = 0. The x- and y-intercepts can be connected to sketch a line.

Horizontal and vertical lines y

Horizontal line A straight line that runs parallel to the x-axis on a Cartesian plane.

x

• The slope (gradient) of a horizontal line is always zero. This is because the rise is 0, so Rise 0 0. Run Run • The equation of any horizontal line is y = a, because b is zero in the equation of a line given by y = a + bx. • The x-axis is a horizontal line, and every point on it has a y-value of 0 so the equation of the x-axis is y = 0.

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y

Vertical line A straight line that runs parallel to the y-axis on a Cartesian plane.

x

• The slope of a vertical line is undefined. The run is 0, leading to division by zero for Rise Rise , which is mathematically undefined. 0 Run

• Vertical lines cannot be written in the slope-intercept form as the slope is undefined. The equation for a vertical line is written in the form x = k where k is the x-intercept of the line. • The y-axis is a vertical line, and every point on it has a x-value of 0 so the equation of the y-axis is x = 0.

Example 3 Digital only example

Explore in the online lesson

mathspace.co

Example 4 Consider the graph: a Determine the equation of the line. 4 3 2 1 –4–3–2–1 –1 –2 –3 –4

y

x 1 2 3 4

Create a strategy

Apply the idea

Horizontal lines have equations of the form y = a, where a is the y-intercept.

The y-intercept of the horizontal line is 4, therefore, a = 4. The equation of the line is y = 4.

b Determine the equation of the line that is parallel to the y-axis and passes through the point (– 8, 3).

Create a strategy

Apply the idea

Any line that is parallel to the y-axis is a vertical line and all vertical lines are in the form of x = k, where k is the x-coordinate of every point on the line.

The x-coordinate of the point (– 8, 3) is – 8, therefore, k = – 8.

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The equation of the line is x = – 8.

Idea summary A horizontal line has a slope (gradient) of 0 with an equation of the form y = a, where a is the y-value of the y-intercept of the line. A vertical line has an undefined slope (gradient) with an equation of the form x = k, where k is the x-value of the x-intercept of the line.

2.05 Practice questions What do you remember? 1

In the equation of a line, y = a + bx, identify what the values of b and a represent.

2

Complete the statement: 2 x , plot ⬚ on the y-axis, then rise ⬚ units up and run ⬚ units to the right 3 which gives the point ⬚, then draw a line passing through the points ⬚ and (3, 6). To graph y

3

4

Match each graph to its corresponding slope description. i

y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

iii

1

1

x

4 3 2 1

y

x

–4 –3 –2 –1 –1 –2 –3 –4

2 3 4

y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

ii

iv

x

4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4

2 3 4

1

2 3 4

1

2 3 4

y

x

a

The slope is positive

b

The slope is negative

c

The slope is zero

d

The slope is undefined

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4

For the equation y = 5x – 2, describe how to determine the: a

Slope

b

y-intercept

c

x-intercept

Practice 5

Sketch on a Cartesian plane: a b

6 Ex 1

Sketch the graph of each equation using the slope and y-intercept: a e

7

Ex 2

8

9

10 Ex 4

y=x+1 1 y x 3 2

y = 3x – 1 3 y x 4 2

b f

y = –x + 4

d

y = – 3x – 4

g

y = – 2x

h

y=x

i

Determine the x-intercept.