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SCHOLAR Study Guide

National 5 Mathematics

Course MaterialsTopic 3: Comparing data sets usingstatistics

Authored by:Margaret Ferguson

Reviewed by:Jillian Hornby

Previously authored by:Eddie Mullan

Heriot-Watt University

Edinburgh EH14 4AS, United Kingdom.

First published 2014 by Heriot-Watt University.

This edition published in 2016 by Heriot-Watt University SCHOLAR.

Copyright © 2016 SCHOLAR Forum.

Members of the SCHOLAR Forum may reproduce this publication in whole or in part foreducational purposes within their establishment providing that no profit accrues at any stage,Any other use of the materials is governed by the general copyright statement that follows.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval systemor transmitted in any form or by any means, without written permission from the publisher.

Heriot-Watt University accepts no responsibility or liability whatsoever with regard to theinformation contained in this study guide.

Distributed by the SCHOLAR Forum.

SCHOLAR Study Guide Course Materials Topic 3: National 5 Mathematics

1. National 5 Mathematics Course Code: C747 75

AcknowledgementsThanks are due to the members of Heriot-Watt University's SCHOLAR team who planned andcreated these materials, and to the many colleagues who reviewed the content.

We would like to acknowledge the assistance of the education authorities, colleges, teachersand students who contributed to the SCHOLAR programme and who evaluated these materials.

Grateful acknowledgement is made for permission to use the following material in theSCHOLAR programme:

The Scottish Qualifications Authority for permission to use Past Papers assessments.

The Scottish Government for financial support.

The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA)curriculum.

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1

Topic 1

Comparing data sets usingstatistics

Contents

3.1 Quartiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3.2 Five-Figure summaries and boxplots . . . . . . . . . . . . . . . . . . . . . . . 12

3.3 Calculating the mean and standard deviation . . . . . . . . . . . . . . . . . . . 20

3.4 Comparing data sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.5 Line of best fit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.6 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2 TOPIC 1. COMPARING DATA SETS USING STATISTICS

Learning objectives

By the end of this topic, you should be able to:

• find the quartiles of a data set;

• identify a five-figure summary;

• calculate the interquartile range and semi-interquartile range;

• construct a boxplot from a data set;

• calculate the mean and standard deviation;

• compare data sets;

• construct a scatter graph;

• determine the equation of a line of best fit.

© HERIOT-WATT UNIVERSITY

TOPIC 1. COMPARING DATA SETS USING STATISTICS 3

1.1 Quartiles

You should already know how to sort a list into order and how to carry out some simplestatistical analysis.

Mean, median, mode and range

Go online

Nine men of varying heights have been assembled for an identity parade.

The tallest man is 2 · 04 m and the smallest man is 1 · 25 m.The difference between them is 2 · 04 − 1 · 25 = 0 · 79 m.The range is the highest score minus the lowest score.

The men are ask to arrange themselves in order of height. Now it is easy to find themiddle height, which is 1 · 80 m.The median is the middle score, once the scores have been put in order.

Everyone is a different height except for the twins who are both 1 · 75 m tall. The readingof 1 · 25 m occurs more often than any other height in this group.The mode is the most common score.



The sum of the heights divided by the number of men gives an average height:(1·25+ 1·67+1·75+1·75+1·80+ 1·85+ 1·90+ 1·98+ 2·05)

9 = 1 · 78 mThe mean is the sum of the scores divided by how many scores there are.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Key point

• The range is the biggest score minus the lowest score.

• The median is the middle score, once the scores have been put in order.

• The mode is the most common score.

• The mean is the sum of the scores divided by how many scores there are.

Key point

The median, mode and mean are all measures of the average score.The range is a measure of the spread of the scores.

Quartiles of a data set

Go onlineQuartiles are used to summarise data. How we calculate the quartiles depends on howmany values there are in the list.

There are four possible outcomes depending on the number of values in the list.

Study each of the techniques shown in the examples below.

Examples

1.

Problem:

We have 11 numbers all in order.

2 2 3 3 4 4 7 7 8 9 9

Solution:

Step 1: Find the median, or middle value, by placing your fingers on 2 and 9 and movingthem inward.2 2 3 3 4 4 7 7 8 9 9

↑ ↑The Median ( or middle quartile) Q2 = 4

Step 2: Use your fingers to find the lower quartile by finding the median of the lower halfof the list.



2 2 3 3 4 4 7 7 8 9 9

↑ ↑2 2 3 3 4 4 7 7 8 9 9

↑ ↑The lower quartile Q1 = 3

Step 3: Use your fingers to find the upper quartile by finding the median of the upperhalf of the list.

2 2 3 3 4 4 7 7 8 9 9

↑ ↑2 2 3 3 4 4 7 7 8 9 9

↑ ↑The upper quartile Q3 = 8

Step 4: Write down the quartiles.

The lower quartile Q1 = 3The median Q2 = 4The upper quartile Q3 = 8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.

Problem:


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

Solution:

Step 1: Find the median, or middle value, by placing your fingers on 1 and 9 and movingthem inward.

Since there is not a value left in the middle you must find the average of the two middlevalues.

The Median (middle quartile) Q2 = (4 + 5) ÷ 2 = 4 · 5Step 2: Use your fingers to find the lower quartile by finding the median of the lower halfof the list.

The lower quartile Q1 = 3




The upper quartile Q3 = 7


The lower quartile Q1 = 3The median Q2 = 4 · 5The upper quartile Q3 = 7

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.

Problem:


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

Solution:


The median (middle value) Q2 = 5


The lower quartile Q1 = (3 + 3) ÷ 2 = 3




The upper quartile Q3 = (7 + 8) ÷ 2 = 7 · 5Step 4: Write down the quartiles.

The lower quartile Q1 = 3The median Q2 = 5The upper quartile Q3 = 7 · 5

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4.

Problem:


1 1 2 4 4 6 6 6 8 9 9 9

Solution:


The median (middle value) Q2 = (6 + 6) ÷ 2 = 6


The lower quartile Q1 = (2 + 4) ÷ 2 = 3




The upper quartile Q3 = (8 + 9) ÷ 2 = 8 · 5Step 4: Write down the quartiles.

The lower quartile Q1 = 3The median Q2 = 6The upper quartile Q3 = 8 · 5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Key point

• The Median (Q2) is the middle value in an ordered list.

• The Lower Quartile (Q1) is the middle of the lower half of the ordered list.

• The Upper Quartile (Q3) is the middle of the upper half of the ordered list.

Examples

1.

Problem:

Arrange the numbers in the list in ascending order and find the quartiles of the data set.

27 42 11 23 37 29 24 13 33 45

13 32 19 24 44 39 15 21 47 26

Solution:

Step 1: Write down the ordered list.

Step 2: Find the median by placing your fingers on 11 and 47 and moving them inward.

Step 3: Find the lower quartile by placing your fingers on 11 and 26 and moving theminward.

Step 4: Find the upper quartile by placing your fingers on 27 and 47 and moving theminward.



. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



2.

Problem:


47 23 39 16 33 13 18 43 22 30 27

12 32 17 40 44 20 47 33 26 13 36

Solution:







. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Top tip

If there are a lot of values to put in order it is a good idea to make a stem and leafdiagram.

Stem and leaf diagrams

Go online

Twenty people in a cinema queue were asked their age, the following was recorded.

32 59 19 11 42 20 27 30 41 38

14 40 27 42 43 48 19 37 42 11

We often need to put such data sets in order.

Note that the tens digits run from 1 to 5. Use these digits to start the rows of a ’table’.We call the column the ’stem’.



Go through the data list and place the units digits in the appropriate row.

The units digits are referred to as ’leaves’. We need to sort the leaves in order placingthe smallest value nearest the stem.

Add a title so that readers know what the table is about and provide an example of howthe data should be read:

The finished article is called a ’stem and leaf’ diagram.

It can be used to:

a) draw conclusions;The 40-49 age group is the least represented in the queue.



b) order the data.12, 13, 14, 14, 17, 19, 23, 26 ,27, 34, 35, 35, 37, 39, 48, 54, 54, 54, 55, 58

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Quartiles of a data set practice

Go online

Q1:


61 39 47 31 55 41 32 59 44 63

55 34 49 61 52 39 65 35 60 38

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q2:


57 32 34 21 45 50 30 48 43 20 39

27 28 57 41 36 58 43 53 34 21 33

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Quartiles of a data set exercise

Go onlineBy arranging the following numbers in ascending order, find the quartiles of the data set.

3 9 10 11 11 13 15 16 20

Q3: What is the median?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q4: What is the lower quartile?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q5: What is the upper quartile?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

By arranging the following numbers in ascending order, find the quartiles of the data set.

113 118 109 112 108 105 104 102 121


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .




60 35 61 52 68 54 48 50 66

43 38 44 34 57 33 61 65 59


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


50 22 52 42 37 43 31

40 35 31 24 31 20 45

20 51 53 47 25 27 34


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1.2 Five-Figure summaries and boxplots

Five-Figure summary

Go online

When we get our exam mark back, it doesn’t really make sense until it is put in context.

Is 59% any good?It is if everyone else got less!It’s not if everyone else got more.

How do we let someone know the context?

We could let them know all the marks. . . but this could be information overload!



We could try and summarise the data. We could tell people the middle score, thequartiles, and the highest and lowest scores.

This would let them see where their 59% stood without having too much data to interpret.They would then know if it was a good or a bad mark.

Such a summary is often referred to as a five-figure summary for obvious reasons.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Example

Problem:

Sort the following list in ascending order and then provide a five-figure summary of thedata.

46 54 65 33 49 59 49 39 60 37 69

43 36 58 43 57 46 49 69 49 58 43

Solution:

Step 1: Sort the data.

Sorted list:

Step 2: Find the median, lower quartile and upper quartile.

Step 3: State the five-figure summary of the data.

Lowest Score . . . 33

Lower Quartile . . . 43

Median (49 + 49) ÷ 2 . . . 49

Upper Quartile . . . 58

Highest Score . . . 69

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



A Five-Figure Summary consists of the following values:

• Lowest Score. . . L

• Lower Quartile. . . Q1

• Median. . . Q2

• Upper Quartile. . . Q3

• Highest Score. . . H

Five-Figure summary practice

Go online

Q15:


37 50 26 45 39 36 45 58 30 46 37

39 57 53 36 37 22 38 42 48 40

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Boxplots

Go online

Five-figure summaries can be neatly shown in a boxplot. Consider the five-figuresummary of 3, 11, 30, 66, 97.

Step 1: Draw a number line.

A number line is drawn whose range includes the highest and lowest score graduatedusing some convenient unit.

Step 2: Mark the 5-figure summary.

Above the number line draw vertical markers to represent the five pieces of summarydata.

Step 3: Create the box.

Use the three middle markers to define a box.



Step 4: Create the whiskers.

Connect the end markers to the box using "whiskers".

Step 5: Add title, labels and units.

Add any details you need to avoid the loss of information, e.g. title, units, actual valuesof the five-figure summary.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Example

Problem:

The marks of eighteen pupils in a test were as follows:

37 55 56 54 58 36 41 47 30

41 41 53 56 61 52 31 43 37

a) Find the median, the quartiles, the maximum and the minimum values, of the dataset.b) Draw a boxplot to illustrate the data.

Solution:






Median (43 + 47) ÷ 2 . . . 45





Step 4: Draw the boxplot.

You should:

1. create a suitable number line;

2. mark up the five-figure summary with vertical markers;

3. draw a box, defined by the three quartiles;

4. join the end markers to the box by whiskers;

5. add a title and the five-figure summary.

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Boxplots practice

Go online

The marks of fifteen pupils in a test were as follows:

45 16 18 33 15 19 30 23

25 45 45 26 32 42 27

Q16: Find the median, the quartiles, the maximum and the minimum values, of the dataset.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q17: Draw a boxplot to illustrate the data.

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Interquartile range

Key point

The interquartile range (IQR) is a descriptive statistic used to summarize thespread of your data.

The Interquartile range (IQR) = Q3 − Q1

A boxplot is a diagrammatical representation of the data and is based on the five-figuresummary.



Key point

Semi-interquartile range is half of the interquartile range.

The semi-interquartile range (SIQR) = Q3 − Q1

2

Top tip

You will need to learn both formulas as they are not given in the formula sheet.

IQR = Q3 − Q1

(SIQR) = Q3 − Q1

2

Example

Problem:

The speeds of 10 cars on a motorway were as follows:

68 71 59 75 70

66 69 72 89 67

1. Find a five-figure summary for this data set.

2. Calculate the interquartile range.

3. Calculate the semi-interquartile range.

Solution:






Median (69 + 70) ÷ 2 . . . 69 · 5Upper Quartile . . . 72




Step 4: State the Interquartile range.

IQR = Q3 − Q1 = 72 − 67 = 5

Step 5: State the Semi-interquartile range.

SIQR = Q3 − Q12 = 72 − 67

2 = 2 · 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Key point

The semi-interquartile range is a measure of the spread of the data.A semi-interquartile range of zero tells us that the data values are all the sameand as the semi-interquartile range gets larger the values are more spread outfrom each other.

Interquartile range practice

Go online

Q18: Nine apples were weighed in grams and recorded below.

99 87 95 102 91 96 100 98 91

Identify the interquartile and semi-interquartile ranges.

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Five-figure summaries and boxplots exercise

Go online

Five-Figure Summary

Q19:


50 23 53 42 38 43 33 40 35 31

24 31 20 46 20 51 55 49 34

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q20: The marks of sixteen pupils in a test were as follows.

40 24 15 36 35 29 30 41

18 17 43 32 12 23 45 39

Find the median, the quartiles, the maximum and the minimum value of this data set.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Boxplots

Q21: Look at the following 17 ordered numbers.

32, 34, 34, 37, 38, 42, 43, 47, 49, 50, 53, 55, 57, 61, 65, 67, 69

This data has a five-figure summary as shown in the table that follows.




Lower Quartile . . . 37 · 5Median . . . 49



Draw a boxplot to illustrate the data.

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Q22: Look at the following set of numbers.

49 24 57 43 45 34 39 31

29 32 24 46 21 52 47

Provide the five-figure summary and create a boxplot to illustrate the data.

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Interquartile and Semi-Interquartile Ranges

In a seed trial 12 sunflowers were measured in metres.

2 · 1 2 · 7 1 · 9 2 · 2 3 · 0 2 · 83 · 4 1 · 9 2 · 7 2 · 2 2 · 5 1 · 9

Q23: Provide the five-figure summary for the data.

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Q24: What are the interquartile and semi-interquartile ranges?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Nine newborn babies were weighed in Kg.

2 · 55 4 · 42 2 · 94 3 · 02 3 · 152 · 86 2 · 94 3 · 09 3 · 45

Q25: Provide the five-figure summary for the data.

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Q26: What are the interquartile and semi-interquartile ranges?

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1.3 Calculating the mean and standard deviation

Standard deviation

Go onlineA group of six people have an average height of 157 cm.

On average, how much does each person differ from the average?

The standard deviation is one way of trying to answer that question.

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Sums and sigma notation

Go online

Look at the table. Each column contains numbers.

Object a b a + b ab a2

3 2 5 6 94 3 7 12 165 7 12 35 256 8 14 48 36

Sum 18 20 38 101 86

The top row tells you how these numbers have been obtained from the first two columns,which are labelled a and b. The bottom row gives the sum of each column.

In maths, we use a special symbol to represent the sum: ΣThis is the Greek letter S and is pronounced sigma.

Σa = 18 The sum of the a column is 18.Σb = 20 The sum of the b column is 20.Σ(a + b) = 38 The sum of the (a + b) column is 38.

Σab = 101 The sum of the ab column is 101.

Σa2 = 86 The sum of the a2 column is 86.

We will use sigma, Σ, in our formula for standard deviation.

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Standard deviation formula (1)

Go online

A formula has been worked out which will let you work out the standard deviation for anyset of numbers.

The formula is easy but it might be difficult to memorise but it will be given to you on theformula sheet in any assessment where you might need it.

It looks like this:

s =

√∑(x − x̄)2

n − 1

It’s actually easy to use. . .

When you work with this formula you will make a table with three columns.

The first column will be called x and Σx is the sum of the x column.x̄ is the mean.n is the amount of x values in the column.

x̄ =∑

xn

The second column will be called x − x̄ and∑

(x − x̄) is the sum of the secondcolumn.∑

(x − x̄) should equal 0 or very close to 0 if x̄ was rounded.

s =

√∑(x − x̄)2

n − 1

The third column will be called (x − x̄)2 and∑

(x − x̄) is the sum of the third column.n − 1 is the amount of x values in the first column less 1.

s =

√∑(x − x̄)2

n − 1

s is the standard deviation. . . the subject of the formula.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Standard deviation formula (2)

Go online

A formula has been worked out which will let you work out the standard deviation for anyset of numbers.

The formula is messy and might be difficult to memorise but it will be given to you on theformula sheet in any assessment where you might need it.

It looks like this:

s =

√∑

x2 − (∑

x)2/nn − 1

It’s actually easy to use. . . but what does the formula mean?

When you work with this you’ll make a table from a list of x values.

These x values will be in a column called x. Σx is the sum of the x column.



You will fill a column with the x values squared. . . this column will be called x2. Σx2 isthe sum of the x2 column.

n is the amount of x values you have.

s is the standard deviation. . . the subject of the formula.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Examples

1.

Problem:

Calculate the mean and standard deviation for this list of numbers.

6 3 5 4 1 7 5 9

Solution:

Step 1: Make a table with three columns.

x x − x̄ (x − x̄)2

6

3

5

4

1

7

5

9

Step 2: Add up the x column and calculate the mean.



x x − x̄ (x − x̄)2

6

3

5

4

1

7

5

9

40

x̄ =∑

xn = 40

8 = 5

Step 3: Subtract the mean from each x value and sum column 2.

x x − x̄ (x − x̄)2

6 6 − 5 = 1

3 3 − 5 = − 2

5 0

4 −1

1 −4

7 2

5 0

9 4

40 0

Step 4: Square each value in column 2 and sum column 3.

x x − x̄ (x − x̄)2

6 6 − 5 = 1 12 = 1

3 3 − 5 = − 2 (−2)2 = 4

5 0 0

4 −1 1

1 −4 16

7 2 4

5 0 0

9 4 16

40 0 42

Step 5: Write out the formula and calculate the standard deviation.

s =

√∑(x − x̄)2

n − 1

s =

√42

8 − 1

s =

√42

7

s =√6

s = 2 · 45 (to 3 s.f.). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



2.

Problem:

Calculate the standard deviation of this sample of digits.

4 8 3 8 1 9 3 9 4

Solution:

Step 1: Make a table with 2 columns and a row for each number and a row for headingsand a row for sums.

Note that in this example n = 9.

Call the first column x and the second one x2.

Enter the x values in the first column.

x x2

4

8

3

8

1

9

3

9

4

Step 2: Square each x-value and put the results in the second column.

x x2

4 16

8 64

3 9

8 64

1 1

9 81

3 9

9 81

4 16

Step 3: Add up each column and put the sums in the bottom row.



x x2

4 16

8 64

3 9

8 64

1 1

9 81

3 9

9 81

4 16

49 341

We now have that Σx = 49 and Σx2 = 341 and we know n = 9.

Step 4: Write out the formula.

s =

√∑

x2 − (∑

x)2/nn − 1

Step 5: Substitute into formula.

s =

√341 − (49)2/9

9 − 1

Step 6: Work it out a bit at a time.

s =

√√√√341 − (49)2/9

9 − 1

= 3 · 05

The standard deviation is 3 · 05 ( to 3 s.f.).

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Key point

The formulae for standard deviation are:

s =

√∑(x − x̄)2

n − 1

s =

√∑

x2 − (∑

x)2/nn − 1

Calculating the mean and standard deviation practice

Go online

Choose the formula you think is easier and try the following question. Make sure youshow all the steps and be careful entering the formula in your calculator.

Q27: Calculate the standard deviation of this sample of digits.

3 6 9 4 5

Give your answer to 1 decimal place.

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Calculating the mean and standard deviation exercise

Go online

Q28:

Calculate the mean and use the formula of your choice to calculate the standarddeviation for the data set below.

23 25 21 21 24 18

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Q29: What is the standard deviation of this random sample of digits?

6 2 4 1 8

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Q30: What is the standard deviation of this random sample of numbers?

3 3 1 4 3 5 1

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Q31:

The number of matches in eight boxes were counted. Calculate the mean and standarddeviation for number of matches.

38 42 36 44 40 39 44 41

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1.4 Comparing data sets

You will already know some techniques for comparing data for example back-to-backstem and leaf diagrams. Boxplots can be used in the same way if they are placedtogether on a single scale but we now have other measures in our toolkit to help.

The mean tells us the average value in a list of numbers and the Range tells us howspread out the numbers are.

The interquartile range is the difference between the upper quartile and the lowerquartile. It gives you an idea of how spread out the values are. It is often better than therange, because an outlier could skew the range so that it is not representative of all thedata.

For example, looking at this data: 5, 6, 6, 6, 7, 7, 8, 8, 8, 25

The five-figure summary is:

Lowest = 5

Q1 = 6

Q2 = 7

Q3 = 8

Highest = 25

The range would be 25 − 5 = 20, but the interquartile range is 8 − 6 = 2. The



interquartile range is the better figure to use, because it describes how close togetherthe majority of the data is, since the range is affected significantly by the outlier 25.

The standard deviation is, like the range, interquartile range and semi-interquartilerange, a measure of the spread of the data values.

The standard deviation is often the preferred measure as it takes into account all thevalues.

The standard deviation tells us how consistent the numbers are. A low value for thestandard deviation tells us that the values are very consistent or close together.

Examples

1.

Problem:

This year the percentage mean and standard deviation for the S1 exam were 59 and7 · 84 respectively. The same exam was used with last year’s S1 and the mean andstandard deviation were 63 and 9 · 91.

Make two comparisons on the two year groups.

Solution:

Step 1: Compare the means

Last year the mean was 63% which is 4% higher so on average pupils did better.

Step 2: Compare the standard deviations

Last year the standard deviation was higher so the scores were more variable last year.

Note: We could have chosen to comment on this year’s exam. i.e. This year the meanwas lower by 4% so on average pupils did worse in this year’s S1 exam. This year thestandard deviation was lower so the scores were more consistent this year.

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2.

Problem:

A bus traveller noted the number of minutes late the bus was arriving at their busstop during July and found that the median was 9 minutes and interquartile range was3 minutes. He wrote to the bus company who assured him that he should find animprovement so he noted the minutes in August also. This time the median was 6minutes and the interquartile range was 5 minutes.

Was there an improvement in August?

Solution:

Step 1: Compare the means

In August the average number of minutes late was reduced because the median waslower by 3 minutes.

Step 2: Compare the IQRs

However, the interquartile range was higher in August so the number of minutes latewere more spread out or more variable, hence the buses were less reliable.

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3.

Problem:

Two sets of four batteries were tested to see how long in hours they lasted in constantuse.

Ultrapower 31 37 30 34

Megalife 38 33 40 29

Calculate the mean and standard deviation for the two types of batteries to determinethe differences between the two types of batteries.

Solution:

Ultrapower

x x− x̄ (x− x̄)2

31 −2 4

37 4 16

30 −3 9

34 1 1

132 0 30

x̄ =Σx

n=

132

4= 33

s =

√Σ(x − x̄)2

n − 1=

√30

4 − 1

s =

√30

3=

√10

s = 3 · 16 (to 3 s.f.)

Megalife

x x2

38 1444

33 1089

40 1600

29 841

140 4974

x̄ =Σx

n=

140

4= 35

s =

√Σx2 − (Σx)2

n

n − 1

s =

√4974 − (140)2

4

4 − 1

s = 4 · 97 (to 3 s.f.)

Megalife last longer on average by 2 hours because the mean was higher but the lifespanis more variable because the standard deviation is higher.

OR



Ultrapower do not last as long as Megalife because the mean is lower by 2 hours.Ultrapower batteries are more consistent because the standard deviation is lower.

Note: you should always compare the means and the standard deviations for any twosets of data.

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4.

Problem:

5 pupils recorded their pulse before and after a PE class.

Before 64 59 62 60 69

After 71 68 76 73 82

Identify a five-figure summary, draw boxplots and calculate the interquartile ranges forbefore and after PE then comment.

Solution:Before AfterLowest = 59 Lowest = 68Q1 = 59 · 5 Q1 = 69 · 5Q2 = 62 Q2 = 73

Q3 = 66 · 5 Q3 = 79

Highest = 69 Highest = 82

IQR = 7 IQR = 9 · 5

It is clear from the boxplots that most of the pupils pulse rates were higher after PEbecause 3/4 of the after boxplot is higher than the whole before box plot.. The pulserates were a little more varied after PE because the IQR is higher.

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Top tip

Commenting on your findings is the hardest thing to do. You should usually beable to make two points.



Comparing data sets exercise

Go online

A sample of petrol prices were taken in Glasgow and Edinburgh.

Compare the mean and standard deviations for the two data sets below by making twovalid comments.

Glasgow: Mean = £1 · 34 Standard deviation = £0 · 96Edinburgh: Mean = £1 · 32 Standard deviation = £1 · 05

Q32: Which city had higher petrol prices on average?

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Q33: Which city’s prices varied most?

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The speed of traffic was recorded in two different places in the high street. A five-figuresummary was obtained for both locations.

Lowest Q1 Median Q3 Highest

Location A 27 30 31 34 41

Location B 29 31 32 33 44

Q34: Which location had the lower average speed?

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Q35: Which location had the smallest spread of speeds?

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Q36: Why did that location have the smallest spread?

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Last year the number of hours of sunshine for one week in February were3, 5, 4, 1, 0, 5, 3.

The same week this year yielded a mean of 3 · 5 hours and a standard deviation of 2 · 3.

Q37: Calculate the mean.

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Q38: Calculate the standard deviation.

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Q39: Were the average hours of sunshine higher this year or last year?

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Q40: In which year was the weather more varied?

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1.5 Line of best fit

Scatter graphs

Go online

If a student scores well in his maths exam, will he score well in his science exam?

The results of seven students have been plotted and we can see that the points lie on avague line.

It’s not really a straight line, but perhaps good enough to say there might be a correlationbetween maths and science marks.

We could imagine a straight line being fitted to the diagram which could be used to makea prediction of a student’s science mark given his maths mark.

The line follows the rough trend of the points. About half of the points are above andhalf are below the line.

We call such a line a best fitting straight line.

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Equation of the best fitting line

Go online

The maths and science marks of seven students have been plotted on a scatter graph.The graph was such that it was felt appropriate that a straight line could be drawn torepresent the relationship.

The line was chosen using the eye as a judge. How was this done?Look at the line, see that if it follows the rough trend of the points and about half thepoints are above and half the points are below the line.

Notice that our line goes through the point (0,10) on the y-axis.

Here is the table of values that was used to create the graph. We can use this to helpus get coordinates for the line too.

maths science54 55

57 56

21 25

59 61

27 38

41 47

52 55



Notice the coordinate (0,10) on the graph.

We have chosen to plot a line though (0,10) and (52,55).

Remember, a straight line has an equation of the form y = mx + c, where c is they-intercept. . . in this case 10, y is the science mark and x is the maths mark.

So we have y = mx + c, where m is the gradient.You will already have met how to find the gradient of a straight line by using,m = vertical

horizontal or m = y2 − y1x2 − x1

So. . . m = 4552 and therefore y = 45

52x + 10

So we have y = mx + 10.When x = 52, y = 55 then 55 = 52 × m + 10Which gives y = 45

52x + 10

Using S instead of y and M instead of x we get a formula connecting the science andmaths mark.Giving, S = 45

52M + 10

If a student scored 35 in a Maths exam, we can use the formula to predict his likelyScience mark.

S = 4552M + 10 so S = 45

52 × 35 + 10 = 40 to the nearest whole number.

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Key point

A Line of Best Fit is drawn using your eye to judge.

• The line follows the rough trend of the points.

• About half of the points are above and half below the line.

• Two points on the line are picked whose coordinates are easy to read.

• These two points help to find the equation of the line.



Examples

1.

Problem:

A large chain of stores records their scarf sales over a period starting at mid-summer.

The scatter diagram shows the sales plotted against days since mid-summer.

The table gives the figures on which the graph is based.

Days sinceMid-Summer

Scarf Sales

30 4500

40 9000

50 1500

70 5500

100 9000

160 17000

200 12000

a) On the scattergraph draw your estimate of the line of best-fit.

b) Work out the equation of this line.

c) Use your equation to estimate the number of scarf sales 150 days after mid-summer.



Solution:

a)

A line is drawn:

• The line follows the rough trend of the points.

• About half of the points are above and half below the line.

• Some points are picked whose coordinates are easy to read.

• (0,2000) is a convenient point on the y-axis.

• From the original table we see that the line passes through (100, 9000).

b)

• The line is of the form y = mx + c

• The line passes through (0, 2000) so the equation has the form y = mx + 2000

• The line also passes through (100, 9000) so m = 9000 − 2000100 − 0 = 7000

100 = 70

• The equation of my line is: y = 70x + 2000 where y is the number of scarvessold x days after mid-summer.

c) To estimate the number of scarf sales 150 days after mid-summer:

• let x = 150

• y = 70 × 150 + 2000 = 12500

As an estimate we would probably say 12500 scarves were sold. Always have a quickcheck of your graph to see if your answer seems reasonable.



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2.

Problem:

A large chain of sweetshops records their ice cream sales over a period starting atmid-summer.The scatter diagram shows the sales plotted against days since mid-summer.

The table gives the figures on which the graph is based.

Days sinceMid-Summer

Ice Cream Sales

30 21000

40 20000

50 19000

70 14000

100 11000

160 5000

200 1000

a) On the scattergraph draw your estimate of the line of best-fit.

b) Work out the equation of this line.

c) Use your equation to estimate the number of ice cream sales 80 days after mid-summer.



Solution:

a)

b)

• The line is of the form y = mx + c

• The line passes through (0, 23000) so the equation has the form y = mx + 23000

• The line passes through (200, 1000) so m = 1000 − 23000200 − 0 = −22000

200 = −110

• The equation of my line is: y = − 110x + 23000 where y is the number of icecreams sold x days after mid-summer.

c) To estimate the number of ice creams sold 80 days after mid-summer:

• let x = 80

• y = − 110 × 80 + 23000 = 14200

As an estimate we would probably say 14200 ice creams were sold. This answer seemsreasonable if we look at the graph.

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3.

Problem:

The height of a sunflower was measured once every week during July and August.

Days after planting (D) 35 40 47 54 60 65 78 83

Plant height in cm (H) 40 44 51 58 63 70 82 90

a) Plot the points and draw the best fitting straight line

b) Determine the equation of the line.

c) Use your line to estimate the height of the sunflower after 120 days.



Solution:

a)Step 1: Draw axes remembering that the top row of your table represents the x-axis andthe bottom row represents the y-axis.Step 2: Plot the 8 points.Step 3: Draw in your best fitting straight line.

b)We cannot identify the y-intercept since the axes do not begin at the origin but we cansee at least 2 points. Choose points which are further apart rather than two which areclose together i.e. (47,51) and (78,82).

m = 82 − 5178 − 47 = 31

31 = 1

To get the equation we can use y − b = m(x − a) with the point (47,51) giving:

y − 51 = 1(x − 47)

y − 51 = x − 47

y = x + 4

We must remember that our axes are not named x and y and change our equationaccordingly giving H = D + 4.

c) To estimate the height of the sunflower 120 days after planting let D = 120

H = 120 + 4

H = 124

As an estimate we would probably expect the sunflower to reach a height of around 124cm.

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Top tip

Remember to check the labels on the axes and use them in the equation of yourline of best fit to gain full marks in assessments.

Line of best fit exercise

Go onlineFor each of the following questions use two points on your line to calculate the gradientand determine the equation of your line. Use your equation to find the estimated value.

Q41:

The scatter graph shows the amount of rainfall (in mm) plotted against the distance eastof Ardrossan one particular day. The table shows the distances and rainfall recorded.

Distance(km)

Rainfall(mm)

24 8 · 620 11 · 412 20

35 1 · 90 27 · 432 1

38 0

Draw a line of best fit on the graph. Mark two points on your line. Work out the equationof your line and use it to find the estimated rainfall in mm that fell 23 km east of Ardrossan.

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Q42:

A large chain of stores records their sales of scarves over a period starting at mid-summer. The scattergraph shows the sales plotted against days since mid-summer.The table shows the sales recorded.

Days Scarf sales46 37200

104 65700

239 109000

170 86200

136 78600

3 20800

87 42000

Draw a line of best fit on the graph. Mark two points on your line. Work out the equationof your line and use it to find the estimated value of the number of scarves sold 200 daysafter mid-summer.

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Q43:

The scattergraph shows the noon-day temperature plotted against the amount ofmetered water used by an office block. The table shows the temperature in degreesCelsius and volume of water in litres that were recorded.

Temperature(◦C)

Water(litres)

14 12700

36 41900

19 25600

10 9210

22 21300

26 32200

1 7250

Draw a line of best fit on the graph. Mark two points on your line. Work out the equationof your line and use it to find the estimated volume of water in litres required when thenoon-day temperature is 11 ◦C.

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Q44:

A large chain of sweet shops records their ice-cream sales over a period starting mid-summer. The scattergraph shows the sales plotted against days since mid-summer.The table shows the number of days that had passed when the sales of ice cream wererecorded.

Days Ice-cream sales243 16200

26 35300

180 20900

205 17000

143 27500

53 29800

223 19800

Draw a line of best fit on the graph. Mark two points on your line. Work out the equationof your line and use it to find the estimated ice cream sales 190 days after mid-summer.

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1.6 Learning points• To make an ordered list of numbers it is often easier to construct a stem and leaf

diagram.

• A five-figure summary requires the lowest value, the lower quartile, the median,the upper quartile and the highest value in an ordered list.

• The median (Q2) is the middle value in an ordered list.

• The lower quartile (Q1) is the middle value of the lower half of the list.

• The upper quartile (Q3) is the middle value of the upper half of the list.

• The interquartile range IQR = Q3 − Q1

• The semi-interquartile range SIQR = Q3 − Q1

2

• Σ means the sum of a list of numbers.

• The mean x̄ = Σxn where n is the number of values in the list.

• The standard deviation s =

√Σ(x − x̄)2

n − 1

or

s =

√Σx2 − (Σx)2

nn − 1

• The mean, median and mode are measures of average.

• The range, interquartile range, semi-interquartile range and standard deviation aremeasures of spread.

• When comparing data sets you should comment on:

◦ the averages;

◦ the spread, consistency or variability;

• When drawing a line of best fit try to make it go through 2 definite points.

• A Line of Best Fit is drawn using your eye to judge:

◦ the line follows the rough trend of the points;

◦ about half of the points are above and half below the line;

◦ two points on the line are picked whose coordinates are easy to read

◦ these two points help to find the equation of the line.

• Use the gradient formula to find the gradient of the line of best fit: m = y2 − y1x2 − x1

• Use the equation formula y = mx + c or y − b = m (x − a) where (a, b) is apoint on the line of best fit.

• Remember to replace x and y in the equation if the x and y axes have been definedwith different letters.



1.7 End of topic test

End of topic 3 test

Go online

Five-Figure Summary

Q45: The marks of sixteen pupils in a test were as follows:

40 24 14 37 34 28 30 41

18 17 43 32 13 22 44 38

Find the median, the quartiles and the maximum and minimum values of this data set.

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Boxplot

Q46: Look at the following set of numbers:

39 16 45 31 29 33 22 31 25 21

16 21 12 37 12 43 50 37 23

Provide the five-figure summary and create a boxplot to illustrate the data.

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Interquartile and Semi-Interquartile Ranges

Q47: The monthly rainfall (mm) recorded at a weather station during 2012 was:

140 63 44 59 63 132

124 129 111 131 135 184

Calculate the interquartile (IQR) and semi-interquartile (SIQR) ranges.

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Standard Deviation

Use either of the formulae for standard deviation.

s =

√∑(x − x̄)2

n − 1 or s =

√∑

x2 − (∑

x)2/nn − 1

For this random sample of digits:

6 9 1 3 1 6 3 9

Q48: Calculate the mean.

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Q49: Calculate the standard deviation.

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Comparing data sets

The monthly hours of sunshine during the summer of 2011 were:

164 254 120

In 2012 the data yielded∑

x = 428 and∑

x2 = 62854

s =

√∑(x − x̄)2

n − 1 or s =

√∑

x2 − (∑

x)2/nn − 1

Q50: What is the mean hours of sunshine for 2011?

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Q51: What is the standard deviation for 2011?

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Q52: What is the mean hours of sunshine for 2012?

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Q53: What is the standard deviation for 2012?

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Q54: During which year were there more hours of sunshine on average?

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Q55: During which year were the hours of sunshine more variable?

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Comparing data sets

The number of cyclists crossing a bridge each hour between 7 a.m. and 6 p.m. wererecorded two days in a row.

Day 1 yielded these numbers.

12 15 9 4 0 6 3 1 8 1 9

Day 2 yielded the following five-figure summary:

Lowest 0

Q1 3

Q2 5

Q3 7

Highest 15

Q56: Identify the median on Day 1.

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Q57: What is the IQR for Day 1?

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Q58: What is the SIQR for Day 1?

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Q59: What is the IQR for Day 2?

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Q60: On which day were the number of cyclists lower on average, day 1 or 2?

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Q61: On which day were the number of cyclists more consistent on average, day 1 or2?

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Best Fitting Line

Q62:

The scatter graph shows the amount of rainfall (in mm) plotted against the distance eastof Ardrossan one particular day. The table shows the distances and rainfall recorded.

Distance(km)

Rainfall(mm)

3 45 · 413 39 · 413 37 · 423 39 · 4

26 · 5 40 · 735 36

31 · 5 34 · 7



Draw a line of best fit on the graph. Mark two points on your line. Work out the equationof your line and use it to find the estimated rainfall (in mm) that fell 20 km east ofArdrossan.

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Q63:

A large chain of stores records their sales of scarves over a period starting at mid-summer. The scattergraph shows the sales plotted against days since mid-summer.The table shows the sales recorded.

Days Scarf sales12 · 5 11750

62 · 5 51750

82 · 5 27750

90 30000

122 · 5 69750

157 · 5 65250

160 76000

Draw a line of best fit on the graph. Mark two points on your line. Work out the equationof your line and use it to find the estimated value of the number of scarves sold 50 daysafter mid-summer.

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48 GLOSSARY

Glossary

outlier

An outlier is a value in an ordered list of data which is either much bigger or muchsmaller than the other values in the list.


ANSWERS: TOPIC 3 49

Answers to questions and activities

3 Comparing data sets using statistics

Quartiles of a data set practice (page 11)

Q1:






The lower quartile Q1 = 38 · 5The median Q2 = 48The upper quartile Q3 = 59 · 5Q2:







50 ANSWERS: TOPIC 3


Quartiles of a data set exercise (page 11)

Q3:

Hint:

• Once the list is in order place your fingers on the lowest number and the highestnumber then move them into the middle to find the median. If there are twonumbers in the middle find the average of them by adding them together anddividing them by 2.

Answer: Median (Q2) = 11

Q4:

Hint:

• The lower quartile is the median of the lower half of the list. Place your fingers onthe lowest number and the one to the left of the middle of the list i.e. the one to leftof the median or the bar.

Answer: Q1 = 9 · 5Q5:

Hint:

• The upper quartile is the median of the upper half of the list. Place your fingers onthe highest number and the one to the right of the middle of the list i.e. the one toright of the median or the bar.

Answer: Q2 = 11 · 5

Q6:

Hint:



Q7:

Hint:


Answer: Q1 = 104 · 5


ANSWERS: TOPIC 3 51

Q8:

Hint:


Answer: Q2 = 115 · 5

Q9:

Hint:



Q10:

Hint:


Answer: Q1 = 43

Q11:

Hint:


Answer: Q3 = 61

Q12:

Hint:



Q13:

Hint:



52 ANSWERS: TOPIC 3

Answer: Q1 = 26

Q14:

Hint:


Answer: Q3 = 46

Five-Figure summary practice (page 14)

Q15:



Step 3: State the 5-figure summary of the data.


Lower Quartile (36 + 37) ÷ 2 . . . 36 · 5Median . . . 39

Upper Quartile (46 + 48) ÷ 2 . . . 47


Boxplots practice (page 16)

Q16:






Median . . . 27




ANSWERS: TOPIC 3 53

Q17:

Interquartile range practice (page 18)

Q18:






Median . . . 96

Upper Quartile . . . 99 · 5Highest Score . . . 102

Step 4: State the Interquartile range.

IQR = Q3 − Q1 = 99 · 5 − 91 = 8 · 5Step 5: State the Semi-interquartile range.

SIQR = Q3 − Q12 = 99·5 − 91

2 = 4 · 25

Five-figure summaries and boxplots exercise (page 18)

Q19:

20 20 23 24 31 31 33 34 35 38 40 42 43 46 49 50 51 53 55

Find the median, lower quartile and upper quartile.20 20 23 24 31 31 33 34 35 38 40 42 43 46 49 50 51 53 55

Give the five-figure summary.



Median . . . 38




54 ANSWERS: TOPIC 3

Q20:


Lower Quartile . . . 20 · 5Median . . . 31

Upper Quartile . . . 39 · 5Highest Score . . . 45

Q21:

Q22:



Median . . . 39



Q23:

Lowest Score . . . 1 · 9Lower Quartile . . . 2 · 0Median . . . 2 · 35Upper Quartile . . . 2 · 75Highest Score . . . 3 · 4

Q24:

IQR = 0 · 75SIQR = 0 · 375

Q25:

Lowest Score . . . 2 · 55Lower Quartile . . . 2 · 90Median . . . 3 · 02Upper Quartile . . . 3 · 3Highest Score . . . 4 · 42


ANSWERS: TOPIC 3 55

Q26:

IQR = 0 · 4SIQR = 0 · 2

Calculating the mean and standard deviation practice (page 25)

Q27:

x x2

3 9

6 36

9 81

4 16

5 25

27 167

We know Σx = 27 and Σx2 = 167 and we know n = 5.

s =

√√√√167 − (27)2/5

5 − 1

=

√167 − 729/5

4

= 2 · 3 (to 1 d.p.)

Calculating the mean and standard deviation exercise (page 26)

Q28:

x̄ = 22

s = 2 · 53Q29:

Steps:

• What is∑

x? 21

• What is∑

x2? 121

• What is n? 5

Answer: s = 2 · 86Q30:

Steps:

• What is∑

x? 20

• What is∑

x2? 70

• What is n? 7


56 ANSWERS: TOPIC 3

Answer: s = 1 · 46Q31:

x̄ = 40 · 5s = 2 · 83

Comparing data sets exercise (page 30)

Q32: Glasgow

Q33: Edinburgh

Q34: Location A

Q35: Location B

Q36: Location B’s IQR was lower.

Q37: 3

Q38: 1·9

Q39: This year

Q40:

Hint:

• Which year had the higher standard deviation?

Answer: This year


ANSWERS: TOPIC 3 57

Line of best fit exercise (page 39)

Q41:

Using the points (0,27·4) and (36,0) we get m = 27·4 − 00 − 36 = −0 · 76

Then using the point (0,27·4) as the y-intercept we get the equation y = −0·76x + 27·4Rainfall 23 km east of Ardrossan: 9 · 92 mm

Q42:

Using the points (46,37200) and (170,86200) we get m = 86200 − 37200170 − 46 = 395


58 ANSWERS: TOPIC 3

Then using the point (46,37200) we get the equation,y − 37200 = 395(x − 46)

y − 37200 = 395x − 18170

y = 395x + 19030

Scarf sales 200 days after mid-summer: 98030

Q43:


Then using the point (15,18000) we get the equation,y − 18000 = 1067(x − 15)

y − 18000 = 1067x − 16005

y = 1067x + 1995

Litres of water required at 11 ◦C: 13732 litres


ANSWERS: TOPIC 3 59

Q44:

Using the points (100,28000) and (200,20000) we get m = 20000 − 28000200 − 100 = −80

Then using the point (100,28000) we get the equation,y − 28000 = −80(x − 100)

y − 28000 = −80x + 8000

y = −80x + 36000

Estimated ice-cream sales 190 days after mid-summer: 20800

End of topic 3 test (page 44)

Q45:

Hint:


• Find the lower and upper quartiles using a similar method.

Answer:Minimum 13

Lower quartile 20

Median 31

Upper quartile 39

Maximum 44


60 ANSWERS: TOPIC 3

Q46:

Minimum 12

Lower quartile 21

Median 29

Upper quartile 37

Maximum 50

Q47:

Steps:

• Put the list in order. 44, 59, 63, 63, 111, 124, 129, 131, 132, 135, 140, 184• What are the median and the lower and upper quartiles? Q 1 = 63, Q2 = 126 ·

5, Q3 = 133 · 5Answer:

• IQR = 133 · 5 − 63 = 70 · 5• SIQR = 70·5

2 = 32 · 25

Q48:

Steps:

• What is∑

x? 38

• What is n? 8

Answer: 4 · 75Q49:

Steps:

• What is∑

x2? 254

Answer: 3 · 24 to 2 d.p.

Q50:

Steps:

• What is∑

x? 538

Answer: 179 · 3Q51:

Steps:


ANSWERS: TOPIC 3 61

• What is∑

x2? 105812

Answer: 68 · 3Q52: 142 · 7Q53:

Hint:

• The information we have for 2012 means we will need to use the larger formula forstandard deviation.

Answer: 29 · 9Q54:

Hint:

• Look at which mean is higher.

Answer: 2011

Q55:

Hint:

• Look at which standard deviation is higher.

Answer: 2011

Q56: 6

Q57:

Hint:

• IQR = Q3 − Q1

Answer: IQR = 9 − 1 = 8

Q58:

Hint:

• SIQR = (Q3 − Q1) ÷ 2

Answer: SIQR = 8 ÷ 2 = 4

Q59: IQR = 7 − 3 = 4

Q60:

Hint:

• Look at which median is lower.

Answer: Day 1

Q61:

Hint:


62 ANSWERS: TOPIC 3

• Look at which IQR is lower.

Answer: Day 2

Q62:

Using the points (0,43) and (40,35) we get m = 35 − 4340 − 0 = −1

5 or − 0 · 2The equation of the line is y = − 0 · 2x + 43

Rainfall 20 km east of Ardrossan: 39 mm

Q63:


The equation of the line is y = 300x + 18000


ANSWERS: TOPIC 3 63

Scarf sales 50 days after mid-summer: 33000


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