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\title{Statistics }

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\begin{document}
\maketitle
In previous books in this series, we have looked at the measures of central tendency, such as the mean and the median.

In this chapter, we discuss two measurements of spread - the interquartile range and standard deviation. The representation of numerical data by boxplots is also introduced.

In our study of statistics up to now, we have often associated one measurement with an item. For example, the height of each person in a class, the number of possessions obtained by a player in a football match or the number of marks obtained by a student in a test.

In the last two sections of this chapter, we look at associating a pair of numbers with an item, for example, the height and weight of a person or the age and salary of an employee. This is called bivariate data.

When a measurement is collected or recorded at successive intervals of time, it is referred to as time-series data. This type of bivariate data is also introduced in this chapter.

\section*{\(18 A\) The median and the interquartile range}
The median has been introduced and discussed in earlier books in this series. We review it here, because it is the measure of central tendency used when working with the interquartile range as a measurement of spread.

\section*{Median}
We often see the median value being used to describe the housing market in a city. The median is the 'middle value' when all values are arranged in numerical order.

Here are 13 numbers in numerical order:

\[
2,2,3,3,3,4,5,11,13,18,18,19,21
\]

This data set has an odd number of values. The middle value is 5 , since it has the same number of values on either side of it. Hence, the median of this data set is 5 .

Here is a set of 12 numbers, arranged in numerical order:

\[
1,3,4,4,5,7,9,11,13,13,19,21
\]

This data set has an even number of values. The middle values are 7 and 9 . We take the average of 7 and 9 to calculate the median.

\[
\begin{aligned}
\text { Median } & =\frac{7+9}{2} \\
& =8
\end{aligned}
\]

Hence, the median of this data set is 8 , even though this value does not occur in the data set.

\section*{Median}
\begin{itemize}
  \item When a data set has an odd number of values and they are arranged in numerical order, the median is the middle value.
  \item When a data set has an even number of values and they are arranged in numerical order, the median is the average of the two middle values.
  \item When a data set with \(n\) items is arranged in numerical order, the median lies in the \(\left(\frac{n+1}{2}\right)^{\text {th }}\) position.
\end{itemize}

\section*{Example 1}
Calculate the median of the data sets.
a 33354329533945
b 57951210

\section*{Solution}
a To locate the median, first put the values in numerical order. This gives: 29333539434553

Median \(=39\)

(continued over page)
b Again, the values are placed in numerical order.

55791012

\[
\begin{aligned}
\text { Median } & =\frac{7+9}{2} \\
& =8
\end{aligned}
\]

\section*{Quartiles and the interquartile range}
The interquartile range(IQR) measures the spread of the middle \(50 \%\) of the data in an ordered data set.

We use the interquartile range to see how closely the data are grouped around the median. When we calculate the interquartile range, we organise the data into quartiles, each containing \(25 \%\) of the data. The word 'quartile' is related to 'quarter'.

Olivia has been playing Sudoku on the internet. Her last 11 games were all rated 'diabolical', and her times, correct to the nearest minute and arranged in ascending order, were:

\(8,12,14,14,16,18,19,19,25,78,523\)

The range of these times is \(523-8=515\).

Clearly the range does not give a clear picture of Olivia's considerable skills, because the last two times, 78 and 523, are outliers. An outlier is a single data value far away from the rest of the data. That is, it is much larger or much smaller than all of the other values. Outliers have a huge influence on the value of both the mean and the range. (In fact, the time of 78 minutes occurred when Olivia left the game running over dinner, and the time of 523 minutes occurred when Olivia left the game running overnight.)

Because of situations like this, the interquartile range is often a better measure of the spread of the data than the range. Here is the procedure for finding it.

Step 1: Find the median. Divide the data into two equal groups. Omit the median (middle value) if there is an odd number of values. In Olivia's case, there are 11 values so, omitting the median 18 , the two groups of 5 are:

\(8,12,14,14,16\) and \(19,19,25,78,523\)

Step 2: The lower quartile is the median of the lower set of values. In Olivia's case, the lower quartile is 14 .

Step 3: The upper quartile is the median of the upper set of values. In Olivia's case, the upper quartile is 25 .

Step 4: The interquartile range is the difference between the two quartiles. In Olivia's case:

Interquartile range \(=25-14\)

\[
=11
\]

Thus, the middle \(50 \%\) of Olivia's times have a spread of 11 minutes.

Notice that the interquartile range is unaffected by the lower quarter and the upper quarter of the values. Hence, the large sizes of two of Olivia's times, when she left the game running to eat dinner and to sleep, do not affect the interquartile range.

The calculations begin slightly differently when there is an even number of results. For example, suppose that Olivia played one more game, which she solved in 22 minutes.

There are now 12 results to arrange in ascending order:

\(8,12,14,14,16,18,19,19,22,25,78,523\)

Step 1: Since there is an even number of results, we divide them into two equal groups of 6. (The median lies 'between' the 6th and 7 th member of the ordered data set. That is, in the \(\frac{12+1}{2}=6.5\) th position.) \(8,12,14,14,16,18\) and \(19,19,22,25,78,523\)

Step 2: The lower quartile is now \(\frac{14+14}{2}=14\).

Step 3: The upper quartile is now \(\frac{22+25}{2}=23 \frac{1}{2}\).

Step 4: The interquartile range is now \(23 \frac{1}{2}-14=9 \frac{1}{2}\).

In this case, the middle \(50 \%\) of Olivia's times have a spread of \(9 \frac{1}{2}\) minutes.

The minimum, maximum, median and the two quartiles are sometimes called the five-number summary. Sometimes the lower quartile is called the first quartile, because it marks the first quarter of the ordered data. The median is then the second quartile, although this term is seldom used. The upper quartile is called the third quartile.

We denote the lower quartile by \(\mathrm{Q}_{1}\) and the upper quartile by \(\mathrm{Q}_{3}\). We sometimes use the abbreviation IQR for the interquartile range.

\section*{Example 2}
Find the interquartile range of the data set:

2619251324232325202823

\section*{Solution}
First arrange in order and locate the median.

\begin{center}
\begin{tabular}{lllllllllll}
13 & 19 & 20 & 23 & 23 & 23 & 24 & 25 & 25 & 26 & 28 \\
\end{tabular}
\end{center}

\[
\text { median }=23
\]

There are 11 data values. The 6th value is \(23,\left(\frac{11+1}{2}=6\right.\) th value \()\), so the median is 23 .

The lower group contains 5 values. The 3 rd value is 20 . So the lower quartile is 20 .

Similarly, the upper quartile is 25 .

Thus, interquartile range \(=25-20\)

\[
=5
\]

That is, the middle \(50 \%\) of data values have a spread of 5 .

\section*{Example 3}
For the stem-and-leaf plot opposite, find the median and the quartiles.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-05(1)}
\end{center}

\section*{Solution}
There are 22 data values. First locate the median to divide the data into two equal groups.

The median lies in the \(\frac{22+1}{2}=11.5\) th position of the ordered set. The 11th value is 36 and the 12 th value is 37 , so the median is 36.5 .

The lower group contains 11 values. The 6th value is 30 . So the lower quartile is 30 .

Similarly, the upper quartile is 47.

\section*{Measures of spread}
\begin{itemize}
  \item The range is the difference between the highest and lowest values in a data set.
  \item The interquartile range measures the spread of the middle \(50 \%\) of the data in an ordered data set.
  \item To calculate the interquartile range, find the difference between the upper quartile \(Q_{3}\) and the lower quartile \(\mathrm{Q}_{1}\).
\end{itemize}

\section*{Exercise 18A}
1 Find the range and interquartile range of each data set.
a 75151013320715
b 85175781057
c 40646794
d 313811118513

2 Locate the median and the quartiles for each of the following stem-and-leaf plots. State the interquartile range for each data set.
a 2201244779
311122466789
401224
\(3 \mid 2\) means 32

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-05}
\end{center}

\(6 \mid 1\) means 61

3 Find the mean, the mode, the median and the interquartile range of this data set.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
Value & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Frequency & 5 & 2 & 0 & 7 & 1 & 8 & 4 & 6 & 0 & 2 & 11 \\
\hline
\end{tabular}
\end{center}

4 Complete the following table for the positions of the median and the quartiles for data sets of 100 and 101 items. (Note: A position of 8.5 means it is between the eighth and ninth data values).

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\begin{tabular}{c}
Number of data \\
items \\
\end{tabular} & \begin{tabular}{l}
Lower quartile \\
position \\
\end{tabular} & Median position & \begin{tabular}{l}
Upper quartile \\
position \\
\end{tabular} \\
\hline
100 &  &  &  \\
\hline
101 &  &  &  \\
\hline
\end{tabular}
\end{center}

5 The stem-and-leaf plot opposite gives the height in centimetres of 20 students in a class.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-06(1)}
\end{center}

172678

1802

\(15 \mid 1\) means 151

a What is the range of the height of students in the class?

b What is the median height of students in the class?

c What is the interquartile range?

6 The stem-and-leaf plot opposite gives the lengths in centimetres of 15 leaves that have fallen from a tree. The

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-06}
\end{center}

7 The following figures are the amounts a family spent on food each week for 13 weeks.

\(\begin{array}{lllll}\$ 148 & \$ 143 & \$ 152 & \$ 149 & \$ 158\end{array}\)

\(\$ 155 \quad \$ 147 \quad \$ 152 \quad \$ 158 \quad \$ 139\)

\(\$ 143 \quad \$ 150 \quad \$ 141\)

a Find the median, upper quartile and lower quartile.

b Find the interquartile range of the amounts spent.

8 Write down two sets of seven whole numbers with minimum data value 3, lower quartile 5, median 10, upper quartile 12 and maximum data value 13 .

9 The median is always between the two quartiles. Is the mean always between the two quartiles? If not, give an example of seven whole numbers where the mean is above the upper quartile and an example where the mean is below the lower quartile.

10 a For a data set, the minimum value is 8 and the range is 27. Find the maximum value.

b For a particular data set, the upper quartile is 25.6 , and the interquartile range is 11.9. Find the lower quartile.

\section*{18B Boxplotis}
A useful way of displaying the maximum value and the minimum value, the upper and lower quartiles and the median of a data set (the five-number summary) is a boxplot.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-07}
\end{center}

The rectangle is called the box.

The horizontal lines from the lower and upper quartiles to the minimum and maximum are called the whiskers. In a boxplot, the box itself indicates the location of the middle \(50 \%\) of the data.

Boxplots are especially useful for large data sets. A boxplot is a visual summary of some of the main features of the data set. Boxplots are also useful for comparing related data sets - see Questions 9, 10 and 11 in Exercise 18B.

\section*{Example 4}
The weights of 20 students are recorded here. The weights are given to the nearest kilogram.

4852545455585861626363646566666769707279

a Find the median, upper quartile, lower quartile and interquartile range.

b Draw a boxplot for this data.

\section*{Solution}
a There are 20 data values. Therefore, the median \(=\frac{63+63}{2}=63 \mathrm{~kg}\) Divide the data into two equal groups of 10 .

48525454555858616263

63646566666769707279

The lower quartile \(=\frac{55+58}{2}=56.5 \mathrm{~kg}\)

The upper quartile \(=\frac{66+67}{2}=66.5 \mathrm{~kg}\)

The interquartile range \(=66.5-56.5\)

\[
=10 \mathrm{~kg}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-07(1)}
\end{center}

\section*{Exercise 18B}
1 The boxplot below shows the price (in \$) of 20 different brands of sports shirts.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-08(2)}
\end{center}

What is the cost of the most expensive and least expensive sports shirt?

2 The boxplot below gives information regarding the annual salaries (in thousands of dollars) of employees in a large company.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-08(1)}
\end{center}

a What is the lowest salary?

b What is the range of the salaries?

c What is the median salary?

d What is the interquartile range?

3 The boxplot below gives information about the marks out of 100 obtained by a group of 40 people on a general knowledge quiz.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-08}
\end{center}

a What was the lowest mark obtained on the quiz?

b What was the median mark obtained on the quiz?

c What was the range of marks?

d What was the interquartile range?

4 Construct a boxplot for the data set given in Exercise 18A, question 2b.

5 The pulse rates of 21 adult females are recorded.

\(\begin{array}{lllllllllllllllllllll}60 & 61 & 67 & 68 & 69 & 70 & 70 & 70 & 73 & 74 & 75 & 75 & 76 & 77 & 77 & 78 & 79 & 80 & 81 & 89 & 90\end{array}\)

a Find the median, upper quartile, lower quartile and interquartile range.

b Draw a boxplot for this data.

6 In a boxplot for a large data set, approximately what percentage of the data set is:

a below the median?

b below the lower quartile?

c in the box?

d in each whisker?

7 In a boxplot, is one whisker always longer than the other?

8 In a boxplot, why is the median not always in the centre of the box?

9 Here are two boxplots drawn on the one scale.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-09}
\end{center}

Which data set has:

a the greater median?

b the greater range?

c the greater interquartile range?

d the greater largest data value?

10 Students in two classes sat the same mathematics test. Their results are shown in the two boxplots below.

Class A

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-09(2)}
\end{center}

a Which class had the higher median mark?

b Which class had the higher interquartile range?

c In which class was the highest mark for the test obtained?

d In which class was the lowest mark for the test obtained?

e Which class did better on the test? Give reasons for your choice. (Class discussion)

11 The ratings for a number of television programs on Channel A, Channel B and Channel C were collated. The information is shown in the boxplots below. (If a program has a rating of 14 , it means that \(14 \%\) of the viewing audience watched that particular program.)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-09(1)}
\end{center}

a Write down the approximate values of the median, quartiles and maximum and minimum values for each channel.

b Which channel has the largest interquartile range?

c If the winning channel is the one with the highest rated program, which channel is the winner? Which is second? Which is third?

d If the winning channel is the one with the largest median, rank the channels.

e Can you find a criterion that makes Channel \(\mathrm{C}\) the winning channel?

\section*{\(18 C\) Boxplots, histograms and outliers}
It is common to use a form of the boxplot that is designed to illustrate any possible outliers in the data. Outliers are unusual, or 'freak', values that differ greatly in magnitude from the majority of data values.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-10}
\end{center}

\(\Delta\) outlier

\begin{itemize}
  \item Any point that is more than 1.5 IQRs away from the end of the box is classified as an outlier. That is, if a data value is greater than \(\mathrm{Q}_{3}+1.5 \times \mathrm{IQR}\) or less than it \(\mathrm{Q}_{1}-1.5 \times \mathrm{IQR}\) is considered to be an outlier. An outlier is indicated by a marker, as shown in the diagram above.
  \item The whiskers end at the highest and lowest data values that lie within 1.5 IQRs from the ends of the box.
\end{itemize}

\section*{Comparing a boxplot to the histogram of the same data}
In ICE-EM Mathematics Year 9 we looked at different shapes of histograms and the distributions of data, and in particular we used the terms symmetric, positively skewed and negatively skewed to describe the shapes.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-10(3)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-10(1)}
\end{center}

Negatively skewed distribution

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-10(2)}
\end{center}

The following examples look at representing data with histograms and boxplots.

\section*{Example 5}
The house prices of 50 houses sold in a town over a period of two years are recorded. The prices are in thousands of dollars.

\(110,110,120,130,140,150,150,170,170,170,180,190,200,210,210,230,270,270\), \(290,310,340,340,340,340,350,360,360,365,365,400,400,400,400,410,430\), \(440,450,460,460,460,460,564,678,678,750,760,904,1320,2350,2350\)

a Find the quartiles, the median and the interquartile range.

b Calculate \(1.5 \times \mathrm{IQR}\).

c Name the outliers.

d Draw a histogram and boxplot of this information. The boxplot should show outliers.

e i Calculate the mean, including the outliers.

ii Calculate the mean, not including the outliers.

\section*{Solution}
a The data has been given in ascending order. There are 50 data values. The median is the mean of the 25th and 26th values.

Median \(=\$ 355000\)

\(\mathrm{Q}_{1}\) is the median of the lower set of 25 values. This is the 13 th value.

\(\mathrm{Q}_{1}=\$ 200000\)

\(\mathrm{Q}_{3}\) is the median of the upper set of 25 values.

\(\mathrm{Q}_{3}=\$ 460000\)

\(\mathrm{IQR}=\$ 260000\)

b \(1.5 \times \mathrm{IQR}=1.5 \times\left(\mathrm{Q}_{3}-\mathrm{Q}_{1}\right)=\$ 390000\)

Hence, a value is an outlier if it is greater than \(460000+390000=\$ 850000\)

or less than \(200000-1.5 \times 260000\)

c The outliers are \(\$ 904000, \$ 1320000, \$ 2350000\) and \(\$ 2350000\).

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-11}
\end{center}

(Note: The right-hand whisker ends with the value \(\$ 760000\) )

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-11(1)}
\end{center}

The classes are \(\$ 100000\) to \(\$ 199000\), \(\$ 200000\) to \(\$ 299000\) etc.

e i Mean with outliers \(=\$ 449300\), to the nearest \(\$ 100\).

ii Mean without outliers \(=\$ 337800\), to the nearest \(\$ 100\).

It could be said that the distribution has a positive skew. The left-hand whisker is short. Most of the values lie in the interval from \(\$ 100000\) to \(\$ 500000\).

\section*{Example 6}
The waiting times in seconds at a ticket counter were as follows:

\(0,0,3,5,5,5,9,10,12,13,16,17,18,18,21,22,23,23,24,24,24,24,24,25\),

\(25,25,26,26,27,28,29,28,29,29,28,30,31,31,31,32,34,34,33,33,33,34,34,33\),

\(34,35,35,35,36,36,37,38,39,38,39,39,38,40,41,41,52\)

a Find \(\mathrm{Q}_{1}\), the median, \(\mathrm{Q}_{3}\) and the \(\mathrm{IQR}\).

b Draw a boxplot, showing outliers.

c Draw a histogram.

d Comment on the shape of the histogram and the boxplot.

\section*{Solution}
a \(\mathrm{Q}_{1}=22.5\), median \(=29, \mathrm{Q}_{3}=34.5, \mathrm{IQR}=\mathrm{Q}_{3}-\mathrm{Q}_{1}=12\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-12}
\end{center}

\(\mathrm{Q}_{3}+1.5 \times \mathrm{IQR}=34.5+1.5 \times 12=52.5\)

\(\mathrm{Q}_{1}-1.5 \times \mathrm{IQR}=22.5-1.5 \times 12=4.5\)

Therefore, the values 0,0 and 3 are considered to be outliers.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-12(1)}
\end{center}

d There is a negative skew. The right-hand whisker is short. The left-hand whisker is longer, indicating a tailing off of the data values. The values 0,0 and 3 are outliers.

\section*{Example 7}
Fifty-four lengths of wire are cut off by a machine. The resulting lengths measured in \(\mathrm{cm}\) are as shown:

103, 104, 105, 106, 106, 106, 107, 107, 107, 107, 107, 108, 108, 108, 108, 108, 108,

\(108,108,109,109,109,109,109,109,109,109,110,110,110,110,110,110,110\),

\(110,110,111,111,111,111,111,111,111,112,112,112,112,113,113,113,113\),

\(114,115,116\)

a Find \(\mathrm{Q}_{1}\), the median, \(\mathrm{Q}_{3}\) and the \(\mathrm{IQR}\).

b Draw a boxplot, showing outliers.

c Draw a histogram.

d Comment on the shape of the histogram and the boxplot.

\section*{Solution}
a \(\mathrm{Q}_{1}=108 \mathrm{~cm}\), median \(=109.5 \mathrm{~cm}, \mathrm{Q}_{3}=111 \mathrm{~cm}\) and \(\mathrm{IQR}=3 \mathrm{~cm}\)

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-13}
\end{center}

c 10

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-13(1)}
\end{center}

d The histogram is symmetric. The whiskers on the boxplot are of equal length. The values \(103 \mathrm{~cm}\) and \(116 \mathrm{~cm}\) are outliers.

\section*{Exercise 18C}
1 The heights, measured in centimetres, of 25 students in a class are:

\begin{center}
\begin{tabular}{lllllllllllll}
170 & 175 & 133 & 153 & 164 & 189 & 143 & 133 & 167 & 145 & 150 & 164 & 169 \\
159 & 177 & 186 & 173 & 164 & 177 & 168 & 142 & 155 & 153 & 167 & 166 &  \\
\end{tabular}
\end{center}

a Find \(\mathrm{Q}_{1}\), the median and \(\mathrm{Q}_{3}\).

b Find the interquartile range.

c Draw a boxplot, showing any outliers.

2 The annual incomes of 30 people, given correct to the nearest \(\$ 1000\), are:

\begin{center}
\begin{tabular}{rrrrrrrr}
54000 & 67000 & 92000 & 78000 & 54000 & 87000 & 102000 & 112000 \\
132000 & 45000 & 256000 & 89000 & 78000 & 98000 & 34000 & 75000 \\
65000 & 100000 & 34000 & 68000 & 79000 & 81000 & 82000 & 103000 \\
21000 & 345000 & 98000 & 67000 & 105000 & 98000 &  &  \\
\end{tabular}
\end{center}

a Find \(\mathrm{Q}_{1}\), the median and \(\mathrm{Q}_{3}\).

b Find the interquartile range.

c Draw a boxplot, showing any outliers.

3 Match each histogram \(\mathbf{a}-\mathbf{c}\) with its box plot \(\mathbf{i}-\) iii and describe the shape of the data distribution.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-14(5)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-14(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-14(4)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-14(1)}
\end{center}

ii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-14}
\end{center}

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-14(2)}
\end{center}

4 Consider the data shown in the stem-and-leaf plot.

\(15 \mid 68\)

1699

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-15}
\end{center}

180013447788
19) 123

a Draw a histogram.

b Find \(\mathrm{Q}_{1}\), the median, \(\mathrm{Q}_{3}\) and the \(\mathrm{IQR}\).

c Draw the boxplot.

d Comment on the shape of the histogram and the boxplot.

5 The lower and upper quartiles for a data set are 116 and 134. Which of the following data values would be classified as an outlier?
a 190
b 60
c 150

6 The speeds of 20 cars measured on a city street were recorded.

\(\begin{array}{llllllllll}40 & 14 & 3 & 26 & 20 & 31 & 42 & 36 & 17 & 24\end{array}\)

\(\begin{array}{llllllllll}28 & 33 & 27 & 29 & 24 & 51 & 11 & 35 & 5 & 24\end{array}\)

a Construct a stem-and-leaf diagram.

b Construct a boxplot.

c Comment on the shape of the distribution of data.

7 The reaction times (in milliseconds) of 20 people are listed here.

\(\begin{array}{llllllllll}38 & 31 & 36 & 39 & 35 & 25 & 35 & 44 & 43 & 44\end{array}\)

\(\begin{array}{llllllllll}46 & 34 & 62 & 22 & 42 & 48 & 31 & 30 & 45 & 40\end{array}\)

a Find the median, \(\mathrm{Q}_{1}, \mathrm{Q}_{3}\) and the interquartile range.

b Construct a boxplot.

c Identify any outliers.

8 The weight loss (in kilograms) of 20 randomly selected people undertaking a special diet over three weeks is:

\(\begin{array}{llllllllll}8 & 5 & 10 & 6 & 6 & 12 & 4 & 5 & 5 & 6\end{array}\)

\(\begin{array}{llllllllll}8 & 13 & 7 & 7 & 7 & 6 & 6 & 4 & 5 & 5\end{array}\)

a Construct a dotplot of the data.

b Construct a boxplot of the data.

c Comment on the shape.

\section*{\(18 D\) The mean and the standard deviation}
\section*{Mean}
The mean of a data set is a measure of its centre. The mean is calculated by adding together all the data values and then dividing the resulting sum by the number of data values.

\[
\text { Mean }=\frac{\text { sum of values }}{\text { number of values }}
\]

A more common name for the mean is 'average'. We use the symbol \(\bar{x}\) to denote the mean.

For a set of data \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\),

\[
\bar{x}=\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}
\]

\section*{Example 8}
A student obtained the following marks in seven tests:

\(43,35,41,29,33,39\) and 42

Calculate the mean mark correct to two decimal places.

\section*{Solution}
\[
\begin{aligned}
\bar{x} & =\frac{43+35+41+29+33+39+42}{7} \\
& \approx 37.43 \quad \text { (Correct to two decimal places. })
\end{aligned}
\]

For larger sets of data, a frequency table can be prepared. Let \(f_{1}\) be the frequency of the data item \(x_{1}\), let \(f_{2}\) be the frequency of the data item \(x_{2}\) and so on. In this case we can write:

\[
\bar{x}=\frac{f_{1} x_{1}+f_{2} x_{2}+\ldots+f_{s} x_{s}}{f_{1}+f_{2}+\ldots+f_{s}}
\]

The numerator is the sum of the data items and the denominator is the number of data items.

\section*{Example 9}
The following information gives the number of children in each of 20 families. Calculate the mean number of children per family.

\begin{center}
\begin{tabular}{c|c}
Number of children \(x_{i}\) & Frequency \(f_{i}\) \\
\hline
0 & 4 \\
1 & 5 \\
2 & 7 \\
3 & 4 \\
\end{tabular}
\end{center}

\section*{Solution}
Add in a column for \(f_{i} x_{i}\).

\begin{center}
\begin{tabular}{c|c|c}
Number of children \(x_{i}\) & Frequency \(f_{i}\) & \(f_{i} x_{i}\) \\
\hline
0 & 4 & 0 \\
1 & 5 & 5 \\
2 & 7 & 14 \\
3 & 4 & 12 \\
\hline
 & Total \(=20\) & Total \(=31\) \\
\hline
\end{tabular}
\end{center}

\(\bar{x}=\frac{31}{20}=1.55\)

It is obviously impossible for a family to have 1.55 children. The mean is not necessarily a member of the data set.

\section*{Standard deviation}
The standard deviation of a set of data is a measure of how far the data values are spread out from the mean. The difference between each data item and the mean is called the deviation of the data value. The sum of the deviations is zero, which will be proved in question 10 of Exercise 18D.

The standard deviation is calculated from the squares of the deviations.

Here are the steps in finding the standard deviation:

\begin{itemize}
  \item Calculate the mean.
  \item Square each of the deviations.
  \item Sum these squares.
  \item Divide the sum of the squares by the number of data values.
  \item Take the square root of the value obtained.
\end{itemize}

This is given by the formula:

\[
\sigma=\sqrt{\frac{\left(x_{1}-\bar{x}\right)^{2}+\left(x_{2}-\bar{x}\right)^{2}+\left(x_{3}-\bar{x}\right)^{2}+\ldots+\left(x_{n}-\bar{x}\right)^{2}}{n}}
\]

where the \(x_{i}\) are the data values, \(\bar{x}\) is the mean and \(n\) is the number of data values.

We will use the Greek letter \(\sigma\) (sigma) to denote the standard deviation of a data set.

\section*{Example 10}
Find the standard deviation, correct to two decimal places, for the data set.

\(5,7,11,13,14\)

\section*{Solution}
\[
\begin{aligned}
\bar{x} & =\frac{5+7+11+13+14}{5} \\
& =10 \\
\sigma^{2} & =\frac{(5-10)^{2}+(7-10)^{2}+(11-10)^{2}+(13-10)^{2}+(14-10)^{2}}{5} \\
& =\frac{25+9+1+9+16}{5} \\
& =\frac{60}{5} \\
& =12
\end{aligned}
\]

Hence, \(\sigma=\sqrt{12} \approx 3.46\) (Correct to two decimal places.)

When calculating the standard deviation from a frequency table, we can use the following formula:

\[
\sigma=\sqrt{\frac{f_{1}\left(x_{1}-\bar{x}\right)^{2}+f_{2}\left(x_{2}-\bar{x}\right)^{2}+f_{3}\left(x_{3}-\bar{x}\right)^{2}+\ldots+f_{s}\left(x_{s}-\bar{x}\right)^{2}}{f_{1}+f_{2}+\ldots+f_{s}}}
\]

When frequencies are taken into account, we can see that this is the same formula as above.

We can calculate the standard deviation with an extended frequency table with five columns. Fill in the first three columns, then calculate \(\bar{x}\). Fill in the other two columns and then calculate \(\sigma\).

\section*{Example 11}
Calculate the mean and standard deviation of the set of values, correct to two decimal places.

\(1,3,4,5,7,3,6,9,9,4,5,2,5,7\)

\section*{Solution}
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(x_{i}\) & \(f_{i}\) & \(f_{i} x_{i}\) & \(\left(x_{i}-\bar{x}\right)\) & \(f_{i}\left(x_{i}-\overline{\boldsymbol{x}}\right)^{\mathbf{2}}\) \\
\hline
1 & 1 & 1 & -4 & 16 \\
\hline
2 & 1 & 2 & -3 & 9 \\
\hline
3 & 2 & 6 & -2 & 8 \\
\hline
4 & 2 & 8 & -1 & 2 \\
\hline
5 & 3 & 15 & 0 & 0 \\
\hline
6 & 1 & 6 & 1 & 1 \\
\hline
7 & 2 & 14 & 2 & 8 \\
\hline
9 & 2 & 18 & 4 & 32 \\
\hline
 & Total = 14 & Total = 70 &  & Total = 76 \\
\hline
\end{tabular}
\end{center}

\[
\begin{aligned}
& \bar{x}=\frac{70}{14}=5 \\
& \sigma=\sqrt{\frac{76}{14}} \\
& \approx 2.33 \quad \text { (Correct to two decimal places. })
\end{aligned}
\]

Note: The sum of the deviations \(x_{i}-\bar{x}\) is zero. Hence, the average of the deviations is not useful.

\section*{Mean and standard deviation}
\begin{itemize}
  \item The mean of a set of data is denoted by \(\bar{x}\).
  \item The standard deviation of a data set is a measure of spread and is denoted by the Greek letter \(\sigma\).
  \item There are two formulas for the standard deviation.
\end{itemize}

\(\sigma=\sqrt{\frac{\left(x_{1}-\bar{x}\right)^{2}+\left(x_{2}-\bar{x}\right)^{2}+\left(x_{3}-\bar{x}\right)^{2}+\ldots+\left(x_{n}-\bar{x}\right)^{2}}{n}}\), when the data is in a list.

\(\sigma=\sqrt{\frac{f_{1}\left(x_{1}-\bar{x}\right)^{2}+f_{2}\left(x_{2}-\bar{x}\right)^{2}+f_{3}\left(x_{3}-\bar{x}\right)^{2}+\ldots+f_{s}\left(x_{s}-\bar{x}\right)^{2}}{f_{1}+f_{2}+\ldots+f_{s}}}\), when the data is in a frequency table.

It is clear that the larger the standard deviation, the more spread out the data are about the mean.

For example, here is a bar chart of the data in Example 11, and also another set of 14 data items where the data are not as spread out but have the same mean.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-19}
\end{center}

\(\bar{x}=5\) and \(\sigma \approx 2.33\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-19(1)}
\end{center}

\(\bar{x}=5\) and \(\sigma \approx 1.25\)

In the following section we will see how the standard deviation may be used to make comparisons between data sets.

\section*{Use of calculators}
Many calculators and spreadsheets have a built-in facility for calculating the standard deviation of a set of data.

To save time, we recommend using this facility for all but the simplest data sets. In particular, if \(\bar{x}\) is not an integer, then calculating \(\sigma\) is very tedious.

It should be noted that in this book we calculate the standard deviation by dividing the sum of the squares of the deviations by \(n\), the number of data items, and taking the square root. There is also another type of standard deviation that is obtained by dividing the sum of the squares of the deviations by \(n-1\), and taking the square root. Many calculators offer both versions. Sometimes they are denoted by symbols such as \(\sigma_{n}\) and \(\sigma_{n-1}\). In this book, we only use \(\sigma_{n}\).

\section*{Exercise 18D}
Give all answers correct to two decimal places unless otherwise specified.

1 During a 13-week football season, the number of kicks obtained by a particular player each week is:

\(18,18,20,26,10,8,21,14,16,14,12\) and 16

Calculate the mean number of kicks obtained by the player.

2 The daily maximum temperature was recorded in two different cities for a week. The results are shown below.

City A: 28, 31, 34, 32, 31, 29, 28

City B: 26, 32, 36, 38, 37, 29, 25

Which city had the greater mean daily maximum temperature?

3 The average of 5 masses is \(67 \mathrm{~kg}\). If a mass of \(25 \mathrm{~kg}\) is added, what is the average of the 6 masses?

4 During a term, a student has an average of 46 marks after the first four tests and his average for the next six tests is 38 marks. What is his average for the ten tests?

5 a Calculate, correct to two decimal places, the mean and standard deviation for the data sets.

\[
\text { i } 2,4,8,10,2,9,3,8,2,2 \quad \text { ii } \quad 3,6,4,5,6,7,3,4,6,6
\]

b Comment on the results from part \(\mathbf{a}\).

6 Complete the following extended frequency table to calculate the mean and standard deviation of the given data set.

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(x_{i}\) & \(f_{i}\) & \(f_{i} x_{i}\) & \(\left(x_{i}-\bar{x}\right)\) & \(f_{i}\left(x_{i}-\bar{x}\right)^{\mathbf{2}}\) \\
\hline
1 & 2 &  &  &  \\
\hline
2 & 7 &  &  &  \\
\hline
3 & 6 &  &  &  \\
\hline
4 & 1 &  &  &  \\
\hline
5 & 2 &  &  &  \\
\hline
6 & 2 &  &  &  \\
\hline
 & Total \(=\) & Total \(=\) &  & Total \(=\) \\
\hline
\end{tabular}
\end{center}

7 Use a calculator to find, correct to two decimal places, the mean and standard deviation for each data set.

a \(3,6,7,5,8,5,10,12,13,12,6,9,12,14,15\)

b \(8,10,12,14,16,17,19,12,11,10,14,16,18,19\)

8 Twenty students sat a test and their results are given in the stem-and-leaf plot below.

\begin{center}
\begin{tabular}{ll|lllll}
 & 1 & 2 & 2 & 8 & 9 &  \\
 & 2 & 2 & 4 & 5 & 6 & 8 \\
\(1 \mid 2\) means 12 & 3 & 0 & 2 & 6 & 8 & 8 \\
 & 4 & 0 & 1 & 2 & 3 & 6 \\
\end{tabular}
\end{center}

a Calculate their mean mark.

b How many students obtained a mark higher than the mean mark?

c Find the standard deviation of their marks.

9 Twenty people completed a test worth 10 marks. Their scores are shown in the frequency table below.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
Score & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Number of people & 0 & 2 & 0 & 1 & 1 & 2 & 4 & 6 & 0 & 2 & 2 \\
\hline
\end{tabular}
\end{center}

a Calculate the mean mark.

b How many students obtained a mark lower than the mean mark?

c Find the standard deviation of their marks.

10 a Prove that the sum of the deviations for the data set \(a, b, c\) is zero.

b Prove that the sum of the deviations of any data set is zero.

\section*{\(18 E\) Interpreting the standard deviation}
Consider the data sets 4, 5, 6, 7, 8 and 2, 4, 6, 8, 10 .

Both the data sets have a mean and median of 6 . However, when we apply the formula for \(\sigma\), it can be observed that the standard deviation for the second data set is \(2 \sqrt{2}\), which is twice the standard deviation of the first data set, \(\sqrt{2}\). This reflects the difference in spread between the two data sets. That is, even though both have evenly distributed values, the spread of data from the mean is twice as great in the second data set as compared to the first.

\section*{Intervals about the mean}
In the following we will look at a 'symmetric' set of data which 'tails off' as you move away from the mean in either direction.

The stem-and-leaf plot on the next page gives the incomes, in thousands of dollars, of 134 people.

\begin{center}
\begin{tabular}{l|l}
0 & 889 \\
1 & 00223 \\
2 & 4444448888888 \\
3 & 11122444466667777788888999 \\
4 & 111112223334444455556677777788999999999 \\
5 & 000001111111122233344444577 \\
6 & 3333366669999 \\
7 & 77899 \\
8 & \(666 \quad 7 \mid 7\) means \(\$ 77000\) \\
\end{tabular}
\end{center}

The mean is 45.1 , the median is 45.5 , and the standard deviation is 16.1 .

We next consider intervals centred on the mean.

\(\bar{x}+\sigma=45.1+16.1=61.2\) and \(\bar{x}-\sigma=45.1-16.1=29.0\)

We can observe from the plot above that there are 92 values between 29 and 61 ; hence, the percentage of values within one standard deviation of the mean is \(68.7 \%\). Also,

\(\bar{x}+2 \sigma=45.1+2 \times 16.1=77.3\) and

\(\bar{x}-2 \sigma=45.1-2 \times 16.1=12.9\)

There are 121 values between 13 and 77 .

Thus, the percentage of values within two standard deviations of the mean is \(90.3 \%\).
\includegraphics[max width=\textwidth, center]{2024_02_22_c6b381e97fdd24dc5428g-22}

We have seen that about \(69 \%\) of the data is within one standard deviation of the mean and about \(90 \%\) of the data is within two standard deviations of the mean.

Histograms similar to this one occur frequently. In most cases like these the median and the mean are very close.

\section*{Example 12}
David plays golf every Friday. He has recorded his score each Friday for five years, and has found that his mean score for all his games is 85 and the standard deviation of his scores is 5.2.

Find the range of scores that lie within:

a one standard deviation of the mean b two standard deviations of the mean

\section*{Solution}
a \(\bar{x}+\sigma=85+5.2=90.2\) and \(\bar{x}-\sigma=85-5.2=79.8\) So the range of scores within one standard deviation of the mean is 80 to 90 .

b \(\bar{x}+2 \sigma=85+10.4=95.4\) and \(\bar{x}-2 \sigma=85-10.4=74.6\) So the range of scores within the two standard deviations of the mean is 75 to 95 .

A remarkable result known as Chebyshev's inequality states that, for any set of data, if we take an interval between \(\bar{x}-k \sigma\) and \(\bar{x}+k \sigma\), then all values can lie outside this interval for \(0<k \leq 1\), but for \(k>1\), at most \(\frac{1}{k^{2}}\) of the data can lie outside this interval.

So, for example, taking \(k=2\), not more than \(\frac{1}{4}\) of the data can be outside this interval.

So at least \(75 \%\) of the data must lie inside this interval.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-23}
\end{center}

\section*{Using the standard deviation to compare data}
To compare values from different data sets with approximately the same shape, it is useful to consider where they are positioned relative to their respective means. This can be achieved by using their respective standard deviations, and calculating where these values lie in terms of the number of standards above or below the mean.

\section*{Example 13}
Gus scored 14 in a maths test and 14 in an English test. The scores of each student in the maths and English classes are listed below. In which test did Gus perform better, relative to the class results?

Maths test: 10, 13, 18, 17, 12, 16, 9, 8, 7, 11, 10, 12

English test: 15, 17, 18, 19, 18, 17, 19, 16, 14, 15, 14, 12

\section*{Solution}
Maths test \(\quad \bar{x}=\frac{143}{12} \approx 11.92, \quad \sigma \approx 3.38\)

English test \(\quad \bar{x} \approx 16.17, \quad \sigma \approx 2.11\)

It can be seen that in the maths test Gus scored about 0.6 of a standard deviation above the mean \(\left(\frac{14-11.92}{3.38} \approx 0.6\right)\) and in the English test Gus scored about 1 standard deviation below the mean \(\left(\frac{14-16.17}{2.11} \approx-1\right)\). So Gus has done better relative to the class in the maths test.

\section*{Exercise 18E}
1 Find the mean and standard deviation of each set of data.

a \(5,6,6,7,8,9,22\)

b \(11,7,8,9,8,10,10\)

c \(1,3,7,9,11,15,17\)

Compare the sets of data using their means and standard deviations.

2 The mean and standard deviation of each set of data is given. Find the range of values that is within:

i one standard deviation of the mean ii two standard deviations of the mean

a \(\bar{x}=35, \sigma=2.5\)

b \(\bar{x}=40, \sigma=5\)

c \(\bar{x}=35, \sigma=8\)

3 The mathematics and English marks for a class of 15 students are given below.

Mathematics: \(12,16,14,19,17,18,15,15,19,20,14,18,19,15,11\)

English: \(\quad 10,13,16,19,20,19,18,16,15,14,17,11,15,18,17\)

a Calculate, correct to two decimal places, the mean and standard deviation for each set of marks.

b If a student scored 16 for the mathematics test and 14 for the English test, which is the better mark relative to the class results?

4 The following table lists the marks of several students on different tests in English and mathematics. Compare the English and mathematics marks of each student.

a

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
 & Mark & Mean & Standard deviation \\
\hline
\multicolumn{4}{|l|}{David} \\
\hline
English & 15 & 17 & 2 \\
\hline
Mathematics & 13 & 17 & 3 \\
\hline
\multicolumn{4}{|l|}{Akira} \\
\hline
English & 42 & 30 & 6 \\
\hline
Mathematics & 39 & 25 & 8 \\
\hline
\multicolumn{4}{|l|}{Katherine} \\
\hline
English & 70 & 75 & 5 \\
\hline
Mathematics & 65 & 70 & 10 \\
\hline
\multicolumn{4}{|l|}{Daniel} \\
\hline
English & 70 & 55 & 9 \\
\hline
Mathematics & 69 & 62 & 7 \\
\hline
\end{tabular}
\end{center}

5 The bar charts of three sets of data are shown.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-25}
\end{center}

a For each set of data, calculate the mean and the standard deviation.

b Add 5 onto each data item in each of i, ii and iii and state the mean and standard deviation of each new set of data.

c Multiply each data item in each of \(\mathbf{i}\), ii and iii by 2 and state the mean and standard deviation of each new set of data.
\includegraphics[max width=\textwidth, center]{2024_02_22_c6b381e97fdd24dc5428g-25(1)}

6 (There is no arithmetic required in the following.)

Make up a list of 10 numbers so that the standard deviation is as large as possible and:
a every number is either 1 or 5
b every number is either 1 or 9
c every number is either 1 or 5 or 9 , and at least two of them are 5

7 Repeat question 6 , but this time so the standard deviation is as small as possible.

8 An employer has 29 employees whose weekly salaries have \(\bar{x}=\$ 429\) and \(\sigma=\$ 1.53\). The employer decides to give a flat \(\$ 100\) raise to every employee.

a What would be the change to the average annual salary paid by the employer?

b Would there be a change in the standard deviation?

c What would be the change in total weekly payments to employees?

\section*{\(18 \mathrm{~F}_{\text {Time-series data }}\)}
A time series is a set of data that has been obtained by taking repeated measurements over time.

Maximum daily temperatures, average weekly wages, quarterly sales figures of a company and annual population of a city are all examples of a time series.

To represent the information obtained in a time series pictorially, a graph is drawn in which:

\begin{itemize}
  \item the horizontal axis represents time
  \item the vertical axis represents the quantity that is being measured at regular intervals
  \item adjacent plotted points are joined by line intervals.
\end{itemize}

\section*{Example 14}
The mean daily maximum temperature was measured each month in a particular city.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Month & Jan & Feb & Mar & Apr & May & Jun & Jul & Aug & Sep & Oct & Nov & Dec \\
\hline
\begin{tabular}{l}
Mean daily \\
max. temp \(\left({ }^{\circ}\right.\) C) \\
\end{tabular} & 29.2 & 28.9 & 28.1 & 26.4 & 23.5 & 21.2 & 20.6 & 21.7 & 23.8 & 25.7 & 27.4 & 28.7 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a time-series plot.

b Briefly comment on the annual variation in daily maximum temperature.

\section*{Solution}
a To construct a time-series plot, the months are placed on the horizontal axis and the vertical axis will represent the mean daily maximum temperature. The points are plotted and joined by lines. The following time-series plot is obtained.

b There is a gradual decrease in the mean daily maximum temperature over the months January, February and March. During April, May and June, the mean daily maximum temperature falls quite

\includegraphics[max width=\textwidth, center]{2024_02_22_c6b381e97fdd24dc5428g-26}
quickly to a minimum during July. For the remainder of the year, there is a steady increase in the mean daily maximum temperature each month.

\section*{Exercise 18F}
1 a Construct a time-series plot for the average rainfall (in \(\mathrm{cm}\) ) in a particular city, which is given in the table below.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Month & Jan & Feb & Mar & Apr & May & Jun & Jul & Aug & Sep & Oct & Nov & Dec \\
\hline
Rainfall (in cm) & 16.2 & 17.5 & 14.2 & 9.1 & 9.6 & 7.1 & 6.2 & 4.1 & 3.3 & 9.3 & 9.6 & 12.6 \\
\hline
\end{tabular}
\end{center}

b Use the time-series plot to write a brief description as to how the rainfall varies in this particular city.

2 The table below gives the annual profit (in \$ million) of a particular company over a 10-year period. Construct a time-series plot of the information.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|}
\hline
Year & 1989 & 1990 & 1991 & 1992 & 1993 & 1994 & 1995 & 1996 & 1997 & 1998 \\
\hline
Profit (\$ million) & 1.2 & 1.8 & 2.4 & 2.2 & 2.6 & 3.1 & 3.2 & 3.4 & 3.6 & 4.0 \\
\hline
\end{tabular}
\end{center}

3 The table below gives the number of births that occurred in a hospital each month for a year.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Month & Jan & Feb & Mar & Apr & May & Jun & Jul & Aug & Sep & Oct & Nov & Dec \\
\hline
Number of births & 52 & 46 & 43 & 40 & 31 & 32 & 26 & 27 & 24 & 20 & 26 & 26 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a time-series plot.

b Briefly describe how the number of births recorded each month changed over the year.

4 The table below gives the position of a particular football team in a competition of 12 teams at the completion of each round throughout the season.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Round & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\
\hline
Position & 10 & 12 & 11 & 9 & 8 & 6 & 5 & 5 & 4 & 5 & 5 \\
\hline
Round & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\
\hline
Position & 6 & 4 & 4 & 3 & 4 & 3 & 5 & 7 & 6 & 9 & 8 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a time-series plot.

b Briefly describe the progress of the team throughout the season.

5 The data below shows the quarterly sales of a department store over a period of three years. The quarters are labelled 1 to 12 in the corresponding time-series graph.

\begin{center}
\begin{tabular}{|c|c|}
\hline
Sales quarter & Sales \$'000 \\
\hline
\(2009-1\) & 45 \\
\hline
\(2009-2\) & 63 \\
\hline
\(2009-3\) & 67 \\
\hline
\(2009-4\) & 43 \\
\hline
\(2010-1\) & 51 \\
\hline
\(2010-2\) & 69 \\
\hline
\(2010-3\) & 75 \\
\hline
\(2010-4\) & 39 \\
\hline
\(2011-1\) & 55 \\
\hline
\(2011-2\) & 71 \\
\hline
\(2011-3\) & 79 \\
\hline
\(2011-4\) & 49 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-27}
\end{center}

a In which quarter of each year are the sales figures the worst?

b In which quarter of each year are the sales figures the best?

c Are the sales figures improving? Compare the sales figures for the first quarter of each year and do the same for the other quarters.

6 The table below gives the quarterly sales figures for a car dealer for the period 2009-2011.

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
Number of sales & Q1 & Q2 & Q3 & Q4 \\
\hline
2009 & 72 & 62 & 90 & 98 \\
\hline
2010 & 87 & 78 & 112 & 111 \\
\hline
2011 & 90 & 84 & 132 & 117 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a time-series plot.

b Briefly describe how the car sales have altered over the given time period.

c Does it appear that the car dealer is able to sell more cars in a particular period each year?

\section*{\(18 G_{\text {Bivariate data }}\)}
We often want to know if there is a relationship between the items in two different data sets.

\begin{itemize}
  \item Is there a relationship between children's ages and their heights?
  \item Is there a relationship between people's heights and weights?
  \item Is there a relationship between students' marks in English and their marks in mathematics?
\end{itemize}

In each of the above, two pieces of information are to be collected from each person in the investigation and then the two data sets are to be compared. When two pieces of information are collected from each subject in an investigation, we are then concerned with bivariate data.

A scatter graph or scatter plot is a type of display that uses coordinates to display values for two variables for a set of data. The data is displayed as a collection of points, each having the value of one variable determining the position of the horizontal coordinate and the value of the other variable determining the position of the vertical coordinate.

\section*{Example 15}
The age (in years) and height (in \(\mathrm{cm}\) ) of a group of people was recorded. The data obtained is shown in the table on the right. Present the information in the table on a scatter plot.

\begin{center}
\begin{tabular}{|l|c|c|}
\hline
Person & Age (years) & Height \((\mathbf{c m})\) \\
\hline
Alan & 12 & 145 \\
\hline
Brianna & 14 & 140 \\
\hline
Chiyo & 15 & 160 \\
\hline
Danielle & 14 & 150 \\
\hline
Ezra & 10 & 130 \\
\hline
Frankie & 11 & 135 \\
\hline
\end{tabular}
\end{center}

\section*{Solution}
The variables under consideration are age and height. The horizontal axis represents the age and the vertical axis represents height. The axes are broken (using the symbol \(\downarrow\) ) to allow us to focus on the data points.

In this scatter plot, it is noted that points towards the top-right of the plot represent individuals who are older and taller. Points in the bottom-right represent individuals who are older but shorter than the rest of the group. The bottom-left of the plot represents people who are younger and shorter, while the top-left

\includegraphics[max width=\textwidth, center]{2024_02_22_c6b381e97fdd24dc5428g-28}
portion of the graph represents individuals who are younger but taller than the rest of the group.

We can see from the general trend of the points, which is upward as we move to the right, that the height of a child increases as the child grows older (for children in this data set).

\section*{Example 16}
The second-hand price and age of a particular model of car are recorded in the table below, and the points plotted on a scatter plot.

\begin{center}
\begin{tabular}{|c|c|}
\hline
\begin{tabular}{c}
Age of car \\
(year) \\
\end{tabular} & \begin{tabular}{c}
Second-hand \\
price(\$) \\
\end{tabular} \\
\hline
1 & 22000 \\
\hline
2 & 19500 \\
\hline
2 & 18700 \\
\hline
3 & 16400 \\
\hline
3 & 17000 \\
\hline
3 & 16800 \\
\hline
4 & 15800 \\
\hline
4 & 15950 \\
\hline
5 & 14800 \\
\hline
6 & 12500 \\
\hline
6 & 12000 \\
\hline
6 & 12800 \\
\hline
7 & 12200 \\
\hline
7 & 11580 \\
\hline
8 & 10500 \\
\hline
8 & 9200 \\
\hline
8 & 8600 \\
\hline
9 & 5700 \\
\hline
10 & 4850 \\
\hline
11 & 4500 \\
\hline
 &  \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-29}
\end{center}

a Describe the points in the top-left of the plot.

b Describe the points in the bottom-right of the plot.

c Describe the trend.

\section*{Solution}
a The top-left of the scatter plot has points corresponding to relatively new second-hand cars with higher prices.

b The bottom-right of the scatter plot has points corresponding to older second-hand cars with lower prices.

c As the age of the car increases the value decreases.

\section*{Exercise 18G}
1 The table below gives the marks obtained by 10 students in a mathematics examination and an English examination.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
Mathematics mark & 72 & 50 & 96 & 58 & 86 & 94 & 78 & 66 & 85 & 78 \\
\hline
English mark & 78 & 64 & 70 & 46 & 88 & 72 & 70 & 62 & 72 & 74 \\
\hline
\end{tabular}
\end{center}

Represent this information on a scatter plot, using the horizontal axis to represent the mathematics marks and the vertical axis to represent the English marks.

Break the axes so that the vertical axis starts near 40 and the horizontal axis starts near 50 .

2 The table below gives the average monthly rainfall, in \(\mathrm{mm}\), and the average number of rainy days per month for twelve different cities in Australia.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Average rainfall (in \(\mathrm{mm}\) ) & 161 & 175 & 142 & 90 & 96 & 71 & 62 & 41 & 33 & 93 & 96 & 126 \\
\hline
Average number of rainy days & 13 & 14 & 14 & 11 & 10 & 7 & 7 & 6 & 7 & 10 & 10 & 12 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a scatter plot. Use the horizontal axis to represent average monthly rainfall and the vertical axis to represent the average number of rainy days per month.

b Give a brief description of the relationship between rainy days and average rainfall.

3 The table below gives the amount of carbohydrates, in grams, and the amount of fat, in grams, in \(100 \mathrm{~g}\) of a number of breakfast cereals.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\hline
Carbohydrates (in g) & 88.7 & 67.0 & 77.5 & 61.7 & 86.8 & 32.4 & 72.4 & 77.1 & 86.5 \\
\hline
Fat (in g) & 0.3 & 1.3 & 2.8 & 7.6 & 1.2 & 5.7 & 9.4 & 10.0 & 0.7 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a scatter plot. Use the \(x\)-axis to represent the amount of carbohydrates and the \(y\)-axis to represent the amount of fat.

b Does there appear to be any relationship between the carbohydrate content and the fat content?

4 The table below gives the IQ of a number of adults and the time, in seconds, for them to complete a simple puzzle.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\hline
IQ & 115 & 118 & 110 & 103 & 120 & 104 & 124 & 116 & 110 \\
\hline
Time (in seconds) & 14 & 15 & 21 & 27 & 11 & 25 & 9 & 16 & 18 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a scatter plot. Use the \(x\)-axis to represent IQ and the \(y\)-axis to represent the time taken to complete the puzzle.

b Is there any trend in the data?

5 The table below gives the number of kicks and the number of handballs obtained by each player in an AFL team in a particular match.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Player & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\
\hline
Number of kicks & 3 & 20 & 7 & 19 & 7 & 6 & 2 & 9 & 7 & 26 & 3 \\
\hline
Number of handballs & 8 & 11 & 11 & 6 & 4 & 6 & 3 & 1 & 3 & 3 & 8 \\
\hline
Player & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\
\hline
Number of kicks & 12 & 17 & 6 & 11 & 14 & 5 & 1 & 21 & 6 & 13 & 4 \\
\hline
Number of handballs & 4 & 5 & 0 & 3 & 8 & 3 & 0 & 11 & 0 & 17 & 11 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a scatter plot. Use the \(x\)-axis to represent the number of kicks and the \(y\)-axis to represent the number of handballs.

b Does your scatter plot support the claim, 'the more kicks a player obtains, the more handballs he gives'? Explain your answer.

6 The table below gives the number of 'goals for' (scored by the team) and the number of 'goals against' (scored by the opposing team) for each team in a soccer competition.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Team & A & B & C & D & E & F & G & H & I & J & K & L \\
\hline
Goals for & 36 & 45 & 22 & 26 & 20 & 59 & 24 & 41 & 23 & 43 & 32 & 41 \\
\hline
Goals against & 31 & 16 & 33 & 26 & 64 & 16 & 53 & 42 & 47 & 21 & 49 & 14 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a scatter plot. Use the \(x\)-axis to represent 'goals for' and the \(y\)-axis to represent 'goals against'.

b Use your scatter plot to answer the following questions.

i Which team is the best team in the competition? Why?

ii Which team is the worst team in the competition? Why?

iii Which of team \(\mathrm{J}\) and team \(\mathrm{H}\) is better? Why?

7 The scatter plot at the right gives information about the height and weight of a number of people. Annabelle's height and weight is represented by the point \(\mathrm{A}\).

Write down the point that represents each of the following people.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-31}
\end{center}

a Barry, who is heavier and taller than Annabelle

b Chandra, who is shorter but heavier than Annabelle

c Dario, who is the same height as Barry but a little heavier

Height (cm)

d Edwina, who is shorter and lighter than Chandra

e Frederick, who is the same weight as Barry but a bit taller

f George, who is the same height as Annabelle but heavier

g Harriet, who is the same weight as Annabelle but shorter

h Ivan, who is the tallest person in the group

8 The scatter plot at the right gives the marks obtained by students in two tests.

John's marks on the tests are represented by the point \(\mathrm{J}\).

Which point represents each of the following students?

a Alex, who got the top mark in both tests

b Bao, who got the top mark in Test 1 but not in Test 2

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-32}
\end{center}

c Charlene, who did better in Test 1 than John, but not as well on Test 2

d Drago, who did not do as well as Charlene on either test

e Eddie, who got the same mark as John for Test 2, but did not do as well as John on Test 1

f Francis, who got the same mark as John for Test 1, but did better than John on Test 2

g Georgina, who got the lowest mark for Test 1

h Harvir, who had the greatest discrepancy between his two marks

9 The test results of a group of 9 students is recorded in the table and plotted on a scatter plot. A line has been drawn through the 'middle of the points'.

\begin{center}
\begin{tabular}{|c|c|}
\hline
Test 1 & Test 2 \\
\hline
53 & 54 \\
\hline
70 & 67 \\
\hline
53 & 55 \\
\hline
81 & 81 \\
\hline
85 & 82 \\
\hline
51 & 51 \\
\hline
52 & 53 \\
\hline
76 & 78 \\
\hline
75 & 77 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-32(1)}
\end{center}

The equation for this line is Test \(2=0.95 \times\) Test \(1+3.85\).

a Use this equation to predict the Test 2 mark of a student if their mark on Test 1 was:
i 53
ii 54
iii 34
iv 84
v 67

b Use this equation to predict the Test 1 mark of a student if their mark on Test 2 was:
i 53
ii 54
iii 34
iv 84
v 67

\section*{\(18 \mathrm{H}_{\text {Line of best fit }}\)}
Consider the four scatter plots below. A trend line or 'line of best fit' has been fitted to each 'by eye'. It is constructed by first noting the general trend, increasing or decreasing. A line (or curve) is then drawn through the middle of the scatter plot following that upwards or downwards trend, with roughly equal number of points above and below the line. The distance points lie from the line must also be taken into account.

I

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-33(3)}
\end{center}

III

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-33(2)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-33}
\end{center}

IV

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-33(1)}
\end{center}

\section*{Observations}
\begin{itemize}
  \item Graphs I and III show an increasing trend whilst graphs II and IV show a decreasing trend.
  \item Graph II shows a strong linear relationship between the variables and all points are in close proximity to the line of best fit. However, graphs I and IV show moderately strong linear relationships between the variables.
  \item Graph III shows a non-linear relationship between variables and a 'curve' of best fit is suggested. The other graphs display a linear relationship.
\end{itemize}

In this section, only linear relationships will be studied. To determine the equation of the line of best fit we draw on skills that were introduced in Chapter 4.

\section*{Example 17}
Consider the scatter plot below showing the relationship between ice-creams sold by vendor during the month of February and maximum temperature for the day.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-34(1)}
\end{center}

a Draw a line of best fit by eye.

b Determine the equation of the line.

c Use the equation to predict the number of ice-creams the vendor will sell on a \(35^{\circ} \mathrm{C}\) day.

d Use the equation to predict the maximum temperature of the day if the vendor sells 58 ice-creams.

\section*{Solution}
a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-34}
\end{center}

Note: Small variations in the placement of the line of best fit is expected using this technique.

(continued over page)
b Use the point-gradient form, \(y-y_{1}=m\left(x-x_{1}\right)\), to find the equation of the line.

Note: The grid lines can assist you to find two points on the line. For improved accuracy, ensure they are not too close together.

Choose \((34,70)\) and \((22,50)\). (Other selections are possible.)

\[
\begin{gathered}
m=\frac{70-50}{34-22}=\frac{20}{12}=\frac{5}{3} \\
y-50=\frac{5}{3}(x-22) \\
y=\frac{5}{3} x+50-\frac{110}{3} \\
y=\frac{5}{3} x+\frac{40}{3}
\end{gathered}
\]

Interpreting this equation in the given context, we get;

Number of ice-creams sold \(=\frac{5}{3} \times\left(\right.\) maximum temperature \(\left.{ }^{\circ} \mathrm{C}\right)+\frac{40}{3}\)

c \(\quad\) Number of ice-creams sold \(=\frac{5}{3} \times 35+\frac{40}{3}\)

\[
=71 \frac{2}{3}
\]

\(\approx 72\) (Round up to the nearest integer.)

d \(58=\frac{5}{3} \times\left(\right.\) maximum temperature \(\left.{ }^{\circ} \mathrm{C}\right)+\frac{40}{3}\)

\(174=5 \times\left(\right.\) maximum temperature \(\left.{ }^{\circ} \mathrm{C}\right)+40 \quad\) (multiplying all terms by 3 )

\(\therefore\) maximum temperature \(=\frac{174-40}{5}=26.8^{\circ} \mathrm{C}\)

\section*{Interpolation versus extrapolation}
When we use the line of best fit to make predictions of values within the range of data already obtained it is called interpolation. In the example above, the predicted number of ice-creams sold, based on a maximum temperature of \(35^{\circ} \mathrm{C}\) was interpolation. This is because \(35^{\circ} \mathrm{C}\) lies between the minimum \(\left(17^{\circ} \mathrm{C}\right)\) and maximum \(\left(38^{\circ} \mathrm{C}\right)\) recorded temperatures. The same can be said for predicting the maximum temperature based on a sale of 58 ice-creams.

Extrapolation is the term used for making predictions outside the range of values already obtained. Extrapolation should be performed with a degree of caution, since there is no guarantee the noted relationship between variables will continue beyond the observed range.

For example, using the equation, number of ice-creams sold \(=\frac{5}{3} \times\left(\right.\) maximum temperature \(\left.{ }^{\circ} \mathrm{C}\right)+\frac{40}{3}\),

to predict the maximum temperature when 20 ice-creams are sold is an act of extrapolation. The predicted maximum temperature of \(4^{\circ} \mathrm{C}\) may not be feasible.

\section*{Lines of best fit by other techniques}
You may have noticed that creating a line of best fit by eye is prone to variation and discrepancy. This is not desirable if we need to be consistent and accurate with fitting a line to data. Fortunately, there are several alternative approaches to drawing a line of best fit. The approach commonly used is called the least squares method.

\section*{Line of best fit}
\begin{itemize}
  \item Drawing a line of best fit by eye consists of tracing the trend of the scatter plot with a straight line, ensuring that there are roughly equal numbers of points above and below the line, with distance of points from the line taken into account.
  \item Once two points have been identified on the straight line, the equation of the line can be determined using the point-gradient form.
  \item Interpolation is making predictions using data that lies within the range of observed values.
  \item Extrapolation is making predictions using data that lies outside the range of observed values. Caution must be used when predicting values based on extrapolation.
\end{itemize}

\section*{Exercise \(18 \mathrm{H}\)}
1 Copy these scatter plots and draw a line of best fit by eye though each.

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-36}
\end{center}

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-36(2)}
\end{center}

ii
\includegraphics[max width=\textwidth, center]{2024_02_22_c6b381e97fdd24dc5428g-36(1)}

2 In the scatter plots in question 1, comment on the following.

a Do the scatter plots display an increasing or decreasing trend?

b What is the strength of the relationships between \(y\) and \(x\) ?

3 Data was collected on 100 adults comparing shoe size and height. Shoe sizes ranged from 6 to 13. An equation relating height (in \(\mathrm{cm}\) ) to shoe size was determined to be:

height \(=127.18+4.84 \times\) shoe size

Use this equation to predict (to the nearest \(\mathrm{cm}\) ) the height of a person whose shoe size is as follows. Are you interpolating or extrapolating?
a size 7
b size 12
c size 14

4 A line of best fit for a scatter plot, relating the weight of a pumpkin \((\mathrm{kg})\) to the number of seeds it contains, was found to pass through the points \((1,300)\) and \((7,540)\). Assume weight is on the \(x\)-axis.

a Find the equation of the line of best fit.

b Use your equation to estimate the number of seeds a pumpkin contains that weighs \(5.2 \mathrm{~kg}\).

c Use your equation to estimate the weight of a pumpkin containing 600 seeds.

5 A class of Year 10 PE students were asked to run a lap of the school's oval. Their times were recorded and compared against their fitness levels, which had been previously analysed and placed on a scale of 1 to 10 . The teacher then drew a line of best fit over the scatter plot as shown.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-37}
\end{center}

a Determine the equation of the line of best fit.

b Use the equation to predict the time it would take a Year \(10 \mathrm{PE}\) student to run a lap of the oval if that student has a fitness level of 3. Leave your answer correct to one decimal place.

c Use the equation to predict the fitness level of a Year \(10 \mathrm{PE}\) student if a lap of the oval is run in 62 seconds.

d Are these predictions examples of interpolation or extrapolation? Explain your answer.

6 State the problems with making predictions using the lines of best in the following scatter plots.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-38}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-38(1)}
\end{center}

7 Consider the time series below, showing a company's profit for consecutive financial years over a 10 year period. 'Year 1' marks the financial year 1988-1989, 'Year 2' marks the financial year 1989-1990, and so on. 'Year 10' marks the financial year 1997-1998.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-38(2)}
\end{center}

Create a line of best fit on the time series and use it to predict the company's profits, to the nearest \$100 000, in the financial year 1998-1999. (Predicting future values in a time series based on previously observed values is called forecasting.) Is your answer an example of interpolation or extrapolation?

\section*{Review exercise}
1 The stem-and-leaf plot on the right gives the times for which a \(11 \mid 5\) class of 26 Year 10 students ran \(100 \mathrm{~m}\).

123469

a What is the range of times to run \(100 \mathrm{~m}\) in the class?

1300026

b What is the median time to run \(100 \mathrm{~m}\) in the class?

c What is the interquartile range?

1401424799

1512455

d Would the median time change if the fastest and slowest times were removed?

2 The 'life' of alkaline batteries is compared through continuous use in a standard product. 40 Grade A and 40 Grade B batteries are tested in this way. Their results are shown in the two boxplots below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-39}
\end{center}

a State the median battery life for the Grade A and Grade B batteries.

b State the range in battery life for the Grade A and Grade B batteries.

c State the interquartile range for the Grade A and Grade B batteries.

d Determine the number of Grade A and Grade B batteries lasting longer than 29 hours.

e Describe the shape of data distributions for the Grade A and Grade B battery life.

f Under what criterion is the Grade B battery 'better' than the Grade A battery in this test?

3 The following data are the speeds of 45 semi-trailers passing a given point on an interstate highway. The speeds are measured in \(\mathrm{km} / \mathrm{h}\).

\begin{center}
\begin{tabular}{llllllllllll}
88 & 90 & 93 & 94 & 95 & 96 & 98 & 100 & 100 & 100 & 100 & 100 \\
101 & 102 & 102 & 102 & 103 & 103 & 103 & 104 & 105 & 106 & 106 & 107 \\
109 & 109 & 110 & 110 & 110 & 112 & 113 & 114 & 116 & 117 & 118 & 120 \\
120 & 121 & 128 & 130 & 130 & 139 & 141 & 144 & 150 &  &  &  \\
\end{tabular}
\end{center}

a Construct a dotplot of the data.

b Construct a boxplot of the data.

c Comment on the shape.

4 The number of times 35 randomly chosen Year 10 students go online in the course of a school day was recorded. The results are shown in the frequency table below.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
Times online & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
Number of students & 8 & 3 & 5 & 6 & 7 & 5 & 0 & 1 \\
\hline
\end{tabular}
\end{center}

a Calculate the mean number of times students in this random sample go online.

b Find the standard deviation of the number of times students go online, correct to two decimal places.

c Find the range of times online that lie within one standard deviation of the mean.

d If every student in this sample went online one more time than what was recorded, determine the effect on the mean and standard deviation.

5 Kathryn scored \(78 \%\) on both her history and mathematics tests. Both tests had a class mean of \(70 \%\), but history had a standard deviation of \(8 \%\) and mathematics had a standard deviation of \(12 \%\). In which test did Kathryn perform better relative to the rest of the class?

6 The table below gives the quarterly sales figures for a Melbourne swimwear shop in the period 2014-2016.

\begin{center}
\begin{tabular}{|l|c|c|c|c|}
\hline
Sales \$'000 & January-March & April-June & July-September & October-December \\
\hline
\(\mathbf{2 0 1 4}\) & 33 & 16 & 5 & 21 \\
\hline
\(\mathbf{2 0 1 5}\) & 35 & 19 & 8 & 26 \\
\hline
\(\mathbf{2 0 1 6}\) & 44 & 22 & 10 & 30 \\
\hline
\end{tabular}
\end{center}

a Represent this information on a time-series plot. (Use numbers 1 to 12 to mark the quarters.)

b In which quarter of each year are the sales figures the best?

c Describe briefly how the quarterly sales figures change over time. Are the sales figures improving?

7 In an all-female class of Year 10 students, the length of each student's tibia (shin bone) and height (in centimetres) was recorded and graphed below. A line of best fit was drawn.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_c6b381e97fdd24dc5428g-40}
\end{center}

a Determine the equation of the line of best fit.

b Use the equation to predict the height of a Year 10 female with tibia length of \(44 \mathrm{~cm}\).

c Use the equation to predict the tibia length of a \(145 \mathrm{~cm}\) tall Year 10 female.

d Are these predictions examples of interpolation or extrapolation? Explain your answer.


\end{document}