Summary Statistics – Box Plots (Std2-2027)

Statistics, STD2 EQ-Bank 03 MC

For the data below, which is the correct five figure summary?

$12,\ \ 16, \ \ 4,\ \ 6, \ \ 4, \ \ 5, \ \ 22, \ \ 20, \ \ 12, \ 8\ $

$4,\ \ 5, \ \ 8,\ \ 16, \ \ 21$
$4,\ \ 5, \ \ 8,\ \ 16, \ \ 22$
$4,\ \ 5, \ \ 10,\ \ 16, \ \ 21$
$4,\ \ 5, \ \ 10,\ \ 16, \ \ 22$

Show Answers Only

$D$

Show Worked Solution

$\text{Ordered data set}\ \rightarrow\ \ 4,\ \ 4, \ \ 5,\ \ 6, \ \ 8, \ \ 12, \ \ 12, \ \ 16, \ \ 20, \ 22\ $

$\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 4\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & \dfrac{8+12}{2}=10\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 22\\
\hline
\end{array}

$\Rightarrow D$

Statistics, STD2 EQ-Bank 02 MC

The results of a test are displayed in the box-and-whisker plot below.

Which of the following statements is false?

The median is 155
The range is 60
The interquartile range is 50
25% of the scores are below 150

Show Answers Only

$C$

Show Worked Solution

$\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 140\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 155\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 190 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 200\\
\hline
\end{array}

$\text{Median}\ =\ 155\ \checkmark$

$\text{Range}\ =\ 200-140=60\ \checkmark$

$\text{IQR}\ =\ 190-150=40\ \text{not}\ 50\ $X

$\text{Q1}\ =\ 150\ \therefore\ 25\%\ \text{of scores lie below 150}\ \checkmark$

$\Rightarrow C$

Statistics, STD2 EQ-Bank 01

A Physics class of 12 students is going on a 4 day excursion by bus.

The students are asked to each pack one bag for the trip. The bags are weighed, and the weights (in kg) are listed in order as follows:

$8,\ \ 9, \ \ 10,\ \ 10, \ \ 15, \ \ 18, \ \ 22, \ \ 25, \ \ 29, \ \ 35, \ \ 38, \ \ 41 $

Use the above data to produce a five number summary for the weights of the bags. (2 marks)
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Using your five number summary from part (a), calculate the interquartile range of the weights. (2 marks)
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a. $\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 8\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 20\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 32 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 41\\
\hline
\end{array}

b. $22$

Show Worked Solution

a. $\text{Five number Summary}$

b.	$\text{IQR}$	$=Q_3-Q_1$
		$=32-10=22$

Statistics, STD2 S1 2025 HSC 28

The heights of students in a class were recorded.

The results for this class are displayed in the cumulative frequency graph shown.

The shortest student in this class is 130 cm and the tallest student is 180 cm.

Construct a box-plot for this class in the space below. (3 marks)

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Show Worked Solution

$Q_1(7.5 \ \text{students })=135$

$Q_3(22.5 \ \text{students })=160$

$\text{Median (15 students )}=140$

Statistics, STD2 S1 2024 GEN1* 6

More than 11 000 athletes from more than 200 countries competed in the Tokyo Summer Olympic Games.

An analysis of the number of athletes per country produced the following five-number summary.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Minimum} \rule[-1ex]{0pt}{0pt}& \textbf{First quartile } & \textbf{Median } & \textbf{Third quartile} & \textbf{Maximum } \\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt}& 5 & 11 & 48 & 613 \\
\hline
\end{array}

Find the smallest number of athletes per country that would display as an outlier on a boxplot of this dataset. (2 marks)

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$C$

Show Worked Solution

$IQR=48-5=43$

$\text{Upper boundary}$	$=Q_3+1.5\times IQR$
	$=48+1.5\times 43$
	$=112.5$

$\therefore\ \text{Smallest number of athletes to show as an outlier = 113.}$

Statistics, STD2 S1 2024 HSC 15 MC

Some data are used to create a box plot shown.

A histogram is created from the same set of data.

Which of these histograms is NOT possible for the given box plot?

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$D$

Show Worked Solution

$\text{By inspection of box plot,}\ \ IQR=\dfrac{2}{3}\times \text{range}$

$\text{ln options A, B and C (given 16 data points):}$

$Q_1\ \text{(4th data point)}\ \rightarrow \text{2nd column}$

$Q_3\ \text{(13th data point)}\ \rightarrow \text{6th column}$

$\text{Since}\ IQR=\dfrac{2}{3}\times \text{range}\ \Rightarrow\ \text{histograms are possible.}$

$\text{ln option D:}$

$Q_1 \rightarrow \text{3rd column}$

$Q_3 \rightarrow \text{5th column}$

$\text{Since}\ IQR=\dfrac{1}{3} \times \text{range}\ \Rightarrow\ \text{not possible.}$

$\Rightarrow D$

♦♦♦ Mean mark 9%.

Statistics, STD2 S1 2024 HSC 28

Flowers were planted in two gardens (Garden A and Garden B).

On a particular day, 25 flowers were randomly selected from each garden and their heights measured in millimetres.

The data are represented in parallel box-plots.

Compare the two datasets by examining the skewness of the distributions, and the measures of central tendency and spread. (3 marks)

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$\text{Skewness comparison:}$

$\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}$

$\text{Measures of central tendency comparison:}$

$\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9$

$\rightarrow\ \text{This means the middle 50% of data points of B are more closely}$

$\text{clustered around the median than A.}$

$\text{Spread comparison:}$

$\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is greater than range (B)}\ \approx 28$

$\rightarrow\ \text{This means the spread of data points of A is wider than the spread}$

$\text{of data points of B.}$

$\text{Overall, flowers from Garden A will tend to be higher than Garden B}$

$\text{flowers as well as exhibiting a larger range of heights.}$

Show Worked Solution

$\text{Skewness comparison:}$

$\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}$

$\text{Measures of central tendency comparison:}$

$\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9$

$\rightarrow\ \text{This means the middle 50% of data points of B are more closely}$

$\text{clustered around the median than A.}$

$\text{Spread comparison:}$

$\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is greater than range (B)}\ \approx 28$

$\rightarrow\ \text{This means the spread of data points of A is wider than the spread}$

$\text{of data points of B.}$

$\text{Overall, flowers from Garden A will tend to be higher than Garden B}$

$\text{flowers as well as exhibiting a larger range of heights.}$

♦ Mean mark 41%.

Statistics, STD2 S1 2016 VCE-G 2*

A weather station records daily maximum temperatures

The five-number summary for the distribution of maximum temperatures for the month of February is displayed in the table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} & \textbf{Temperature (°C)} \\
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} Q_1 \rule[-1ex]{0pt}{0pt} & 21 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 25 \\
\hline
\rule{0pt}{2.5ex} Q_3 \rule[-1ex]{0pt}{0pt} & 31 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 39 \\
\hline
\end{array}

Use the five-number summary above to construct a boxplot on the grid below. (1 mark)

Show, using calculations, that there are no outliers in the dataset. (2 marks)

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Show Worked Solution

b. `IQR = Q_3-Q_1=31-21=10`

`text(Lower fence)`	`= Q_1-1.5 xx IQR`
	`= 21-1.5 xx 10`
	`= 6`

`text(Upper fence)`	`= Q_3 + 1.5 xx IQR`
	`= 31 + 1.5 xx 10`
	`= 46`

`text{Since 6 < 16 (minimum) and 46 > 39 (maximum)}`

`=>\ \text{There are no outliers.}`

Statistics, STD2 S1 2019 VCE-G 2*

The five-number summary below was determined from the sleep time, in hours, of a sample of 59 types of mammals.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ \textbf{Statistic} \rule[-1ex]{0pt}{0pt} & \textbf{Sleep time (hours)} \\
\hline
\rule{0pt}{2.5ex} \text{minimum} \rule[-1ex]{0pt}{0pt} & \text{2.5} \\
\hline
\rule{0pt}{2.5ex} \text{first quartile} \rule[-1ex]{0pt}{0pt} & \text{8.0} \\
\hline
\rule{0pt}{2.5ex} \text{median} \rule[-1ex]{0pt}{0pt} & \text{10.5} \\
\hline
\rule{0pt}{2.5ex} \text{third quartile} \rule[-1ex]{0pt}{0pt} & \text{13.5} \\
\hline
\rule{0pt}{2.5ex} \text{maximum} \rule[-1ex]{0pt}{0pt} & \text{20.0} \\
\hline
\end{array}

Show with calculations, that a boxplot constructed from this five-number summary will not include outliers. (2 marks)

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Construct the boxplot below. (1 mark)

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`text(Proof (See Worked Solution))`

Show Worked Solution

a. `IQR = Q_3-Q_1 = 13.5-8.0 = 5.5`

`text(Lower fence)`	`= Q_1-1.5 xx IQR`
	`= 8-1.5 xx 5.5`
	`= -0.25`

`text(Upper fence)`	`= Q_3 + 1.5 xx IQR`
	`= 13.5 + 1.5 xx 5.5`
	`= 21.75`

`text(S) text(ince) \ -0.25 < 2.5 \ text{(minimum value) and} \ 21.75 > 20.0 \ text{(maximum value)}`

`=> \ text(no outliers)`

Statistics, STD2 S1 2022 HSC 15 MC

The cumulative frequency graph shows the distribution of the number of movie downloads made by 100 people in one month.

Which box-plot best represents the same data as displayed in the cumulative frequency graph?

Show Answers Only

`C`

Show Worked Solution

`text{1st quartile}\ ~~ 3`

`text{Median}\ ~~ 6`

`text{3rd quartile}\ ~~ 7`

`=>C`

♦♦♦ Mean mark 30%.

Statistics, STD2 S1 2019 HSC 39

Two netball teams, Team A and Team B, each played 15 games in a tournament. For each team, the number of goals scored in each game was recorded.

The frequency table shows the data for Team A.

The data for Team B was analysed to create the box-plot shown.

Compare the distributions of the number of goals scored by the two teams. Support your answer with the construction of a box-plot for the data for Team A. (5 marks)

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`text(See Worked Solution)`

Show Worked Solution

`text(Team A: High = 28, Low = 19,)\ Q_1 = 23, Q_3 = 27,\ text{Median = 26}`

`text(Team A’s distribution is negatively skewed while)`

♦♦ Mean mark 28%.

`text(Team B’s distribution is slightly positively skewed.)`

`text(The standard deviation of Team A’s distribution is)`

`text(smaller than Team B, as both its IQR and range is)`

`text(smaller.)`

`text(Team B is a more successful team at scoring goals)`

`text(as each value in its 5-point summary is higher than)`

`text(Team A’s equivalent value.)`

Statistics, STD2 S1 SM-Bank 2 MC

A dataset has the following five-number summary.

If the range of the dataset is 8, what is the minimum value of the dataset?

Show Answers Only

`D`

Show Worked Solution

`text(Range)`	`=\ text{Max}-text{Min}`
`8`	`= 15-text{Min Value}`
`:.\ text{Min}`	`= 15-8`
	`=7`

`=> D`

Statistics, STD2 S1 SM-Bank 1

Write down the five-number summary for the dataset

`3, \ 7, \ 8, \ 11, \ 13, \ 18.` (2 marks)

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`text(Minimum value:)`	`3`
`text(First quartile:)`	`7`
`text(Median:)`	`(11 + 8)/2 = 9.5`
`text(Third quartile:)`	`13`
`text(Maximum value:)`	`18`

Show Worked Solution

`text(Minimum value:)`	`3`
`text(First quartile:)`	`7`
`text(Median:)`	`(11 + 8)/2 = 9.5`
`text(Third quartile:)`	`13`
`text(Maximum value:)`	`18`

Statistics, STD2 S1 2017 HSC 1 MC

The box-and-whisker plot for a set of data is shown.

What is the median of this set of data?

Show Answers Only

`C`

Show Worked Solution

`text(Median = 30)`

`=> C`

Statistics, STD2 S1 2016 HSC 22 MC

The box-and-whisker plots show the results of a History test and a Geography test.

In History, 112 students completed the test. The number of students who scored above 30 marks was the same for the History test and the Geography test.

How many students completed the Geography test?

Show Answers Only

`=> C`

Show Worked Solution

`text{In History} \ => \ text{Q}_3 = 30\ \text{marks}`

`:.\ text{Scoring over 30}\ = 25text(%) xx 112 = 28\ \text{students}`

`text{In Geography} \ => \ text{Median}\ = 30\ \text{marks}`

`:.\ text{Students completing Geography}\ =2 xx 28 = 56\ \text{students}`

`=> C`

Statistics, STD2 S1 2016 HSC 19 MC

A soccer referee wrote down the number of goals scored in 9 different games during the season.

`2, \ 3, \ 3, \ 3, \ 5, \ 5, \ 8, \ 9, \ ...`

The last number has been omitted. The range of the data is 10.

What is the five-number summary for this data set?

`2, 3, 5, 8.5, 12`
`2, 3, 5, 8.5, 10`
`2, 3, 5, 8, 12`
`2, 3, 5, 8, 10`

Show Answers Only

`=> A`

Show Worked Solution

`text{Since range is 10} \ => \ text{Last data point = 12}`

`text{Q}_1 = 3`

`text{Q}_3 = (8 + 9)/2 = 8.5`

`text(Median = 5)`

`=> A`

♦ Mean mark 46%.

Statistics, STD2 S1 2015 HSC 6 MC

The times, in minutes, that a large group of students spend on exercise per day are presented in the box‑and‑whisker plot.

What percentage of these students spend between 40 minutes and 60 minutes per day on exercise?

Show Answers Only

`C`

Show Worked Solution

`text{Q}_1 = 40, \ text(Median) = 60`

`:.\ text(% Students between 40 and 60)`

`= 50text{%}-25text{%}`

`=25 text{%}`

`=>C`

Statistics, STD2 S1 2006 HSC 24c

The heights of the 60 members of a choir were recorded. These results were grouped and then displayed as a cumulative frequency histogram and polygon.

The shortest person in the choir is 140 cm and the tallest is 190 cm.

Draw an accurate box-and-whisker plot to represent the data. (3 marks)

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`text{See Worked Solutions}`

Show Worked Solution

`text(Low) = 140`

`text(High) = 190`

`text(Median) = 150\ \ \ \ text{(# People = 30)}`

`Q_1 = 145\ \ \ \ text{(# People = 15)}`

`Q_3 = 170\ \ \ \ text{(# People = 45)}`

`text(Box and Whisker)`

Statistics, STD2 S1 2005 HSC 22 MC

Two groups of people were surveyed about their weekly wages. The results are shown in the box-and-whisker plots.

Which of the following statements is true for the people surveyed?

The same percentage of people in each group earned more than $325 per week.
Approximately 75% of people under 21 years earned less than $350 per week.
Approximately 75% of people 21 years and older earned more than $350 per week.
Approximately 50% of people in each group earned between $325 and $350 per week.

Show Answers Only

`B`

Show Worked Solution

`text{Option A: 50% of Under 21 group earned over $325 and 75%}`

`text{of Over 21 group did. NOT TRUE.}`

`text{Option B: 75% of Under 21 group earned below $350 is TRUE.}`

`text{Options C and D: can both be proven to be untrue using their}`

`text{median and quartile values.}`

`=> B`

Statistics, STD2 S1 2004 HSC 12 MC

This box-and-whisker plot represents a set of scores.

What is the interquartile range of this set of scores?

Show Answers Only

`C`

Show Worked Solution

`text{Q}_1 = 8, \ text{Q}_3 = 11`

`text{IQR}`	`= text{Q}_3-text{Q}_1`
	`= 11-8`
	`= 3`

`=> C`

Statistics, STD2 S1 2008 HSC 10 MC

The marks for a Science test and a Mathematics test are presented in box-and-whisker plots.

Which measure must be the same for both tests?

Mean
Range
Median
Interquartile range

Show Answers Only

`D`

Show Worked Solution

`text(IQR)=text(Upper Quartile)-text(Lower Quartile)`

`text{In both box plots, IQR = 3 intervals (against bottom scale)}`

`=> D`

Statistics, STD2 S1 2014 HSC 29c

Terry and Kim each sat twenty class tests. Terry’s results on the tests are displayed in the box-and-whisker plot shown in part (i).

Kim’s 5-number summary for the tests is 67, 69, 71, 73, 75.

Draw a box-and-whisker plot to display Kim’s results below that of Terry’s results. (1 mark)
What percentage of Terry’s results were below 69? (1 mark)

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Terry claims that his results were better than Kim’s. Is he correct?

Justify your answer by referring to the summary statistics and the skewness of the distributions. (4 marks)

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`text(See Worked Solutions)`
`text(50%)`
`text(See Worked Solutions)`

Show Worked Solution

♦ Mean mark 39%

ii. `text(50%)`

iii. `text(Terry’s results are more positively skewed than)`

♦♦ Mean mark 29%
COMMENT: Examiners look favourably on using language of location in answers, particularly the areas they have specifically pointed students towards (skewness in this example).

`text(Kim’s and also have a higher limit high.)`

`text(However, Kim’s results are more consistent,)`

`text(showing a tighter IQR. They also have a)`

`text(significantly higher median than Terry’s and)`

`text(are evenly skewed.)`

`:.\ text(Kim’s results were better.)`

Statistics, STD2 S1 2010 HSC 27b

The graphs show the distribution of the ages of children in Numbertown in 2000 and 2010.

In 2000 there were 1750 children aged 0–18 years.

How many children were aged 12–18 years in 2000? (1 mark)

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The number of children aged 12–18 years is the same in both 2000 and 2010.

How many children aged 0–18 years are there in 2010? (1 mark)

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Identify TWO changes in the distribution of ages between 2000 and 2010. In your answer, refer to measures of location or spread or the shape of the distributions. (2 marks)

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What would be ONE possible implication for government planning, as a consequence of this change in the distribution of ages? (1 mark)

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i. `875`

ii. `3500`

iii. `text{Changes in distribution (include 2 of the following):}`

`text(the lower quartile age is lower in 2010)`
`text(the median is lower in 2010)`
`text(the upper quartile age is lower in 2010)`
`text(the interquartile range is greater in 2010)`
`text(2010 is positively skewed while 2000 is negatively)`

iv. `text(Implication for government planning:)`

`text(Since the children are getting younger in 2010,)`

`text(Approve and build more childcare facilities)`
`text(Build more school and public playgrounds)`

Show Worked Solution

i. `text{Since the median = 12 years}`

♦ Mean mark (i) 45%

`=>\ text{50% of children are aged 12–18 years}`

`:.\ text{Children aged 12–18}\ = 50\text{%}\ xx 1750 = 875`

♦♦ Mean mark (ii) 25%

ii. `text{Upper quartile (2010) = 12 years}`

`text{Children in upper quartile = 875 (from part (i))}`

`:.\ text{Children aged 0–18}\ =4 xx 875= 3500`

iii. `text{Changes in distribution (include 2 of the following):}`

♦ Mean mark (iii) 35%
MARKER’S COMMENT: A number of students incorrectly identified “positive” skew as “negative” skew here.

`text(the lower quartile age is lower in 2010)`
`text(the median is lower in 2010)`
`text(the upper quartile age is lower in 2010)`
`text(the interquartile range is greater in 2010)`
`text(2010 is positively skewed while 2000 is negatively)`

iv. `text(Implication for government planning:)`

♦ Mean mark (iv) 46%
MARKER’S COMMENT: Answers should reflect the 1 mark allocation.

`text(Since the children are getting younger in 2010,)`

`text(Approve and build more childcare facilities)`
`text(Build more school and public playgrounds)`

Statistics, STD2 S1 2009 HSC 26a

In a school, boys and girls were surveyed about the time they usually spend on the internet over a weekend. These results were displayed in box-and-whisker plots, as shown below.

Find the interquartile range for boys. (1 mark)

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What percentage of girls usually spend 5 or less hours on the internet over a weekend? (1 mark)

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Jenny said that the graph shows that the same number of boys as girls usually spend between 5 and 6 hours on the internet over a weekend.

Under what circumstances would this statement be true? (1 mark)

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`4`
`text(75% of girls spend 5 hours or less)`
`text(5-6 hours for girls accounts for 25% of all girls.)`
`text(5-6 hours for boys accounts for 25% of all boys,)`
`text(as median to upper quartile is 25%.)`
`=>\ text(This will be the same number only if the number of)`
`text(all girls surveyed equals the number of boys surveyed.)`

Show Worked Solution

i.	`text(Interquartile range)`	`= 6` `- 2`
		`= 4`

♦♦ Mean mark part ii: 31%

ii.	`text(Upper quartile = 5`
	`:.\ text(75% of girls spend 5 or less hours)`

♦♦♦ Mean mark part iii: 9%

iii.	`text(5-6 hours for girls accounts for 25% of all girls.)`
	`text(5-6 hours for boys accounts for 25% of all boys,)`
	`text{(median to the upper quartile represents 25%.)}`
	`=>\ text(This will only be the same number if the number of)`
	`text(all girls surveyed equals the number of boys surveyed.)`

Statistics, STD2 S1 2012 HSC 28d

The test results in English and Mathematics for a class were recorded and displayed in the box-and-whisker plots.

What is the interquartile range for English? (1 mark)

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Compare and contrast the two data sets by referring to the skewness of the distributions and the measures of location and spread. (3 marks)

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i. `text{IQR}_text{(English)}\ = 80-50=30`

ii. `text(Skewness)`

`text(English has greater negative skew)`
`text(Maths is more normally distributed)`

`text(Location and Spread)`

`text(English has a range of 85, Maths has 40.)`
`text{English has larger IQR than Maths (30 vs 15)}`
`text{Maths’ median (75) is higher than English (70)}`
`text{Same upper quartile marks (80)}`
`text(English has highest and lowest individual mark)`

Show Worked Solution

i. `text{IQR}_text{(English)}\ = 80-50=30`

♦ Mean mark (ii) 35%
MARKER’S COMMENT: Markers are looking for students to use the correct language of location and spread such as mean, median, interquartile range, standard deviation and skewness.

ii. `text(Skewness)`

`text(English has greater negative skew)`
`text(Maths is more normally distributed)`

`text(Location and Spread)`

`text(English has a range of 85, Maths has 40.)`
`text{English has larger IQR than Maths (30 vs 15)}`
`text{Maths’ median (75) is higher than English (70)}`
`text{Same upper quartile marks (80)}`
`text(English has highest and lowest individual mark)`

Statistics, STD2 S1 2011 HSC 7 MC

A set of data is displayed in this box-and-whisker plot.

Which of the following best describes this set of data?

Symmetrical
Positively skewed
Negatively skewed
Normally distributed

Show Answers Only

`B`

Show Worked Solution

`text{Since the median (155) is closer to the lower quartile (150) and range}`

`text{low (140) than the upper quartile (190) and range high (200), it is}`

`text{positively skewed.}`

`=>B`

♦ Mean mark 47%.

b.	\(\text{IQR}\)	\(=Q_3-Q_1\)
		\(=32-10=22\)

\(\text{Upper boundary}\)	\(=Q_3+1.5\times IQR\)
	\(=48+1.5\times 43\)
	\(=112.5\)