SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Statistics, STD1 S3 2025 HSC 15

A researcher is using the statistical investigation process to investigate a possible relationship between average number of minutes per day a person spends watching television, and the average number of minutes per day the person spends exercising.

  1. State the statistical question being posed.   (1 mark)

Participants were asked to record the number of minutes they spent watching television each day and the number of minutes they spent exercising each day. The averages for each participant were recorded and graphed, and a line of best fit was included.
 

  1. From the graph, identify the dependent variable.   (1 mark)

  2. Describe the bivariate dataset in terms of its form and direction.   (2 marks)

  3. The points \((0, 70)\) and \((60, 10)\) lie on the line of best fit. By first plotting these points on the graph, find the gradient and the \(y\)-intercept of the line of best fit.   (3 marks)
  4. Explain why it is NOT appropriate to extrapolate the line of best fit to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)
Show Answers Only

a.    \(\text{How does the average daily time spent watching television}\)

\(\text{relate to the average daily time spent exercising?}\)
 

b.    \(\text{Dependent variable: Average minutes per day exercising, or }y.\)
 

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d.   \(\text{Gradient}=-1\)
 

e.    \(\text{The extrapolation of the graph past 70 minutes produces}\)

\(\text{negative average minutes per day exercising (impossible).}\)

Show Worked Solution

a.    \(\text{How does the average daily time spent watching television}\)

\(\text{relate to the average daily time spent exercising?}\)
 

b.    \(\text{Dependent variable:}\)

\(\text{Average minutes per day exercising, or }y.\)
 

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d. 


 

\(y-\text{intercept = 70}\)

\(\text{Gradient}=\dfrac{\text{rise}}{\text{run}}=\dfrac{-60}{60}=-1\)
 

e.    \(\text{The extrapolation of the graph past 70 minutes produces}\)

\(\text{negative average minutes per day exercising (impossible).}\)

Statistics, STD1 S3 2023 HSC 19

The scatterplot shows the number of ice-creams sold, \(y\), at a shop over a ten-day period, and the temperature recorded at 2 pm on each of these days.
 

  1. The data are modelled by the equation of the line of best fit given below.

\(y=0.936 x-8.929\), where \(x\) is the temperature.

  1. Sam used a particular temperature with this equation and predicted that 23 ice-creams would be sold.
  2. What was the temperature used by Sam, to the nearest degree?  (2 marks)

  3. In using the equation to make the prediction in part (a), was Sam interpolating or extrapolating? Justify your answer.  (2 marks)

Show Answers Only

a.    \(34^{\circ}\text{ (nearest degree)}\)

b.    \(\text{See worked solutions}\)

Show Worked Solution

a.             \(y\) \(=0.936x-8.929\)
\(23\) \(=0.936x-8.929\)
\(0.936x\) \(=23+8.929\)
\(x\) \(=\dfrac{31.921}{0.936}\)
  \(=34.112\ldots ^{\circ}\)
  \(= 34^{\circ}\text{ (nearest degree)}\)

♦♦ Mean mark (a) 31%.

b.     \(\text{Sam is extrapolating as 34°C is outside the range of data}\)

\(\text{points shown on the graph (i.e. temp between 0 and 30°C).}\)


♦♦ Mean mark (b) 33%.

Statistics, STD1 S3 2022 HSC 23

A teacher surveyed the students in her Year 8 class to investigate the relationship between the average number of hours of phone use per day and the average number of hours of sleep per day.

The results are shown on the scatterplot below.
 

  1. The data for two new students, Alinta and Birrani, are shown in the table below. Plot their results on the scatterplot.  (2 marks)

\begin{array} {|l|c|c|}
\hline
  & \textit{Average hours of} & \textit{Average hours of} \\ & \textit{phone use per day} & \textit{sleep per day} \\
\hline
\rule{0pt}{2.5ex} \text{Alinta} \rule[-1ex]{0pt}{0pt} & 4 & 8 \\
\hline
\rule{0pt}{2.5ex} \text{Birrani} \rule[-1ex]{0pt}{0pt} & 0 & 10.5 \\
\hline
\end{array}

  1. By first fitting the line of best fit by eye on the scatterplot, estimate the average number of hours of sleep per day for a student who uses the phone for an average of 2 hours per day.  (2 marks)

Show Answers Only
  1.  
  2. 9 hours (see LOBF in diagram above)
Show Worked Solution

a.     \(\text{New data points are marks with a × on the diagram below.}\)
 

b.   \(\text{9 hours (see LOBF in diagram above)}\)

Statistics, STD1 S3 2022 HSC 16

A concert organiser is interested in the relationship between the distance from the stage, in metres, and the loudness of the sound measured in decibels.

The data the concert organiser collected is shown on the graph.
 

  1. Is the relationship between distance and loudness linear or non-linear?  (1 mark)

  2. Based on this dataset, at approximately what distance from the stage would the sound be at 90 decibels?  (1 mark)

Show Answers Only
  1. `text{Non-linear}`
  2. `text{6 metres}`
Show Worked Solution

a.   The graph is not a straight line, therefore, non-linear.

b.   Distance =  6 metres.

→ Intersection of line of best fit and horizontal line at decibels (y-axis) = 90


♦ Mean mark part (b) 46%.

Statistics, STD1 S3 2021 HSC 18

People are placed into groups to complete a puzzle. There are 9 different groups.

The table shows the number of people in each group and the amount of time, in minutes, for each group to complete the puzzle.

\begin{array} {|l|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Number of people} \rule[-1ex]{0pt}{0pt} & 2 & 2 & 3 & 5 & 5 & 7 & 7 & 7 & 8 \\
\hline
\rule{0pt}{2.5ex} \textit{Time taken (min)} \rule[-1ex]{0pt}{0pt} & 28 & 30 & 26 & 19 & 21 & 12 & 13 & 11 & 8 \\
\hline
\end{array}

  1. Complete the scatterplot by adding the last four points from the table.  (2 marks)
     
       
  2. Add a line of best fit by eye to the graph in part (a).  (1 mark)
  3. The graph in part (a) shows the association between the time to complete the puzzle and the number of people in the group.
  4. Identify the form (linear or non-linear), the direction and the strength of the association.  (3 marks)

  5. Calculate the mean of the time taken to complete the puzzle for the three groups of size 7 observed in the dataset.  (1 mark)

Show Answers Only
  1.  
       
  2.  
       
  3. `text(The association is linear, negative and strong.)`
  4. `12\ text(minutes)`
Show Worked Solution

a.

b.


 

c.    `text(Form: linear)`

♦ Mean mark (c) 50%.

`text{Direction: negative}`

`text{Strength: strong}`
 

d. `text{Mean time (7 people)}` `= (12 + 13 + 11)/3`
    `= 12\ text(minutes)`

Statistics, STD1 S3 2020 HSC 22

A group of students sat a test at the end of term. The number of lessons each student missed during the term and their score on the test are shown on the scatterplot.
 


 

  1. Describe the strength and direction of the linear association observed in this dataset.  (2 marks)

  2. Calculate the range of the test scores for the students who missed no lessons.  (1 mark)

  3. Draw a line of the best fit in the scatterplot above.  (1 mark)
  4. Meg did not sit the test. She missed five lessons.

     

    Use the line of the best fit drawn in part (c) to estimate Meg's score on this test. (1 mark)

  5. John also did not sit the test and he missed 16 lessons.

     

    Is it appropriate to use the line of the best fit to estimate his score on the test? Briefly explain your answer. (1 mark)

Show Answers Only

a.    \(\text{Strength : strong}\)

\(\text{Direction : negative} \)

b.    \(\text{Range}\ = \text{high}-\text{low}\ = 100-80=20\)
 

c.   

d. 


 

e.    \(\text{John’s missed days are too extreme and the LOBF is not}\)

\(\text{appropriate. The model would estimate a negative score for}\)

\(\text{John which is impossible.}\)

Show Worked Solution

a.    \(\text{Strength : strong}\)

\(\text{Direction : negative} \)

♦ Mean mark (a) 45%.
♦♦ Mean mark (b) 31%.

b.    \(\text{Range}\ = \text{high}-\text{low}\ = 100-80=20\)
 

c.   

d. 


 
\(\therefore\ \text{Meg’s estimated score = 40}\)
 

e.    \(\text{John’s missed days are too extreme and the LOBF is not}\)

\(\text{appropriate. The model would estimate a negative score for}\)

\(\text{John which is impossible.}\)

♦ Mean mark (e) 38%.

Statistics, STD1 S3 2019 HSC 27

A set of bivariate data is collected by measuring the height and arm span of eight children. The graph shows a scatterplot of these measurements.
 

  1. On the graph, draw a line of best fit by eye.  (1 mark)

  2. Robert is a child from the class who was absent when the measurements were taken. He has an arm span of 147 cm. Using your line of best fit from part (a), estimate Robert’s height.  (1 mark)

Show Answers Only
  1.   
  2. `text(Robert’s height ≈ 151.1 cm)`
Show Worked Solution

a.     
       

♦ Mean mark (a) 38%.

b.   `text(Robert’s height ≈ 151.1 cm)`

`text{(Answers can vary slightly depending on line of best fit drawn).}`

Algebra, STD2 A2 2017 HSC 3 MC

The graph shows the relationship between infant mortality rate (deaths per 1000 live births) and life expectancy at birth (in years) for different countries.
 

What is the life expectancy at birth in a country which has an infant mortality rate of 60?

  1. 68 years
  2. 69 years
  3. 86 years
  4. 88 years
Show Answers Only

\(A\)

Show Worked Solution

\(\text{When infant mortality rate is 60, life expectancy}\)

\(\text{at birth is 68 years (see below).}\)
 

\(\Rightarrow A\)

Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2000.
 


  1. According to the graph, what is the life expectancy of a person born in 1932?  (1 mark)

  2. With reference to the value of the gradient, explain the meaning of the gradient in this context.  (2 marks)

Show Answers Only
  1. `text(68 years)`
  2. `text(After 1900, life expectancy increases 0.25 years for each later year someone is born.)`
Show Worked Solution

i.    \(\text{68 years}\)

ii.    \(\text{Using (1900,60), (1980,80):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{80-60}{1980-1900}\)
  \(= 0.25\)

 
\(\text{After 1900, life expectancy increases by 0.25 years for}\)

\(\text{each year later that someone is born.}\)

♦♦ Mean mark (ii) 33%.

Statistics, STD2 S4 2015 HSC 28e

The shoe size and height of ten students were recorded.

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Shoe size} \rule[-1ex]{0pt}{0pt} & \text{6} & \text{7} & \text{7} & \text{8} & \text{8.5} & \text{9.5} & \text{10} & \text{11} & \text{12} & \text{12} \\
\hline \rule{0pt}{2.5ex} \text{Height} \rule[-1ex]{0pt}{0pt} & \text{155} & \text{150} & \text{165} & \text{175} & \text{170} & \text{170} & \text{190} & \text{185} & \text{200} & \text{195} \\
\hline
\end{array}

  1. Complete the scatter plot AND draw a line of fit by eye.  (2 marks)
     
     
  2. Use the line of fit to estimate the height difference between a student who wears a size 7.5 shoe and one who wears a size 9 shoe.  (1 mark)

  3. A student calculated the correlation coefficient to be 1 for this set of data. Explain why this cannot be correct.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `13\ text{cm  (or close given LOBF drawn)}
  3. `text(A correlation co-efficient of 1 would)`
    `text(mean that all data points occur on the)`
    `text(line of best fit which clearly isn’t the case.)`
Show Worked Solution

i.    
      2UG 2015 28e Answer

ii.   `text{Shoe size 7½ gives a height estimate of 162 cm (see graph)}`

`text{Shoe size 9 gives a height estimate of 175 cm (see graph)}`

`:.\ text(Height difference)` `= 175-162`
  `= 13\ text{cm  (or close given LOBF drawn)}`

 

iii.   `text(A correlation co-efficient of 1 would mean)`

♦ Mean mark (c) 39%.

`text(that all data points occur on the line of best)`

`text(fit which clearly isn’t the case.)`

Statistics, STD2 S4 2006 HSC 27b

Each member of a group of males had his height and foot length measured and recorded. The results were graphed and a line of fit drawn.
 

  1. Why does the value of the `y`-intercept have no meaning in this situation?  (1 mark)

  2. George is 10 cm taller than his brother Harry. Use the line of fit to estimate the difference in their foot lengths.  (1 mark)

  3. Sam calculated a correlation coefficient of  −1.2  for the data. Give TWO reasons why Sam must be incorrect.  (2 marks)

Show Answers Only
  1. `text(The y-intercept occurs when)\ x = 0.\ text(It has`
    `text(no meaning to have a height of 0 cm.)`
  2. `text(A 10 cm height difference means George should)`
    `text(have a 3 cm longer foot.)`
  3. `text(A correlation co-efficient must be between –1 and 1.)`
    `text(Foot length is positively correlated to a person’s)`
    `text(height and therefore can’t be a negative value.)`
Show Worked Solution

i.  `text(The y-intercept occurs when)\ x = 0.\ text(It has)`

`text(no meaning to have a height of 0 cm.)`

 

ii.  `text(A 20 cm height difference results in a foot length)`

`text(difference of 6 cm.)`
 

`:.\ text(A 10 cm height difference means George should)`

`text(have a 3 cm longer foot.)`

 

iii.  `text(A correlation co-efficient must be between –1 and 1.)`

`text(Foot length is positively correlated to a person’s)`

`text(height and therefore isn’t a negative value.)`

Statistics, STD2 S4 2009 HSC 28b

The height and mass of a child are measured and recorded over its first two years. 

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Height (cm), } H \rule[-1ex]{0pt}{0pt} & \text{45} & \text{50} & \text{55} & \text{60} & \text{65} & \text{70} & \text{75} & \text{80} \\
\hline \rule{0pt}{2.5ex} \text{Mass (kg), } M \rule[-1ex]{0pt}{0pt} & \text{2.3} & \text{3.8} & \text{4.7} & \text{6.2} & \text{7.1} & \text{7.8} & \text{8.8} & \text{10.2} \\
\hline
\end{array}

This information is displayed in a scatter graph. 
 

  1. Describe the correlation between the height and mass of this child, as shown in the graph.   (1 mark)

  2. A line of best fit has been drawn on the graph.

     

    Find the equation of this line.   (2 marks)

Show Answers Only
  1. `text(The correlation between height and)`

     

    `text(mass is positive and strong.)`

  2. `M = 0.23H-8`
Show Worked Solution

i.  `text(The correlation between height and)`

♦ Mean mark 48%. 

`text(mass is positive and strong.)`

 

ii.  `text(Using)\ \ P_1(40, 1.2)\ \ text(and)\ \ P_2(80, 10.4)`

♦♦♦ Mean mark 18%. 
MARKER’S COMMENT: Many students had difficulty due to the fact the horizontal axis started at `H= text(40cm)` and not the origin.
`text(Gradient)` `= (y_2-y_1)/(x_2-x_1)`
  `= (10.4-1.2)/(80-40)`
  `= 9.2/40`
  `= 0.23`

 

`text(Line passes through)\ \ P_1(40, 1.2)`

`text(Using)\ \ \ y-y_1` `= m(x-x_1)`
`y-1.2` `= 0.23(x-40)`
`y-1.2` `= 0.23x-9.2`
`y` `= 0.23x-8`

 
`:. text(Equation of the line is)\ \ M = 0.23H-8`

Statistics, STD2 S4 2013 HSC 28b

Ahmed collected data on the age (`a`) and height (`h`) of males aged 11 to 16 years.

He created a scatterplot of the data and constructed a line of best fit to model the relationship between the age and height of males.
 

  1. Determine the gradient of the line of best fit shown on the graph.   (1 mark)

  2. Explain the meaning of the gradient in the context of the data.   (1 mark)

  3. Determine the equation of the line of best fit shown on the graph.  (2 marks)

  4. Use the line of best fit to predict the height of a typical 17-year-old male.   (1 mark)

  5. Why would this model not be useful for predicting the height of a typical 45-year-old male?   (1 mark)

Show Answers Only
  1. `text(Gradient = 6)`
  2. `text(Males should grow 6 cm per)`

     

    `text(year between the ages 11-16.)`

  3. `h = 6a + 80`
  4. `text(182 cm)`
  5. `text(People slow and eventually stop growing)`

  6.  

    `text(after they become adults.)`

Show Worked Solution

i.    `text{Gradient}\ =(176-146)/(16-11)=30/5=6`
 

ii.   `text{Males should grow 6cm per year between the}`

`text{ages 11–16.}`
 

♦♦ Mean marks of 38%, 26% and 25% respectively for parts (i)-(iii).
MARKER’S COMMENT: Interpreting gradients has been consistently examined in recent history and almost always poorly answered. 

iii.   `text{Gradient = 6,  Passes through (11, 146)}`

`y-y_1` `=m(x-x_1)`
`h-146` `=6(a-11)`
`:. h` `=6a-66+146`
  `=6a + 80`

 

iv.   `text{Substitue}\ \ a=17\ \ \text{into equation from part (iii):}`

`h=(6 xx 17) +80=182`

`:.\ text{A typical 17 year old is expected to be 182cm.}`
  

v.    `text(People slow and eventually stop growing)`
  `text(after they become adults.)`