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Statistics, STD2 S4 2025 HSC 17

The scatter plot shows a bivariate dataset, where \(x\) is the independent variable and \(y\) is the dependent variable. 
 

 

The points \( (0,14) \) and \( (5,4)\) lie on the line of best fit.

Plot the points \( (0,14) \) and \( (5,4) \) on the graph and hence find the equation of the line of best fit.   (3 marks)

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\( y=-2 x+14 \)

Show Worked Solution

\(\text{Gradient (LOBF)}\ =\dfrac{y_2-y_1}{x_4-x_1}=\dfrac{4-14}{5-0}=-2 \)

\(\text{Find equation of line}\ \ m=-2 \ \ \text{through}\ (0,14):\)

\( y-y_1\) \(=m(x-x_1)\)  
\( y-14\) \(=-2(x-0)\)  
\( y\) \(=-2 x+14 \)  

Statistics, STD2 S4 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)

                               
  2. 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)

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  2. `text{9 hours (see LOBF in diagram above)}`
Show Worked Solution

a.   

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

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
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\(A\)

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)

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  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)

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  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)

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  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 2014* HSC 30b

The scatterplot shows the relationship between expenditure per primary school student, as a percentage of a country’s Gross Domestic Product (GDP), and the life expectancy in years for 15 countries.
 

 
 

  1. For the given data, the correlation coefficient,  `r`, is 0.83. What does this indicate about the relationship between expenditure per primary school student and life expectancy for the 15 countries?   (1 mark)

  2. For the data representing expenditure per primary school student,  `Q_L`  is 8.4 and  `Q_U`  is 22.5.

     

    What is the interquartile range?   (1 mark)

  3. Another country has an expenditure per primary school student of 47.6% of its GDP.

     

    Would this country be an outlier for this set of data? Justify your answer with calculations.   (2 marks)

  4. On the scatterplot, draw the least-squares line of best fit  `y = 1.29x + 49.9`.    (2 marks)

  5. Using this line, or otherwise, estimate the life expectancy in a country which has an expenditure per primary school student of 18% of its GDP.   (1 mark)

  6. Why is this line NOT useful for predicting life expectancy in a country which has expenditure per primary school student of 60% of its GDP?   (1 mark)

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  1. `text(It indicates there is a strong positive)`

     

    `text(correlation between the two variables.)`

  2. `14.1`
  3. `text(Yes, because it’s > 43.65%)`
  4.  
  5. `73.1\ text(years)`
  6. `text(At 60% GDP, the line predicts a life expectancy)`

  7.  

    `text(of 127.3. This line of best fit is only accurate)`

  8.  

    `text(in a lower range of GDP expediture.)`

Show Worked Solution
i. `text(It indicates there is a strong positive)`
  `text(correlation between the two variables)`

 

ii. `text(IQR)` `= Q_U\ – Q_L`
    `= 22.5\ – 8.4`
    `= 14.1`

 

♦ Mean mark 35% 

iii.  `text(An outlier on the upper side must be more than)` 

`Q_u\ +1.5xxIQR`

`=22.5+(1.5xx14.1)`

`=\ text(43.65%)`

`:.\ text(A country with an expenditure of 47.6% is an outlier).`

 

iv.  

v.  `text(Life expectancy) ~~ 73.1\ text{years (see dotted line)}`

♦♦ Mean mark 39%

 

`text(Alternative Solution)`

`text(When)\ x=18`

`y=1.29(18)+49.9=73.12\ \ text(years)`

  

♦♦♦ Mean mark 0%. The toughest question on the 2014 paper.
COMMENT: Examiners regularly ask students to identify and comment on outliers where linear relationships break down.
vi.   `text(At 60% GDP, the line predicts a life)`
  `text(expectancy of 127.3. This line of best)`
  `text(fit is only predictive in a lower range)`
  `text(of GDP expenditure.)`

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)

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  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)

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  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.)`