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Bivariate Data, SM-Bank 009

The height (in cm) and foot length (in cm) for each of eight Year 12 students were recorded and displayed in the scatterplot below.

A line of best fit has been fitted to the data as shown.
 

  1. Determine the predicted foot size of a student who is 176 centimetres tall.   (1 mark)
  2. Calculate the gradient of the line of best fit, giving your answer correct to two decimal places.   (2 marks)
  3. What is the equation of the line of best fit?   (2 marks)
Show Answers Only

i.    \(\text{Height 176 cm}\ \ \Rightarrow\ \ \text{Foot length = 27 cm} \)

ii.    \(1.29 \)

iii.    \(\textit{height}\ = 141.2 + 1.29 \times \textit{foot length} \)

Show Worked Solution

i.    \(\text{Height 176 cm}\ \ \Rightarrow\ \ \text{Foot length = 27 cm} \)
 

ii.    \(\text{LOBF passes through (20, 167) and (34, 185):}\)

\(\text{Gradient}\ = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{185-167}{34-20} = \dfrac{18}{14} = 1.285… = 1.29 \)
 

iii.   \(\text{Equation}\ \ \Rightarrow \ \ \text{Gradient = 1.29, passes through (20, 167)} \)

\(y-y_1 \) \(=m(x-x_1) \)  
\(y-167\) \(=1.29(x-20) \)  
\(y\) \(=1.29x + 141.2\)  

 
\(\textit{height}\ = 141.2 + 1.29 \times \textit{foot length} \)

Bivariate Data, SM-Bank 008 MC

A line of best fit has been fitted to the scatterplot above to enable distance, in kilometres, to be predicted from time, in minutes.

The equation of this line is closest to

  1. distance `= 3.5 + 1.6 ×`time
  2. time `= 3.5 + 1.6 ×`distance
  3. distance `= 1.6 + 3.5 ×`time
  4. time `= 1.8 + 3.5 ×`distance
Show Answers Only

`A`

Show Worked Solution

`text{Line passes through  (0, 3.5) and (50, 82)`

\(\text{Gradient}\ = \dfrac{y_2-y_1}{x_2-x_1} \approx \dfrac{82-3.5}{50-0} \approx 1.57 \)
 

`text{Distance is the dependent variable}\ (y)\ \text{and the}`

`y text(-intercept is approximately 3.5.)`

`=> A`

Bivariate Data, SM-Bank 003

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

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

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

  5.  

    `text(after they become adults.)`

Show Worked Solution

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

ii.   `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`
♦♦ Mean marks of 38% and 25% respectively for parts (i)-(ii).
COMMENT: Choose extreme points for calculating gradient.

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

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

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

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

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, 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`