The table gives the relationship between the height, in metres, of a plant and the number of days it is left to grow. Number of days (x) 10 20 30 40 50 60 7...
Question 1 Report
The table gives the relationship between the height, in metres, of a plant and the number of days it is left to grow.
| Number of days (x) |
10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
| Height (y) | 1.0 | 1.1 | 1.2 | 1.4 | 1.6 | 1.8 | 2.0 | 2.3 |
(a) Using a scale of 2 cm to represent 0.5 units on the y- axis and 2cm to 10 units on the x- axis, draw a scatter diagram for the information.
(b) Find \(\bar{x}\), the mean of x, and \(\bar{y}\), the mean of y, and plot \((\bar{x}, \bar{y})\) on the diagram.
(c) Draw the line of best fit to pass through \((\bar{x}, \bar{y})\) and \((10, 1)\).
(d) From graph, find the :
(i) equation of the line of best fit ; (ii) height of plant in 75 days.
Answer Details
(a)–(c) Scatter diagram, mean point and required line of best fit
(b) Calculate the mean coordinates.
The \(x\)-values have total \(360\), so
\[ \bar{x}=\frac{360}{8}=45 \]The heights total \(12.4\), not \(11.4\):
\[ \bar{y}=\frac{1.0+1.1+1.2+1.4+1.6+1.8+2.0+2.3}{8} =\frac{12.4}{8}=1.55 \]Therefore the mean point to plot is \((45,1.55)\).
(c) Line of best fit. The instruction requires a straight line through both \((45,1.55)\) and \((10,1)\). This is the red line on the graph.
(d)(i) Equation of the line.
\[ \text{gradient}=\frac{1.55-1}{45-10} =\frac{0.55}{35} =\frac{11}{700} \approx0.0157 \]Using \((10,1)\):
\[ y-1=\frac{11}{700}(x-10) \] \[ y=\frac{11}{700}x+\frac{59}{70} \]So an appropriate equation is
\[ \boxed{y\approx0.0157x+0.843} \](d)(ii) Height after 75 days.
\[ y=0.0157(75)+0.843\approx2.02 \] \[ \boxed{\text{The plant's height is approximately }2.0\text{ m after 75 days.}} \]The alternative equation \(y=0.017x+0.78\) is not consistent with the stated construction because it does not pass through \((10,1)\): it gives \(y=0.95\) when \(x=10\). For this question, use the two points explicitly specified for the line.
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