The table shows the marks obtained by some candidates in Physics (y) and Mathematics (x) tests. Mathematics 43 46 48 39 30 60 8 45 40 Physics 54 53 63 30 44...
Question 1 Report
The table shows the marks obtained by some candidates in Physics (y) and Mathematics (x) tests.
| Mathematics | 43 | 46 | 48 | 39 | 30 | 60 | 8 | 45 | 40 |
| Physics | 54 | 53 | 63 | 30 | 44 | 75 | 20 | 33 | 49 |
(a)(i) Represent this information on a scatter diagram.
(ii) Find \(\bar{x}\) and \(\bar{y}\), the mean of x and y respectively.
(iii) Draw the line of best fit to pass through (x, y).
(b) Find the equation of the line in a(iii).
(c) Use your equation in (b) to find, correct to one decimal place, the mark in Physics for a candidate who scored 28 in Mathematics.
Answer Details
(a)(i) Scatter diagram. Using a scale of 2 cm to 10 marks on each axis, the nine pairs \((43,54),(46,53),(48,63),(39,30),(30,44),(60,75),(8,20),(45,33),(40,49)\) are plotted with Mathematics \((x)\) on the horizontal axis and Physics \((y)\) on the vertical axis. The line of best fit required in part (a)(iii) is drawn on the same diagram.
(a)(ii) The means.
\[\bar{x}=\frac{43+46+48+39+30+60+8+45+40}{9}=\frac{359}{9}=39.9\] \[\bar{y}=\frac{54+53+63+30+44+75+20+33+49}{9}=\frac{421}{9}=46.8\]So the mean point is \((\bar{x},\bar{y})=(39.9,\,46.8)\).
(a)(iii) Line of best fit. A single straight line is drawn through the mean point \((39.9,46.8)\), oriented so that the points are balanced roughly equally above and below it (shown as the straight line on the scatter diagram above).
(b) Equation of the line. Set out the sums needed for the least-squares gradient:
| \(x\) | \(y\) | \(xy\) | \(x^{2}\) |
| 43 | 54 | 2322 | 1849 |
| 46 | 53 | 2438 | 2116 |
| 48 | 63 | 3024 | 2304 |
| 39 | 30 | 1170 | 1521 |
| 30 | 44 | 1320 | 900 |
| 60 | 75 | 4500 | 3600 |
| 8 | 20 | 160 | 64 |
| 45 | 33 | 1485 | 2025 |
| 40 | 49 | 1960 | 1600 |
| \(\sum x=359\) | \(\sum y=421\) | \(\sum xy=18379\) | \(\sum x^{2}=15979\) |
With \(n=9\), the gradient is
\[b=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}=\frac{9(18379)-359(421)}{9(15979)-359^{2}}=\frac{165411-151139}{143811-128881}=\frac{14272}{14930}=0.96.\]The line passes through \((\bar{x},\bar{y})\), so using \(y-\bar{y}=b(x-\bar{x})\):
\[y-46.8=0.96(x-39.9)\] \[y=0.96x-38.3+46.8\] \[y=0.96x+8.6\](c) Physics mark when Mathematics \(=28\).
\[y=0.96(28)+8.6=26.9+8.6=35.5\]The predicted mark in Physics for a candidate who scored 28 in Mathematics is 35.5 (to one decimal place).
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