The table shows the marks obtained by some candidates in Physics (y) and Mathematics (x) tests.
(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.
| 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.
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Correct Answer: Option n
Explanation:
(a)(i) and (iii) Scale: On each axis, 2 cm represents 10 marks.
(ii) \(\bar{x} = \frac{43 + 46 + 48 + 39 + 30 + 60 + 8 + 45 + 40}{9} = \frac{359}{9} = 39.89 \approxeq 39.9\)
\(\bar{y} = \frac{54 + 53 + 63 + 30 + 44 + 75 + 20 + 33 + 48}{9} = \frac{421}{9} = 46.78 \approxeq 46.8\)
\((\bar{x}, \bar{y}) = (39.9, 46.8)\)
(b)
The equation of the line :
\(y - \bar{y} = m(x - \bar{x})\)
\(m = \frac{\sum {xy} - N(\bar{x})(\bar{y})}{\sum {x^{2}} - N(\bar{x})^{2}}\)
= \(\frac{18379 - 9 \times 39.9 \times 46.8}{15979 - (39.9)^{2}} = \frac{18379 - 16805.88}{15979 - 9 \times (39.9)^{2}}\)
= \(\frac{1573.12}{1650.91} \approxeq 0.95\)
Substituting in the equation, \(y - \bar{y} = m(x - \bar{x})\)
\(y - 46.8 = 0.95(x - 39.9)\)
= \(y = 0.95x + 8.895\)
(c) y = 0.95x + 8.895
when x = 28,
\(y = (0.95 \times 28) + 8.895\)
= \(26.6 + 8.895 \approxeq 35.5 marks\).
For a student who scored 28 in Mathematics, his mark in Physics is 35.5.
| Mathematics (x) | 43 | 46 | 48 | 39 | 30 | 60 | 8 | 45 | 40 |
| Physics (y) | 54 | 53 | 63 | 30 | 44 | 75 | 20 | 33 | 49 |
(a)(i) and (iii) Scale: On each axis, 2 cm represents 10 marks.
(ii) \(\bar{x} = \frac{43 + 46 + 48 + 39 + 30 + 60 + 8 + 45 + 40}{9} = \frac{359}{9} = 39.89 \approxeq 39.9\)
\(\bar{y} = \frac{54 + 53 + 63 + 30 + 44 + 75 + 20 + 33 + 48}{9} = \frac{421}{9} = 46.78 \approxeq 46.8\)
\((\bar{x}, \bar{y}) = (39.9, 46.8)\)
(b)
| \(x\) | \(y\) | \(xy\) | \(x^{2}\) |
| 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 |
| Total | 18379 | 15979 |
The equation of the line :
\(y - \bar{y} = m(x - \bar{x})\)
\(m = \frac{\sum {xy} - N(\bar{x})(\bar{y})}{\sum {x^{2}} - N(\bar{x})^{2}}\)
= \(\frac{18379 - 9 \times 39.9 \times 46.8}{15979 - (39.9)^{2}} = \frac{18379 - 16805.88}{15979 - 9 \times (39.9)^{2}}\)
= \(\frac{1573.12}{1650.91} \approxeq 0.95\)
Substituting in the equation, \(y - \bar{y} = m(x - \bar{x})\)
\(y - 46.8 = 0.95(x - 39.9)\)
= \(y = 0.95x + 8.895\)
(c) y = 0.95x + 8.895
when x = 28,
\(y = (0.95 \times 28) + 8.895\)
= \(26.6 + 8.895 \approxeq 35.5 marks\).
For a student who scored 28 in Mathematics, his mark in Physics is 35.5.