S1 June 2005 Q3
3. A long distance lorry driver recorded the distance travelled, \(m\) miles, and the amount of fuel used, \(f\) litres, each day. Summarised below are data from the driver’s records for a random sample of 8 days.
The data are coded such that \(x = m - 250\) and \(y = f - 100\).
\[\Sigma x = 130 \qquad \Sigma y = 48 \qquad \Sigma xy = 8880 \qquad S_{xx} = 20\,487.5\](a) Find the equation of the regression line of \(y\) on \(x\) in the form \(y = a + bx\). (6)
(b) Hence find the equation of the regression line of \(f\) on \(m\). (3)
(c) Predict the amount of fuel used on a journey of 235 miles. (1)
| Scheme | Marks |
|---|---|
| \(S_{xy} = 8880 - \dfrac{130 \times 48}{8} = (8100)\) | B1 |
| \(S_{xx} = 20487.5\) | |
| \(b = \dfrac{s_{xy}}{s_{xx}} = \dfrac{8100}{20487.5} = 0.395363\ldots\) | M1 A1 |
| \(a = \dfrac{48}{8} - (0.395363\ldots)\dfrac{130}{8} = -0.424649\ldots\) | M1 A1 |
| \(y = -0.425 + 0.395x\) | B1ft |
| (6) |
Notes
B1 may be implied
M1 A1 allow use of their \(S_{xy}\) for M; awrt 0.395
M1 A1 allow use of their \(b\) for M; awrt −0.425
B1ft 3s.f.
Special case answer only B0 M0 B1 M0 B1 B1 (fully correct 3sf)
( \(\equiv\) to B0 M0 A1 M0 A1 B1 on the epen)
| Scheme | Marks |
|---|---|
| \(f - 100 = -0.424649\ldots + 0.395\ldots(m - 250)\) | M1 A1ft |
| \(f = 0.735 + 0.395m\) | A1 |
| (3) |
Notes
M1 A1ft subst \(f - 100\) & \(m - 250\)
A1 3 s.f.
| Scheme | Marks |
|---|---|
| \(m = 235 \Rightarrow f = 93.64489\ldots\) | B1 |
| (1) | |
| (10 marks) |
Notes
B1 awrt 93.6/93.7