S1 January 2011 Q4
4. A farmer collected data on the annual rainfall, \(x\) cm, and the annual yield of peas, \(p\) tonnes per acre.
The data for annual rainfall was coded using \(v = \dfrac{x - 5}{10}\) and the following statistics were found.
\[S_{vv} = 5.753 \qquad S_{pv} = 1.688 \qquad S_{pp} = 1.168 \qquad \bar{p} = 3.22 \qquad \bar{v} = 4.42\]| Scheme | Marks |
|---|---|
| \(b = \dfrac{1.688}{5.753} = 0.293\) | M1A1 |
| \(a = 3.22 - 4.42 \times 0.293 = 1.9231\ldots\) | M1 |
| \(p = 1.92 + 0.293v\) | A1 |
| (4) |
Notes
Can ignore (a) and (b) labels here
1st M1 for a correct expression for \(b\). \(\dfrac{1.688}{1.168}\) is M0
1st A1 for awrt 0.29
2nd M1 for use of \(a = \bar{p} - b\bar{v}\) follow through their value of \(b\)(or even just the letter \(b\))
2nd A1 for a complete equation with \(a\) = awrt 1.92 and \(b\) = awrt 0.293
\(y\) or \(p\) = 1.92 + 0.293\(x\) is A0
Correct answer with no working is 4/4
| Scheme | Marks |
|---|---|
| \(v = \dfrac{85 - 5}{10} = 8\) | M1 |
| \(p = 1.92 + 0.293 \times 8 = 4.3\) (awrt 4.3) | A1 |
| (2) | |
| (6 marks) |
Notes
M1 for an attempt to find the value of \(v\) when \(x\) = 85 ( at least 2 correct terms in \(\pm\dfrac{85 - 5}{10}\) )
or for an attempt to find an equation for \(p\) in terms of \(x\) and using \(x\) = 85
Attempt at equation of \(p\) in \(x\) requires \(p = 1.92 + 0.293\dfrac{(x - 5)}{10}\)
A1 for awrt 4.3 (award when first seen and apply ISW)
N.B. \(p = 1.92 + 0.293 \times 85\) (o.e.) is M0A0