S1 June 2013 (R) Q3
3. An agriculturalist is studying the yields, \(y\) kg, from tomato plants. The data from a random sample of 70 tomato plants are summarised below.
| Yield (\(y\) kg) | Frequency (f) | Yield midpoint (\(x\) kg) |
|---|---|---|
| \(0 \leqslant y \lt 5\) | 16 | 2.5 |
| \(5 \leqslant y \lt 10\) | 24 | 7.5 |
| \(10 \leqslant y \lt 15\) | 14 | 12.5 |
| \(15 \leqslant y \lt 25\) | 12 | 20 |
| \(25 \leqslant y \lt 35\) | 4 | 30 |
(You may use \(\sum \mathrm{f}x = 755\) and \(\sum \mathrm{f}x^2 = 12\,037.5\))
A histogram has been drawn to represent these data.
The bar representing the yield \(5 \leqslant y \lt 10\) has a width of 1.5 cm and a height of 8 cm.
| Scheme | Marks |
|---|---|
| Width = \(2\times 1.5\) = 3 (cm) | B1 |
| Area = \(8\times 1.5 = 12\) cm\(^2\) Frequency = 24 so 1 cm\(^2\) = 2 plants (o.e.) | M1 |
| Frequency of 12 corresponds to area of 6 so height = 2 (cm) | A1 |
| (3) |
Notes
M1 for forming a relationship between area and no. of plants or their width\(\times\)their height = 6
A1 for height of 2 (cm). Make sure the 2 refers to height and not plants!
| Scheme | Marks |
|---|---|
| \([Q_2 =]\ (5 +)\ \dfrac{19}{24}\times 5\) or (use of \((n + 1)\)) \((5 +)\dfrac{19.5}{24}\times 5\) | M1 |
| \(= 8.9583\ldots\) awrt 8.96 or \(9.0625\ldots\) awrt 9.06 | A1 |
| (2) |
Notes
M1 for a suitable fraction \(\times 5\) (ignore end points)
A1 for awrt 8.96 (or \(\frac{215}{24}\) or \(8\frac{23}{24}\)) or 9.06 ( or \(\frac{145}{16}\) or \(9\frac{1}{16}\)) if using \((n + 1)\)
| Scheme | Marks |
|---|---|
| \([\bar{x} =]\ \dfrac{755}{70}\) or awrt 10.8 | B1 |
| \([\sigma_x =]\ \sqrt{\dfrac{12037.5}{70} - \bar{x}^2} = \sqrt{55.6326\ldots}\) | M1A1ft |
| = awrt 7.46 (Accept \(s\) = awrt 7.51) | A1 |
| (4) |
Notes
B1 for a correct mean. Accept exact fraction or awrt 10.8
M1 for a correct expression for \(\sigma\) or \(\sigma^2\). Condone mixed up labelling- ft their mean
A1ft for a correct expression – ft their mean but must have square root
A1 for awrt 7.46 (use of \(s\) = awrt 7.51). Condone correct working and answer called variance.
| Scheme | Marks |
|---|---|
| \(\bar{x} \gt Q_2\) | B1ft |
| So positive skew | dB1 |
| (2) |
Notes
1st B1ft for a correct comparison of their \(\bar{x}\) and their \(Q_2\)
ALT Allow use of a formula for skewness that involves \((\bar{x} - Q_2)\) or use of quartiles but must have correct values
NB \(Q_1 = 5.31,\ Q_3 = 14.46\) (awrt 14.5), \(Q_3 - Q_2 \approx 5.5,\ Q_2 - Q_1 \approx 3.7/6\)
2nd dB1 Dependent on a suitable reason for concluding “positive skew”. “correlation” is B0
| Scheme | Marks |
|---|---|
| \(\bar{x} + \sigma \approx 18.3\) so number of plants is e.g. \(\dfrac{(25 - \text{"}18.3\text{"})}{10}\times 12\ (+4)\) (o.e.) | M1 |
| \(= 12.04\) so 12 plants | A1 |
| (2) | |
| (13 marks) |
Notes
M1 for a suitable expression involving some interpolation (condone missing 4 so accept awrt 8)
Condone use of end points of 25.5 and 14.5 in their interpolation expressions.
A1 for 12 (condone awrt 12). Answer only 2/2