Higher November 2024 Paper 2 Q2
2 Here is a grouped frequency table.
| Value, \(v\) | Frequency | Midpoint | |
|---|---|---|---|
| \(0 \leqslant v \lt 10\) | 16 | 5 | |
| \(10 \leqslant v \lt 20\) | 22 | 15 | |
| \(20 \leqslant v \lt 30\) | 13 | 25 | |
| \(30 \leqslant v \lt 40\) | 9 | 35 | |
| Total = 60 |
Work out an estimate of the mean value. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Any two of \(16 \times 5\) or 80 \(22 \times 15\) or 330 \(13 \times 25\) or 325 \(9 \times 35\) or 315 | M1 | implied by 1050 |
| (their 80 + their 330 + their 325 + their 315) \(\div\) 60 | M1dep | oe must be sum of four numbers condone missing final bracket |
| 17.5 or \(\dfrac{1050}{60}\) | A1 | oe value eg \(17\dfrac{1}{2}\) |
Additional guidance
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Ignore simplification or conversion attempt after correct answer seen | |
| Answer 17 or 18 with 17.5 seen | M2A1 |
| 17.5 in working with \(10 \lt v \leqslant 20\) on answer line | M2A0 |
| 17.5 then answer doubled | M2A0 |