Higher November 2018 Paper 2 Q6
6 Here is some information about 20 trains leaving a station.
| Number of minutes late, \(t\) | Number of trains | Midpoint | |
|---|---|---|---|
| \(0 \leqslant t < 5\) | 12 | ||
| \(5 \leqslant t < 10\) | 7 | ||
| \(10 \leqslant t < 15\) | 1 | ||
| \(t \geqslant 15\) | 0 |
(a) Work out an estimate of the mean number of minutes late. [3 marks]
(b) The station manager looks at the information in more detail.
| Number of minutes late, \(t\) | Number of trains |
|---|---|
| \(0 \leqslant t < 2\) | 12 |
| \(2 \leqslant t < 4\) | 0 |
| \(4 \leqslant t < 6\) | 7 |
| \(6 \leqslant t < 8\) | 0 |
| \(8 \leqslant t < 10\) | 0 |
| \(10 \leqslant t < 12\) | 1 |
He works out an estimate of the mean using this information.
How does his estimate compare with the answer to part (a)?
Tick one box. [1 mark]
- Higher than part (a)
- Same as part (a)
- Lower than part (a)
- Not possible to tell
| Answer | Mark | Comments |
|---|---|---|
| \(2.5 \times 12\) or 30 and \(7.5 \times 7\) or 52.5 and 12.5 \((\times 1)\) or 95 | M1 | allow one incorrect midpoint or \([2, 3] \times 12\) and \([7, 8] \times 7\) and \([12, 13]\) \((\times 1)\) ignore \(t \geqslant 15\) row |
| \(\dfrac{\text{their } 30 + \text{their } 52.5 + \text{their } 12.5}{12 + 7 + 1}\) or \(95 \div 20\) | M1dep | \(t \geqslant 15\) product must be 0 if seen condone bracket error seen eg \(30 + 52.5 + 12.5 \div 20\) |
| 4.75 | A1 | accept 4.8 or 5 if full working shown using correct midpoints |
Additional guidance
| Two correct from 30, 52.5 and 12.5 implies the first mark and could be used to score up to M2 | M1 |
| Midpoints used in the ranges [2, 3], [7, 8] and [12, 13] must be seen eg \(2.5 \times 12\) and \(7 \times 7\) and 12 \((\times 1)\) or \(3 \times 12\) and \(7 \times 7\) and 13 \((\times 1)\) NB These could be used to score up to M2 | M1 |
| Correct products seen in the table but a different method shown in the working lines eg \(20 \div 4 = 5\) | M0 |
| Answer | Mark | Comments |
|---|---|---|
| Lower than part (a) | B1 |