S1 January 2012 Q4
4. The marks, \(x\), of 45 students randomly selected from those students who sat a mathematics examination are shown in the stem and leaf diagram below.
| Mark | Totals | |
|---|---|---|
| 3 | 6 9 9 | (3) |
| 4 | 0 1 2 2 3 4 | (6) |
| 4 | 5 6 6 6 8 | (5) |
| 5 | 0 2 3 3 4 4 | (6) |
| 5 | 5 5 6 7 7 9 | (6) |
| 6 | 0 0 0 0 1 3 4 4 4 | (9) |
| 6 | 5 5 6 7 8 9 | (6) |
| 7 | 1 2 3 3 | (4) |
Key (3|6 means 36)
For these students \(\sum x = 2497\) and \(\sum x^2 = 143\,369\)
The mean and standard deviation of the marks of all the students who sat the examination were 55 and 10 respectively. The examiners decided that the total mark of each student should be scaled by subtracting 5 marks and then reducing the mark by a further 10%.
| Scheme | Marks |
|---|---|
| 60 | B1 |
| (1) |
Notes
B1 60 only
| Scheme | Marks |
|---|---|
| \(Q_1 = 46\) | B1 |
| \(Q_2 = 56\) | B1 |
| \(Q_3 = 64\) | B1 |
| (3) |
Notes
Award each B1 for correct answer only in this order.
| Scheme | Marks |
|---|---|
| mean = 55.48…. or \(\dfrac{2497}{45}\) awrt 55.5 | B1 |
| sd \(= \sqrt{\dfrac{143369}{45} - \left(\dfrac{2497}{45}\right)^2}\) | M1 |
| = 10.342… (\(s\) = 10.459..) anything which rounds to 10.3 (or s = 10.5) | A1 |
| (3) |
Notes
M1 for use of correct formula, including square root. Correct answers with no working B1M1A1.
| Scheme | Marks |
|---|---|
| Mean < median < mode or \(Q_2 - Q_1 \gt Q_3 - Q_2\) with or without their numbers or median closer to upper quartile (than lower quartile) or (mean-median)/sd <0; | B1 |
| negative skew; | B1dep |
| (2) |
Notes
B1 any correct comparison of a pair of mean, median and mode using their values.
B1 for ‘negative skew’ or allow (almost) symmetrical dependent upon correct reason.
| Scheme | Marks |
|---|---|
| mean = \((55 - 5) \times 0.9\) | M1 |
| = 45 | A1 |
| sd = \(10 \times 0.9\) | M1 |
| = 9 | A1 |
| (4) | |
| (13 marks) |
Notes
M1 for (55 or 55.5 - 5)\(\times\)0.9
A1 for the correct answer only.
M1 for (10 or 10.3 or 10.5) )\(\times\)0.9
A1 for the correct answer only.