S1 January 2006 Q1
1. Over a period of time, the number of people \(x\) leaving a hotel each morning was recorded. These data are summarised in the stem and leaf diagram below.
| Number leaving | \(3 \mid 2\) means 32 | Totals |
|---|---|---|
| 2 | 7 9 9 | (3) |
| 3 | 2 2 3 5 6 | (5) |
| 4 | 0 1 4 8 9 | (5) |
| 5 | 2 3 3 6 6 6 8 | (7) |
| 6 | 0 1 4 5 | (4) |
| 7 | 2 3 | (2) |
| 8 | 1 | (1) |
For these data,
(a) write down the mode, (1)
(b) find the values of the three quartiles. (3)
Given that \(\Sigma x = 1335\) and \(\Sigma x^2 = 71\,801\), find
(c) the mean and the standard deviation of these data. (4)
One measure of skewness is found using
\[\frac{\text{mean} - \text{mode}}{\text{standard deviation}}.\](d) Evaluate this measure to show that these data are negatively skewed. (2)
(e) Give two other reasons why these data are negatively skewed. (4)
| Scheme | Marks |
|---|---|
| Mode is 56 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Q}_1 = 35, \mathrm{Q}_2 = 52, \mathrm{Q}_3 = 60\) | B1,B1,B1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\bar{x} = \dfrac{1335}{27} = 49.\dot{4}\) or \(49\tfrac{4}{9}\) | B1 |
| \(\sigma^2 = \dfrac{71801}{27} - \left(\dfrac{1335}{27}\right)^2 = 214.5432\ldots\) | M1A1ft |
| \(\sigma = 14.6\) or \(14.9\) | A1 |
| (4) |
Notes
B1 exact or awrt 49.4
A1 awrt 14.6(5) or 14.9
| Scheme | Marks |
|---|---|
| \(\dfrac{49.4 - 56}{14.6} = -0.448\) | M1A1 |
| (2) |
Notes
A1 awrt range −0.44 to −0.46
| Scheme | Marks |
|---|---|
| For negative skew; Mean<median<mode (49.4<52<56 not required) | M1 A1 |
| \(\mathrm{Q}_3 - \mathrm{Q}_2 \lt \mathrm{Q}_2 - \mathrm{Q}_1\) 8 and 17 | M1 A1ft |
| (4) | |
| (14 marks) |
Notes
M1 2 compared correctly
A1 3 compared correctly
Accept other valid reason eg. 3(mean-median)/sd as alt for M1A1