| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Standard combined mean and SD |
| Difficulty | Moderate -0.3 This is a standard two-part question on combined means and standard deviations requiring straightforward application of formulas. Part (i) uses weighted averages (routine), while part (ii) requires using the relationship between variance and sum of squares, which is a standard technique explicitly guided by the question structure. The calculations are mechanical with no conceptual challenges beyond A-level S1 curriculum. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| For multiplying by 10 and 13 respectively and dividing by 23 | M1 | |
| For correct answer | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| For correct substitution from recognisable formula with or without sq rt | M1 | |
| For correct \(\sum x_m^2\) (can be rounded) | A1 | |
| For correct \(\sum x_m^2\) (can be rounded) | A1 | |
| For using 23 and their answer to (i) in correct formula | M1 | |
| For correct answer | A1 | 5 marks |
**(i)** $(41.2 \times 10 + 46.3 \times 13) / 23 = 44.1$
| For multiplying by 10 and 13 respectively and dividing by 23 | M1
| For correct answer | A1 | 2 marks
**(ii)** $15.1^2 = \frac{\sum x_m^2}{10} - 41.2^2$; $\sum x_m^2 = 19254.5$; $12.7^2 = \frac{\sum x_m^2}{13} - 46.3^2$; $\sum x_m^2 = 29964.74$; Total $\sum = 49219.24$; $sd = \sqrt{\left(\frac{49219.24}{23} - 44.1^2\right)} = 14.0$
| For correct substitution from recognisable formula with or without sq rt | M1
| For correct $\sum x_m^2$ (can be rounded) | A1
| For correct $\sum x_m^2$ (can be rounded) | A1
| For using 23 and their answer to (i) in correct formula | M1
| For correct answer | A1 | 5 marks4 A group of 10 married couples and 3 single men found that the mean age $\bar { x } _ { w }$ of the 10 women was 41.2 years and the standard deviation of the women's ages was 15.1 years. For the 13 men, the mean age $\bar { x } _ { m }$ was 46.3 years and the standard deviation was 12.7 years.\\
(i) Find the mean age of the whole group of 23 people.\\
(ii) The individual women's ages are denoted by $x _ { w }$ and the individual men's ages by $x _ { m }$. By first finding $\Sigma x _ { w } ^ { 2 }$ and $\Sigma x _ { m } ^ { 2 }$, find the standard deviation for the whole group.
\hfill \mbox{\textit{CAIE S1 2005 Q4 [7]}}