| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Standard combined mean and SD |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for combining means and standard deviations from two groups. Part (i) requires weighted mean calculation, and part (ii) requires finding SD from summations then combining groups using the standard formula. Both are routine textbook exercises with no conceptual challenges beyond formula recall and careful arithmetic. |
| Spec | 2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{15.5 \times 12 + 910}{12 + 20}\) | M1 | Unsimplified total age divided by *their* total members (not 12, 20 or 2) |
| \(= 34.25\) or \(34\tfrac{1}{4}\) (years) | A1 | Correct exact answer (isw rounding), oe (34 years 3 months) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Considering Juniors: variance \(= \dfrac{\sum x^2}{12} - 15.5^2 = 1.2^2\) | M1 | \(\dfrac{\sum x^2}{k} - 15.5^2 = 1.2^2\), \(k = 12\) or 20 |
| \(\sum x^2 = 2900.28\) | A1 | Answer wrt 2900 |
| \(\sum z^2 = \sum x^2 + \sum y^2 = 2900.28 + 42850 = 45750\); Variance \(= \dfrac{\sum z^2}{32} - \mu^2 = \dfrac{\textit{their}\ 45750}{12+20} - (\textit{their}\ 34.25)^2\ (= 256.63)\) | M1 | *Their* \(45750 > 42850\) (not 85700 or rounding to \(1.8 \times 10^9\)) in correct variance or std deviation formula (\(\sum x^2\) and addition may not be seen) |
| \(\text{s d} = 16.0(2)\) | A1 | Correct final answer, condone 16.03 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{15.5 \times 12 + 910}{12 + 20}$ | **M1** | Unsimplified total age divided by *their* total members (not 12, 20 or 2) |
| $= 34.25$ or $34\tfrac{1}{4}$ (years) | **A1** | Correct exact answer (isw rounding), oe (34 years 3 months) |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Considering Juniors: variance $= \dfrac{\sum x^2}{12} - 15.5^2 = 1.2^2$ | **M1** | $\dfrac{\sum x^2}{k} - 15.5^2 = 1.2^2$, $k = 12$ or 20 |
| $\sum x^2 = 2900.28$ | **A1** | Answer wrt 2900 |
| $\sum z^2 = \sum x^2 + \sum y^2 = 2900.28 + 42850 = 45750$; Variance $= \dfrac{\sum z^2}{32} - \mu^2 = \dfrac{\textit{their}\ 45750}{12+20} - (\textit{their}\ 34.25)^2\ (= 256.63)$ | **M1** | *Their* $45750 > 42850$ (not 85700 or rounding to $1.8 \times 10^9$) in correct variance or std deviation formula ($\sum x^2$ and addition may not be seen) |
| $\text{s d} = 16.0(2)$ | **A1** | Correct final answer, condone 16.03 |5 The Quivers Archery club has 12 Junior members and 20 Senior members. For the Junior members, the mean age is 15.5 years and the standard deviation of the ages is 1.2 years. The ages of the Senior members are summarised by $\Sigma y = 910$ and $\Sigma y ^ { 2 } = 42850$, where $y$ is the age of a Senior member in years.\\
(i) Find the mean age of all 32 members of the club.\\
(ii) Find the standard deviation of the ages of all 32 members of the club.\\
\hfill \mbox{\textit{CAIE S1 2018 Q5 [6]}}