| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Construct back-to-back stem-and-leaf from raw data |
| Difficulty | Easy -1.2 This is a straightforward data handling question requiring routine statistical calculations: constructing a stem-and-leaf diagram from ordered data, reading off median and IQR from 15 values, and combining summary statistics for two datasets. All techniques are standard S1 procedures with no problem-solving or conceptual challenges beyond careful arithmetic. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Falcons | 38 | 39 | 42 | 44 | 46 | 48 | 50 | 51 | 52 | 56 | 58 | 59 | 64 | 69 | 76 |
| Kites | 32 | 40 | 40 | 45 | 47 | 48 | 52 | 54 | 58 | 59 | 59 | 60 | 61 | 63 | 65 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct stem (3,4,5,6,7) | B1 | Correct stem, ignore extra values (not in reverse, not split). If split stem-and-leaf used, remaining B marks available |
| Falcons: \(9\ 8 \ | 3\); \(8\ 6\ 4\ 2 \ | 4\); \(9\ 8\ 6\ 2\ 1\ 0 \ |
| Kites: \(\ | 3\ | 2\); \(\ |
| Key: \(1\ | 5\ | 4\) means 51 minutes for Falcons and 54 minutes for Kites |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Median \(= 51\) [minutes] | B1 | Accept Q2, must be identified |
| \([\text{IQR} =]\ 59 - 44\) | M1 | \(58 \leqslant \text{UQ} \leqslant 64 - 42 \leqslant \text{LQ} \leqslant 46\). Implied if both quartile values stated and appropriate IQR calculated accurately |
| \(= 15\) [minutes] | A1 | WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[\text{mean} = \frac{792+783}{30} = \frac{1575}{30} =\right] 52.5,\ 52\frac{1}{2},\ \frac{105}{2}\) | B1 | \(\frac{1575}{30}\) is not acceptable for this mark |
| \(\sum' x^{2'} = 85727\); \(\text{sd}^2 = \left[\text{Variance} =\right]\left[\frac{(43504+42223)}{30} - \left(\frac{792+783}{30}\right)^2\right]\); \(\frac{85727}{30} - \left(\frac{1575}{30}\right)^2\ [= 101.3167]\) | M1 | Accept unsimplified variance formula. FT *their* mean. Ignore any square root leading to sd |
| \(\sigma = (\sqrt{101.3167} =)\ 10.1\) | A1 | AWRT. Must be identified e.g. sd, s, std d, \(\sigma\). Condone 'short' square root signs. If M1 not awarded, SC B1 for \(\sigma = \sqrt{101.3167}\) or \(\sqrt{\frac{6079}{60}} \approx 10.1\) |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct stem (3,4,5,6,7) | B1 | Correct stem, ignore extra values (not in reverse, not split). If split stem-and-leaf used, remaining B marks available |
| Falcons: $9\ 8 \| 3$; $8\ 6\ 4\ 2 \| 4$; $9\ 8\ 6\ 2\ 1\ 0 \| 5$; $9\ 4 \| 6$; $6 \| 7$ | B1 | Correct Falcons labelled on left, leaves in order right to left, lined up vertically, no commas or punctuation |
| Kites: $\|3\| 2$; $\|4\| 0\ 0\ 5\ 7\ 8$; $\|5\| 2\ 4\ 8\ 9\ 9$; $\|6\| 0\ 1\ 3\ 5$; $\|7\|$ | B1 | Correct Kites labelled on same diagram, leaves in order, lined up vertically, no commas or punctuation. Penalise each error only once |
| Key: $1\|5\|4$ means 51 minutes for Falcons and 54 minutes for Kites | B1 | Correct key, need both team names and 'mins' at least once, or in leaf headings/title |
**Total: 4 marks**
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Median $= 51$ [minutes] | B1 | Accept Q2, must be identified |
| $[\text{IQR} =]\ 59 - 44$ | M1 | $58 \leqslant \text{UQ} \leqslant 64 - 42 \leqslant \text{LQ} \leqslant 46$. Implied if both quartile values stated and appropriate IQR calculated accurately |
| $= 15$ [minutes] | A1 | WWW |
**Total: 3 marks**
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\text{mean} = \frac{792+783}{30} = \frac{1575}{30} =\right] 52.5,\ 52\frac{1}{2},\ \frac{105}{2}$ | B1 | $\frac{1575}{30}$ is not acceptable for this mark |
| $\sum' x^{2'} = 85727$; $\text{sd}^2 = \left[\text{Variance} =\right]\left[\frac{(43504+42223)}{30} - \left(\frac{792+783}{30}\right)^2\right]$; $\frac{85727}{30} - \left(\frac{1575}{30}\right)^2\ [= 101.3167]$ | M1 | Accept unsimplified variance formula. FT *their* mean. Ignore any square root leading to sd |
| $\sigma = (\sqrt{101.3167} =)\ 10.1$ | A1 | AWRT. Must be identified e.g. sd, s, std d, $\sigma$. Condone 'short' square root signs. If M1 not awarded, **SC B1** for $\sigma = \sqrt{101.3167}$ or $\sqrt{\frac{6079}{60}} \approx 10.1$ |
**Total: 3 marks**
---6 Teams of 15 runners took part in a charity run last Saturday. The times taken, in minutes, to complete the course by the runners from the Falcons and the runners from the Kites are shown in the table.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | l | l | l | l | l | l | l | l | }
\hline
Falcons & 38 & 39 & 42 & 44 & 46 & 48 & 50 & 51 & 52 & 56 & 58 & 59 & 64 & 69 & 76 \\
\hline
Kites & 32 & 40 & 40 & 45 & 47 & 48 & 52 & 54 & 58 & 59 & 59 & 60 & 61 & 63 & 65 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Draw a back-to-back stem-and-leaf diagram to represent this information, with the Falcons on the left-hand side.
\item Find the median and the interquartile range of the times for the Falcons.\\
Let $x$ and $y$ denote the times, in minutes, of a runner from the Falcons and a runner from the Kites respectively.
It is given that
$$\sum x = 792 , \quad \sum x ^ { 2 } = 43504 , \quad \sum y = 783 , \quad \sum y ^ { 2 } = 42223 .$$
\item Find the mean and the standard deviation of the times taken by all 30 runners from the two teams.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q6 [10]}}