| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Identify appropriate measure with outliers |
| Difficulty | Easy -1.3 This is a straightforward question testing basic understanding of mean vs median with an obvious outlier (19.4). Parts (a) and (b) require only simple calculations with given data, and part (c) asks for standard textbook reasoning about outliers affecting the mean. No problem-solving or novel insight required. |
| Spec | 2.02f Measures of average and spread |
| 4.1 | 4.2 | 4.4 | 4.5 | 4.6 | 4.8 | 5.0 | 5.2 | 5.3 | 5.4 |
| 5.5 | 5.8 | 6.0 | 6.2 | 6.3 | 6.4 | 6.6 | 6.8 | 6.9 | 19.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\frac{123.4}{20} = \right] 6.17\) | B1 | Accept 6 m 17 cm, \(\frac{1234}{200}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{10\text{th} + 11\text{th}}{2} = \frac{5.4 + 5.5}{2} = 5.45\) (m) | B1 | Accept 5 m 45 cm |
| Answer | Marks | Guidance |
|---|---|---|
| The mean is unduly influenced by an extreme value, 19.4. | B1 | Comment must be within context |
## Question 2:
**Part 2(a):**
$\left[\frac{123.4}{20} = \right] 6.17$ | B1 | Accept 6 m 17 cm, $\frac{1234}{200}$
**Part 2(b):**
$\frac{10\text{th} + 11\text{th}}{2} = \frac{5.4 + 5.5}{2} = 5.45$ (m) | B1 | Accept 5 m 45 cm
**Part 2(c):**
The mean is unduly influenced by an extreme value, 19.4. | B1 | Comment must be within context
---2 Twenty children were asked to estimate the height of a particular tree. Their estimates, in metres, were as follows.
\begin{center}
\begin{tabular}{ r r r r r r r r r r }
4.1 & 4.2 & 4.4 & 4.5 & 4.6 & 4.8 & 5.0 & 5.2 & 5.3 & 5.4 \\
5.5 & 5.8 & 6.0 & 6.2 & 6.3 & 6.4 & 6.6 & 6.8 & 6.9 & 19.4 \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the mean of the estimated heights.
\item Find the median of the estimated heights.
\item Give a reason why the median is likely to be more suitable than the mean as a measure of the central tendency for this information.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q2 [3]}}