| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw histogram then estimate mean/standard deviation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard S1 techniques: drawing a histogram with unequal class widths (requiring frequency density calculation) and computing mean/standard deviation from grouped data using midpoints. Both are routine textbook exercises with no problem-solving or conceptual challenges beyond careful arithmetic. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Time in minutes | \(101 - 120\) | \(121 - 130\) | \(131 - 135\) | \(136 - 145\) | \(146 - 160\) |
| Frequency | 18 | 48 | 34 | 32 | 18 |
| Answer | Marks | Guidance |
|---|---|---|
| Class width | 20 | 10 |
| Frequency density | 0.9 | 4.8 |
| M1 | At least 4 frequency densities calculated by \(\frac{f}{cw}\) e.g. \(\frac{18}{20}\) (condone \(\frac{f}{cw \pm 0.5}\) if unsimplified). Accept unsimplified, may be read from graph using their scale, no lower than \(1\text{cm} = 1\text{ fd}\) | |
| A1 | All bar heights correct on graph (no FT), using suitable linear scale with at least 3 values indicated, no lower than \(1\text{cm} = 1\text{ fd}\) | |
| B1 | Bar ends at \(120.5, 130.5, 135.5, 145.5, 160.5\). 5 bars drawn with horizontal linear scale, no lower than \(1\text{cm} = 10\text{ min}\), with at least 3 values indicated. \(100 \leq \text{horizontal scale} \leq 160\) | |
| B1 | Axes labelled frequency density (fd), time (\(t\)) and minutes (min, m) oe, or appropriate title. (Axes may be reversed) | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Midpoints: \(110.5, 125.5, 133, 140.5, 153\) | B1 | At least 4 correct mid-points seen, may be by data table or used in formula |
| \(\text{Mean} = \frac{18\times110.5+48\times125.5+34\times133+32\times140.5+18\times153}{150} = \frac{1989+6024+4522+4496+2754}{150}\) | M1 | Correct formula for mean using midpoints \(\pm0.5\), condone 1 midpoint error within class |
| \(= 131.9\) | A1 | Accept \(132\), \(131\frac{9}{10}\), or \(\frac{1319}{10}\). Must be identified |
| \(\text{Variance} = \frac{18\times110.5^2+48\times125.5^2+34\times133^2+32\times140.5^2+18\times153^2}{150}-(their\ 131.9)^2\) | M1 | Appropriate variance formula with their 5 midpoints within class. If correct midpoints: \(\left\{\frac{3200+41400+194400+157300+153600}{150}\ or\ \frac{2630272.5}{150}\right\} -\{131.9^2\ or\ 17397.61\}\) |
| \([=137.54]\), Standard deviation \(= 11.7\) | A1 | \(11.7277448\ldots\) to at least 3SF. Accept \(11.6 \leqslant \sigma < 11.95\) |
| Total: 5 marks |
## Question 4(a):
| Class width | 20 | 10 | 5 | 10 | 15 |
| Frequency density | 0.9 | 4.8 | 6.8 | 3.2 | 1.2 |
| M1 | At least 4 frequency densities calculated by $\frac{f}{cw}$ e.g. $\frac{18}{20}$ (condone $\frac{f}{cw \pm 0.5}$ if unsimplified). Accept unsimplified, may be read from graph using their scale, no lower than $1\text{cm} = 1\text{ fd}$
| A1 | All bar heights correct on graph (no FT), using suitable linear scale with at least 3 values indicated, no lower than $1\text{cm} = 1\text{ fd}$
| B1 | Bar ends at $120.5, 130.5, 135.5, 145.5, 160.5$. 5 bars drawn with horizontal linear scale, no lower than $1\text{cm} = 10\text{ min}$, with at least 3 values indicated. $100 \leq \text{horizontal scale} \leq 160$
| B1 | Axes labelled frequency density (fd), time ($t$) and minutes (min, m) oe, or appropriate title. (Axes may be reversed)
| **4** |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoints: $110.5, 125.5, 133, 140.5, 153$ | **B1** | At least 4 correct mid-points seen, may be by data table or used in formula |
| $\text{Mean} = \frac{18\times110.5+48\times125.5+34\times133+32\times140.5+18\times153}{150} = \frac{1989+6024+4522+4496+2754}{150}$ | **M1** | Correct formula for mean using midpoints $\pm0.5$, condone 1 midpoint error within class |
| $= 131.9$ | **A1** | Accept $132$, $131\frac{9}{10}$, or $\frac{1319}{10}$. Must be identified |
| $\text{Variance} = \frac{18\times110.5^2+48\times125.5^2+34\times133^2+32\times140.5^2+18\times153^2}{150}-(their\ 131.9)^2$ | **M1** | Appropriate variance formula with their 5 midpoints within class. If correct midpoints: $\left\{\frac{3200+41400+194400+157300+153600}{150}\ or\ \frac{2630272.5}{150}\right\} -\{131.9^2\ or\ 17397.61\}$ |
| $[=137.54]$, Standard deviation $= 11.7$ | **A1** | $11.7277448\ldots$ to at least 3SF. Accept $11.6 \leqslant \sigma < 11.95$ |
| **Total: 5 marks** | | |
---4 The times, to the nearest minute, of 150 athletes taking part in a charity run are recorded. The results are summarised in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Time in minutes & $101 - 120$ & $121 - 130$ & $131 - 135$ & $136 - 145$ & $146 - 160$ \\
\hline
Frequency & 18 & 48 & 34 & 32 & 18 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Draw a histogram to represent this information.\\
\includegraphics[max width=\textwidth, alt={}, center]{e8c2b51e-d788-4917-829e-1b056a24f520-08_1493_1397_936_415}
\item Calculate estimates for the mean and standard deviation of the times taken by the athletes.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q4 [9]}}