| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 6 |
| 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 grouped data question requiring a histogram with unequal class widths (testing frequency density understanding) and a standard mean calculation using midpoints. Both are routine S1 techniques with no problem-solving or conceptual challenges beyond basic application. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Speed \(\left( \mathrm { km } \mathrm { h } ^ { - 1 } \right)\) | \(10 - 29\) | \(30 - 39\) | \(40 - 49\) | \(50 - 59\) | \(60 - 89\) |
| Frequency | 10 | 24 | 30 | 14 | 12 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5 \quad 2.4 \quad 3 \quad 1.4 \quad 0.4\) | M1 | At least 3 frequency densities calculated (frequency \(\div\) class width); e.g. \(\left(\frac{10}{20}, \frac{10}{19} \text{ or } \frac{10}{19.5}\right)\) may be read from graph using *their* scale; 3SF or exact |
| All heights correct on graph | A1 | |
| Bar ends of \(9.5, 29.5, 39.5, 59.5, 89.5\) | B1 | |
| Axes labelled: Frequency density (fd) and speed/km h\(^{-1}\) (or appropriate title). Linear scales \(9.5 \leqslant\) horizontal axis \(\leqslant 89.5\), \(0 \leqslant\) vertical axis \(\leqslant 3\), 5 bars with no gaps | B1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{19.5\times10 + 34.5\times24 + 44.5\times30 + 54.5\times14 + 74.5\times12}{their\ 90} = \frac{195+828+1335+763+894}{90}\) | M1 | Uses at least 4 midpoint attempts (e.g. \(19.5 \pm 0.5\)). Allow unsimplified expression. |
| \(= \frac{4015}{90}\) or \(\frac{803}{18}\) | ||
| \(44\frac{11}{18}\) or \(44.6\) (km h\(^{-1}\)) | A1 | Final answer not an improper fraction. NFWW |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5 \quad 2.4 \quad 3 \quad 1.4 \quad 0.4$ | **M1** | At least 3 frequency densities calculated (frequency $\div$ class width); e.g. $\left(\frac{10}{20}, \frac{10}{19} \text{ or } \frac{10}{19.5}\right)$ may be read from graph using *their* scale; 3SF or exact |
| All heights correct on graph | **A1** | |
| Bar ends of $9.5, 29.5, 39.5, 59.5, 89.5$ | **B1** | |
| Axes labelled: Frequency density (fd) and speed/km h$^{-1}$ (or appropriate title). Linear scales $9.5 \leqslant$ horizontal axis $\leqslant 89.5$, $0 \leqslant$ vertical axis $\leqslant 3$, 5 bars with no gaps | **B1** | |
| **Total: 4** | | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{19.5\times10 + 34.5\times24 + 44.5\times30 + 54.5\times14 + 74.5\times12}{their\ 90} = \frac{195+828+1335+763+894}{90}$ | M1 | Uses at least 4 midpoint attempts (e.g. $19.5 \pm 0.5$). Allow unsimplified expression. |
| $= \frac{4015}{90}$ or $\frac{803}{18}$ | | |
| $44\frac{11}{18}$ or $44.6$ (km h$^{-1}$) | A1 | Final answer not an improper fraction. NFWW |
---3 The speeds, in $\mathrm { km } \mathrm { h } ^ { - 1 }$, of 90 cars as they passed a certain marker on a road were recorded, correct to the nearest $\mathrm { km } \mathrm { h } ^ { - 1 }$. The results are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Speed $\left( \mathrm { km } \mathrm { h } ^ { - 1 } \right)$ & $10 - 29$ & $30 - 39$ & $40 - 49$ & $50 - 59$ & $60 - 89$ \\
\hline
Frequency & 10 & 24 & 30 & 14 & 12 \\
\hline
\end{tabular}
\end{center}
(i) On the grid, draw a histogram to illustrate the data in the table.\\
\includegraphics[max width=\textwidth, alt={}, center]{5307cf3d-3d3a-441a-83d7-4adad917e168-04_1594_1198_657_516}\\
(ii) Calculate an estimate for the mean speed of these 90 cars as they pass the marker.\\
\hfill \mbox{\textit{CAIE S1 2019 Q3 [6]}}