| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate using histogram bar dimensions |
| Difficulty | Standard +0.3 This is a standard S1 histogram question requiring understanding of frequency density and area relationships. Part (a) is trivial recall. Part (b) involves systematic algebraic manipulation using histogram properties (area = frequency, frequency density = height) with unequal class widths, which is routine for S1. Part (c) is mechanical drawing. The question requires careful bookkeeping but no novel insight beyond standard histogram techniques. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency |
| Queuing time (mins) | \(0 -\) | \(10 -\) | \(15 -\) | \(20 -\) | \(30 -\) | \(40 - 60\) |
| Number of visitors | \(15\) | \(24\) | \(x\) | \(13\) | \(10\) | \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) 10 | B1; M1 | |
| (b) 10–15 has area \(9.6 \text{ cm}^2\), so 2.5 visitors : 1 \(\text{cm}^2\) so \(36 \text{ cm}^2 = 90\) visitors. \(62 + x + y = 90\) giving \(x + y = 28\). Also \(\frac{x}{7.5} = 10 \times \frac{y}{10}\) so \(x = 2.5y\). Hence \(x = 20, y = 8\) | M1 A1 A1 M1 A1 A1 | |
| (c) Frequency densities: 1.5, 4.8, 4, 1.3, 1, 0.4. Histogram drawn | B2 B3 | Total: 13 marks |
(a) 10 | B1; M1 |
(b) 10–15 has area $9.6 \text{ cm}^2$, so 2.5 visitors : 1 $\text{cm}^2$ so $36 \text{ cm}^2 = 90$ visitors. $62 + x + y = 90$ giving $x + y = 28$. Also $\frac{x}{7.5} = 10 \times \frac{y}{10}$ so $x = 2.5y$. Hence $x = 20, y = 8$ | M1 A1 A1 M1 A1 A1 |
(c) Frequency densities: 1.5, 4.8, 4, 1.3, 1, 0.4. Histogram drawn | B2 B3 | Total: 13 marksThe length of time, in minutes, that visitors queued for a tourist attraction is given by the following table, where, for example, '$20 -$' means from 20 up to but not including 30 minutes.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Queuing time (mins) & $0 -$ & $10 -$ & $15 -$ & $20 -$ & $30 -$ & $40 - 60$ \\
\hline
Number of visitors & $15$ & $24$ & $x$ & $13$ & $10$ & $y$ \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item State the upper class boundary of the first class. [1 mark]
\end{enumerate}
A histogram is drawn to represent this data. The total area under the histogram is $36$ cm$^2$. The '$10 -$' bar has width $1$ cm and height $9.6$ cm. The '$15 -$' bar is ten times as high as the '$40 - 60$' bar.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the values of $x$ and $y$. [7 marks]
\item On graph paper, construct the histogram accurately. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q4 [13]}}