| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate frequency density from frequency |
| Difficulty | Moderate -0.8 This is a straightforward histogram question requiring basic understanding that frequency density = frequency/class width, and that bar height represents frequency density. Part (a) involves a simple proportion calculation, and part (b) uses the fundamental property that total area equals total frequency. Both parts are routine applications of standard S1 histogram concepts with no problem-solving insight required. |
| Spec | 2.02b Histogram: area represents frequency |
| Group | \(0 - 10\) | \(10 - 20\) | \(20 - 25\) | \(25 - 30\) | \(30 - 50\) | \(50 - 100\) |
| Frequency | \(2 x\) | \(3 x\) | \(5 x\) | \(6 x\) | \(2 x\) | \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(3x\) corresponds to \(24 \text{ cm}^2\), so \(5x\) corresponds to \(40 \text{ cm}^2\) | M1 A1 | |
| Width = \(2\) cm, so height = \(20\) cm | M1 A1 | |
| (b) Area = \(19x = 19 \times 8 = 152 \text{ cm}^2\) | M1 A1 A1 | Total: 7 marks |
(a) $3x$ corresponds to $24 \text{ cm}^2$, so $5x$ corresponds to $40 \text{ cm}^2$ | M1 A1 |
Width = $2$ cm, so height = $20$ cm | M1 A1 |
(b) Area = $19x = 19 \times 8 = 152 \text{ cm}^2$ | M1 A1 A1 | **Total: 7 marks**\begin{enumerate}
\item A histogram is to be drawn to represent the following grouped continuous data:
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | | c | c | c | c | c | c | }
\hline
Group & $0 - 10$ & $10 - 20$ & $20 - 25$ & $25 - 30$ & $30 - 50$ & $50 - 100$ \\
\hline
Frequency & $2 x$ & $3 x$ & $5 x$ & $6 x$ & $2 x$ & $x$ \\
\hline
\end{tabular}
\end{center}
The ' $10 - 20$ ' bar has height 6 cm and width 4 cm . Calculate\\
(a) the height of the ' $20 - 25$ ' bar,\\
(b) the total area under the histogram.\\
\hfill \mbox{\textit{Edexcel S1 Q1 [7]}}