| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate frequency density from frequency |
| Difficulty | Easy -1.8 This is a straightforward recall question requiring only the formula frequency density = frequency ÷ class width. Part (i) is a single-step calculation (8 ÷ 0.2 = 40), part (ii) is routine histogram drawing, and part (iii) is basic probability (60/68). No problem-solving or conceptual insight needed—purely mechanical application of definitions. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency |
| Length of car \(( x\) metres \()\) | Frequency | Frequency density |
| \(2.80 \leqslant x < 3.00\) | 17 | 85 |
| \(3.00 \leqslant x < 3.10\) | 24 | 240 |
| \(3.10 \leqslant x < 3.20\) | 19 | 190 |
| \(3.20 \leqslant x < 3.40\) | 8 | \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 8/0.2 = 40\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| *(histogram)* | M1 | Uniform linear scales from at least \(2.8\) to \(3.4\) on the \(x\)-axis and \(0\) to \(240\) on the \(y\)-axis, both axes labelled; accept m or length on \(x\)-axis |
| B1 | Correct widths, no \(0.05\)s, no gaps | |
| B1 ft | 3 | Four bars correct, ft on their (i), consistent with their vertical labelling, heights within \(\frac{1}{2}\) small square |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{17+24+19}{17+24+19+8}\) | M1 | For three terms in numerator and 4 terms in denominator (can be implied). NB fd.s ie \(85+240+190/555\) get M1 A0 |
| \(= 60/68\) or \(0.882\) | A1 | 2 |
## Question 2:
### Part (i)
$a = 8/0.2 = 40$ | B1 | **1** |
### Part (ii)
*(histogram)* | M1 | Uniform linear scales from at least $2.8$ to $3.4$ on the $x$-axis and $0$ to $240$ on the $y$-axis, both axes labelled; accept m or length on $x$-axis
| B1 | Correct widths, no $0.05$s, no gaps
| B1 ft | **3** | Four bars correct, ft on their (i), consistent with their vertical labelling, heights within $\frac{1}{2}$ small square
### Part (iii)
$\frac{17+24+19}{17+24+19+8}$ | M1 | For three terms in numerator and 4 terms in denominator (can be implied). NB fd.s ie $85+240+190/555$ get M1 A0
$= 60/68$ or $0.882$ | A1 | **2** | For correct answer, a.e.f.
---2 The lengths of cars travelling on a car ferry are noted. The data are summarised in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
Length of car $( x$ metres $)$ & Frequency & Frequency density \\
\hline
$2.80 \leqslant x < 3.00$ & 17 & 85 \\
\hline
$3.00 \leqslant x < 3.10$ & 24 & 240 \\
\hline
$3.10 \leqslant x < 3.20$ & 19 & 190 \\
\hline
$3.20 \leqslant x < 3.40$ & 8 & $a$ \\
\hline
\end{tabular}
\end{center}
(i) Find the value of $a$.\\
(ii) Draw a histogram on graph paper to represent the data.\\
(iii) Find the probability that a randomly chosen car on the ferry is less than 3.20 m in length.
\hfill \mbox{\textit{CAIE S1 2004 Q2 [6]}}