| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Complete frequency table from histogram only |
| Difficulty | Moderate -0.8 This is a straightforward histogram interpretation question requiring students to use frequency density = frequency/class width to find missing frequencies, then apply standard S1 techniques (interpolation for median, skewness comparison). All steps are routine textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the multi-part nature. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread |
| Height \(( h\) cm) | \(5 \leqslant h < 9\) | \(9 \leqslant h < 13\) | \(13 \leqslant h < 15\) | \(15 \leqslant h < 17\) | \(17 \leqslant h < 25\) |
| Frequency | 32 | 152 | 120 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(2 \times 36 = 72\) and \(8 \times 4 = 32\) | M1 A1 (2) | For any equivalent method to find either frequency. May be implied by either correct frequency, or by two frequencies which add to 104, or by a correct scale on the fd axis with at least 3 labels. A1: For 72 and 32 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \([13] + \frac{(204-184)}{120} \times 2\) or \([13] + \frac{(204.5-184)}{120} \times 2\) | M1 | For any equivalent method to find the median, e.g. \(\frac{Q_2 - 13}{15-13} = \frac{204-184}{304-184}\) or \(\frac{15-Q_2}{Q_2-13} = \frac{304-204}{204-184}\). Allow working downwards \([15] - \frac{(304-204)}{120} \times 2\) |
| \(= \frac{40}{3} =\) awrt 13.3 | A1 (2) | awrt 13.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| Symmetrically distributed / No skew as the mean \(\approx\) median | B1 (1) | For a correct identification of skew (symmetric/no skew or slight negative skew) with a correct supporting reason. Condone mean \(<\) median so negative skew. Allow use of "their median" in the comparison provided "their median" \(\approx\) 13.2. Allow \(Q_1 =\) awrt 10.8 or awrt 10.9 and \(Q_3 =\) awrt 15.1 and \(Q_2 - Q_1 > Q_3 - Q_2\) so negative skew. Comments referring only to the diagram (being symmetrical therefore no skew) send to review. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(\frac{32}{4} + 152 + \frac{120}{2} [= 220]\) | M1 | For a correct method to find the number of plants between 8cm and 14cm (may be implied by sight of 220) |
| \(\frac{220}{408} \times \frac{219}{407}\) | M1 | For \(\frac{n}{408} \times \frac{n-1}{407}\) or \(\left(\frac{n}{408}\right)^2\) with \(n = 210, n, 230\) |
| \(\frac{365}{1258}\) or \(0.2901\ldots\) awrt 0.29 | A1 (3) | awrt 0.29; may see \(\frac{3025}{10404}\) from \(\left(\frac{220}{408}\right)^2\) |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $2 \times 36 = 72$ and $8 \times 4 = 32$ | M1 A1 (2) | For any equivalent method to find either frequency. May be implied by either correct frequency, or by two frequencies which add to 104, or by a correct scale on the fd axis with at least 3 labels. A1: For 72 and 32 |
## Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $[13] + \frac{(204-184)}{120} \times 2$ or $[13] + \frac{(204.5-184)}{120} \times 2$ | M1 | For any equivalent method to find the median, e.g. $\frac{Q_2 - 13}{15-13} = \frac{204-184}{304-184}$ or $\frac{15-Q_2}{Q_2-13} = \frac{304-204}{204-184}$. Allow working downwards $[15] - \frac{(304-204)}{120} \times 2$ |
| $= \frac{40}{3} =$ awrt 13.3 | A1 (2) | awrt 13.3 |
## Part (c)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Symmetrically distributed / No skew as the mean $\approx$ median | B1 (1) | For a correct identification of skew (symmetric/no skew or slight negative skew) with a correct supporting reason. Condone mean $<$ median so negative skew. Allow use of "their median" in the comparison provided "their median" $\approx$ 13.2. Allow $Q_1 =$ awrt 10.8 or awrt 10.9 **and** $Q_3 =$ awrt 15.1 **and** $Q_2 - Q_1 > Q_3 - Q_2$ so negative skew. Comments referring only to the diagram (being symmetrical therefore no skew) send to review. |
## Part (d)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\frac{32}{4} + 152 + \frac{120}{2} [= 220]$ | M1 | For a correct method to find the number of plants between 8cm and 14cm (may be implied by sight of 220) |
| $\frac{220}{408} \times \frac{219}{407}$ | M1 | For $\frac{n}{408} \times \frac{n-1}{407}$ or $\left(\frac{n}{408}\right)^2$ with $n = 210, n, 230$ |
| $\frac{365}{1258}$ or $0.2901\ldots$ awrt 0.29 | A1 (3) | awrt 0.29; may see $\frac{3025}{10404}$ from $\left(\frac{220}{408}\right)^2$ |
**Total: 8 marks**\begin{enumerate}
\item The histogram below shows the distribution of the heights, to the nearest cm , of 408 plants.\\
\includegraphics[max width=\textwidth, alt={}, center]{86446ce3-496a-4f02-9566-9b207bac9efa-02_1001_1473_340_296}\\
(a) Use the histogram to complete the following table.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Height $( h$ cm) & $5 \leqslant h < 9$ & $9 \leqslant h < 13$ & $13 \leqslant h < 15$ & $15 \leqslant h < 17$ & $17 \leqslant h < 25$ \\
\hline
Frequency & 32 & 152 & 120 & & \\
\hline
\end{tabular}
\end{center}
(b) Use interpolation to estimate the median.
The mean height of these plants is 13.2 cm correct to one decimal place.\\
(c) Describe the skew of these data. Give a reason for your answer.
Two of these plants are chosen at random.\\
(d) Estimate the probability that both of their heights are between 8 cm and 14 cm
\hfill \mbox{\textit{Edexcel S1 2024 Q1 [8]}}