| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate single values from cumulative frequency graph |
| Difficulty | Easy -1.2 This is a straightforward cumulative frequency question requiring standard reading from a graph (median, quartiles), drawing a box plot, converting between cumulative and grouped frequencies, and calculating means. All techniques are routine S1 procedures with no problem-solving or novel insight required—easier than average A-level. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Distance (metres) | 200 | 400 | 600 | 800 | 1000 | 1200 |
| Cumulative frequency | 20 | 64 | 118 | 150 | 169 | 176 |
| Distance ( \(d\) metres) | Frequency |
| \(0 < d \leqslant 200\) | 20 |
| \(200 < d \leqslant 400\) | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Median distance \(= 88^\text{th}\) value \(= 480\) | M1, A1 | Within 5; cao |
| Lower Quartile \(= 44^\text{th}\) value \(= 320\) | B1 | |
| Upper Quartile \(= 132^\text{nd}\) value \(= 680\) | B1 | |
| Interquartile range \(= 680 - 320 = 360\) | M1 | ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Box-and-whisker plot drawn | G1 | Basic idea |
| Linear scale 0–1200 | G1 | Linear 0–1200 |
| Box including median (accurate), whiskers to 0 and 1200 | G1 | Box including median (accurate) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Classes: \(0 < d \leq 200\): 20; \(200 < d \leq 400\): 44; \(400 < d \leq 600\): 54; \(600 < d \leq 800\): 32; \(800 < d \leq 1000\): 19; \(1000 < d \leq 1200\): 7 | M1, M1 | Correct classes; correct frequencies |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Midpoints: 100, 300, 500, 700, 900, 1100; \(fx\): 2000, 13200, 27000, 22400, 17100, 7700; total \(\sum fx = 89400\), \(n=176\) | M1, M1 | midpoints; \(fx\) |
| Estimate of mean \(= 507.95\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mid point of first class now 150; total increase of 1000; new estimate of mean \(= 513.6\) | M1, A1 | 150 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The point \((0,0)\) would move to \((100,0)\) | E1, E1 | point \((0,0)\); point \((100,0)\) |
## Question 7:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Median distance $= 88^\text{th}$ value $= 480$ | M1, A1 | Within 5; cao |
| Lower Quartile $= 44^\text{th}$ value $= 320$ | B1 | |
| Upper Quartile $= 132^\text{nd}$ value $= 680$ | B1 | |
| Interquartile range $= 680 - 320 = 360$ | M1 | ft |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Box-and-whisker plot drawn | G1 | Basic idea |
| Linear scale 0–1200 | G1 | Linear 0–1200 |
| Box including median (accurate), whiskers to 0 and 1200 | G1 | Box including median (accurate) |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Classes: $0 < d \leq 200$: 20; $200 < d \leq 400$: 44; $400 < d \leq 600$: 54; $600 < d \leq 800$: 32; $800 < d \leq 1000$: 19; $1000 < d \leq 1200$: 7 | M1, M1 | Correct classes; correct frequencies |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Midpoints: 100, 300, 500, 700, 900, 1100; $fx$: 2000, 13200, 27000, 22400, 17100, 7700; total $\sum fx = 89400$, $n=176$ | M1, M1 | midpoints; $fx$ |
| Estimate of mean $= 507.95$ | A1 | |
### Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| Mid point of first class now 150; total increase of 1000; new estimate of mean $= 513.6$ | M1, A1 | 150 |
### Part (vi)
| Answer | Mark | Guidance |
|--------|------|----------|
| The point $(0,0)$ would move to $(100,0)$ | E1, E1 | point $(0,0)$; point $(100,0)$ |
---7 The cumulative frequency graph below illustrates the distances that 176 children live from their primary school.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Distance from school}
\includegraphics[alt={},max width=\textwidth]{b35b2b3b-0d26-4a35-b4d2-110bf270d5dc-4_1073_1571_580_340}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Use the graph to estimate, to the nearest 10 metres,\\
(A) the median distance from school,\\
(B) the lower quartile, upper quartile and interquartile range.
\item Draw a box and whisker plot to illustrate the data.
The graph on page 4 used the following grouped data.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Distance (metres) & 200 & 400 & 600 & 800 & 1000 & 1200 \\
\hline
Cumulative frequency & 20 & 64 & 118 & 150 & 169 & 176 \\
\hline
\end{tabular}
\end{center}
\item Copy and complete the grouped frequency table below describing the same data.
\begin{center}
\begin{tabular}{ | c | c | }
\hline
Distance ( $d$ metres) & Frequency \\
\hline
$0 < d \leqslant 200$ & 20 \\
\hline
$200 < d \leqslant 400$ & \\
\hline
& \\
\hline
& \\
\hline
& \\
\hline
& \\
\hline
\end{tabular}
\end{center}
\item Hence estimate the mean distance these children live from school.
It is subsequently found that none of the 176 children lives within 100 metres of the school.
\item Calculate the revised estimate of the mean distance.
\item Describe what change needs to be made to the cumulative frequency graph.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2005 Q7 [12]}}