| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate single values from cumulative frequency graph |
| Difficulty | Easy -1.3 This is a straightforward statistics question testing standard cumulative frequency graph skills: reading quartiles, drawing a box plot, converting between cumulative and grouped frequency, and calculating means. All techniques are routine A-level procedures with no problem-solving or novel insight required. The most challenging part (v-vi) involves a simple adjustment to the mean calculation, but this is still mechanical application of formulas. |
| Spec | 2.02f Measures of average and spread |
| 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 | Marks | Guidance |
| Median distance = \(88^{th}\) value = 480 | M1 | Within 5 |
| A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Lower Quartile = \(44^{th}\) value = 320 | B1 | |
| Upper Quartile = \(132^{nd}\) value = 680 | B1 | |
| Interquartile range = \(680 - 320 = 360\) | M1 | ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Box plot drawn | G1 | Basic idea |
| Scale linear \(0 - 1200\) | G1 | Linear \(0 - 1200\) |
| Box including median accurate | G1 | Box including median (accurate) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 < d \leq 200\): frequency 20 | M1 | Correct classes |
| \(200 < d \leq 400\): frequency 44 | ||
| \(400 < d \leq 600\): frequency 54 | M1 | Correct frequencies |
| \(600 < d \leq 800\): frequency 32 | ||
| \(800 < d \leq 1000\): frequency 19 | ||
| \(1000 < d \leq 1200\): frequency 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mid-points: 100, 300, 500, 700, 900, 1100 | M1 | mid points |
| \(fx\): 2000, 13200, 27000, 22400, 17100, 7700 | M1 | fx |
| \(\sum f = 176\), \(\sum fx = 89400\) | ||
| Estimate of mean \(= \frac{89400}{176} = 507.95\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mid point of first class now 150, total increase of 1000 | M1 | 150 |
| New estimate of mean \(= 513.6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The point \((0,0)\) would move to \((100,0)\) | E1 | point \((0,0)\) |
| E1 | point \((100,0)\) |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Median distance = $88^{th}$ value = 480 | M1 | Within 5 |
| | A1 | cao |
### Part (i) B
| Answer | Marks | Guidance |
|--------|-------|----------|
| Lower Quartile = $44^{th}$ value = 320 | B1 | |
| Upper Quartile = $132^{nd}$ value = 680 | B1 | |
| Interquartile range = $680 - 320 = 360$ | M1 | ft |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Box plot drawn | G1 | Basic idea |
| Scale linear $0 - 1200$ | G1 | Linear $0 - 1200$ |
| Box including median accurate | G1 | Box including median (accurate) |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 < d \leq 200$: frequency 20 | M1 | Correct classes |
| $200 < d \leq 400$: frequency 44 | | |
| $400 < d \leq 600$: frequency 54 | M1 | Correct frequencies |
| $600 < d \leq 800$: frequency 32 | | |
| $800 < d \leq 1000$: frequency 19 | | |
| $1000 < d \leq 1200$: frequency 7 | | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mid-points: 100, 300, 500, 700, 900, 1100 | M1 | mid points |
| $fx$: 2000, 13200, 27000, 22400, 17100, 7700 | M1 | fx |
| $\sum f = 176$, $\sum fx = 89400$ | | |
| Estimate of mean $= \frac{89400}{176} = 507.95$ | A1 | |
### Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mid point of first class now 150, total increase of 1000 | M1 | 150 |
| New estimate of mean $= 513.6$ | A1 | |
### Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| The point $(0,0)$ would move to $(100,0)$ | E1 | point $(0,0)$ |
| | E1 | point $(100,0)$ |
---2 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]{b4bf1bd0-f85d-42b7-ad15-6672387bb208-2_998_1466_566_367}
\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 Q2 [17]}}