| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw histogram then find median/quartiles from cumulative frequency |
| Difficulty | Moderate -0.8 This is a straightforward S1 statistics question requiring standard histogram construction with unequal class widths (calculating frequency densities) and median estimation from grouped data using linear interpolation. Both are routine textbook procedures with no problem-solving or novel insight required, making it easier than average. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread |
| Distance \(( d\) miles \()\) | \(0 \leqslant d < 50\) | \(50 \leqslant d < 100\) | \(100 \leqslant d < 200\) | \(200 \leqslant d < 400\) |
| Frequency | 360 | 400 | 307 | 133 |
| Answer | Marks | Guidance |
|---|---|---|
| Distance | freq | width |
| 0– | 360 | 50 |
| 50– | 400 | 50 |
| 100– | 307 | 100 |
| 200–400 | 133 | 200 |
| M1, A1 CAO | for fds; accept any suitable unit for fd such as freq per 50 miles | |
| Histogram with correct bars | L1, W1, H1 | L1 linear scales on both axes and label; W1 width of bars; H1 height of bars; 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Median = 600th distance | B1 | for \(600^{\text{th}}\) |
| Estimate \(= 50 + \frac{240}{400} \times 50 = 50 + 30 = 80\) | M1, A1 CAO | M1 for attempt to interpolate; Total: 8 marks |
# Question 5:
## Part (i):
| Distance | freq | width | f dens |
|----------|------|-------|--------|
| 0– | 360 | 50 | 7.200 |
| 50– | 400 | 50 | 8.000 |
| 100– | 307 | 100 | 3.070 |
| 200–400 | 133 | 200 | 0.665 |
| M1, A1 CAO | for fds; accept any suitable unit for fd such as freq per 50 miles
Histogram with correct bars | L1, W1, H1 | L1 linear scales on both axes and label; W1 width of bars; H1 height of bars; **5 marks**
## Part (ii):
Median = 600th distance | B1 | for $600^{\text{th}}$
Estimate $= 50 + \frac{240}{400} \times 50 = 50 + 30 = 80$ | M1, A1 CAO | M1 for attempt to interpolate; **Total: 8 marks**
---5 The frequency table below shows the distance travelled by 1200 visitors to a particular UK tourist destination in August 2008.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Distance $( d$ miles $)$ & $0 \leqslant d < 50$ & $50 \leqslant d < 100$ & $100 \leqslant d < 200$ & $200 \leqslant d < 400$ \\
\hline
Frequency & 360 & 400 & 307 & 133 \\
\hline
\end{tabular}
\end{center}
(i) Draw a histogram on graph paper to illustrate these data.\\
(ii) Calculate an estimate of the median distance.
\hfill \mbox{\textit{OCR MEI S1 2009 Q5 [8]}}