| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Cumulative frequency graph construction then interpretation |
| Difficulty | Easy -1.2 This is a straightforward cumulative frequency question requiring standard plotting from a frequency table, reading off median and quartiles from the graph, calculating IQR, and identifying skewness type. All steps are routine S1 techniques with no problem-solving or novel insight required, making it easier than average. |
| Spec | 2.02a Interpret single variable data: tables and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Bar chart with labelled linear scales | G1 | Labelled linear scales |
| Correct height of lines | G1 | Height of lines |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Negative (skewness) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Sigma fx = 123\), mean \(= \frac{123}{25} = 4.92\) | B1 | |
| \(S_{xx} = 681 - \frac{123^2}{25} = 75.84\) | M1 | For \(S_{xx}\) attempted |
| \(\text{M.s.d} = \frac{75.84}{25} = 3.034\) | A1 | FT their 4.92 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total for 25 days is 123; total for 31 days is 155 | M1 | \(31 \times 5 - 25 \times \text{their } 4.92\) |
| Total for next 6 days is 32, mean \(= 5.33\) | A1 | FT their 123 |
| [2] |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Bar chart with labelled linear scales | G1 | Labelled linear scales |
| Correct height of lines | G1 | Height of lines |
| | **[2]** | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Negative (skewness) | B1 | |
| | **[1]** | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Sigma fx = 123$, mean $= \frac{123}{25} = 4.92$ | B1 | |
| $S_{xx} = 681 - \frac{123^2}{25} = 75.84$ | M1 | For $S_{xx}$ attempted |
| $\text{M.s.d} = \frac{75.84}{25} = 3.034$ | A1 | FT their 4.92 |
| | **[3]** | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total for 25 days is 123; total for 31 days is 155 | M1 | $31 \times 5 - 25 \times \text{their } 4.92$ |
| Total for next 6 days is 32, mean $= 5.33$ | A1 | FT their 123 |
| | **[2]** | |
---3 Every day, George attempts the quiz in a national newspaper. The quiz always consists of 7 questions. In the first 25 days of January, the numbers of questions George answers correctly each day are summarised in the table below.
(i) On the insert, draw a cumulative frequency diagram to illustrate the data.\\
(ii) Use your graph to estimate the median length of journey and the quartiles.
Hence find the interquartile range.\\
(iii) State the type of skewness of the distribution of the data.
\hfill \mbox{\textit{OCR MEI S1 Q3 [8]}}