| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Outliers from box plot or summary statistics |
| Difficulty | Easy -1.2 This is a straightforward box plot interpretation question requiring only basic recall of definitions (skewness, IQR, outlier rule of 1.5×IQR) and simple arithmetic. Part (iii) requires contextual reasoning but no mathematical sophistication. Significantly easier than average A-level questions. |
| Spec | 2.02b Histogram: area represents frequency2.02h Recognize outliers |
| Answer | Marks | Guidance |
|---|---|---|
| Positive skewness | B1 | 1 |
| Answer | Marks |
|---|---|
| Lower limit \(8.0 - 1.5 \times 2.3 = 4.55\) | M1 |
| Upper limit \(10.3 + 1.5 \times 2.3 = 13.75\) | M1 |
| Lowest value is 7 so no outliers at lower end | A1 |
| Highest value is 17.6 so at least one outlier at upper end | A1 |
| Answer | Marks |
|---|---|
| E.g. minimum wage means no very low values | E1 one comment relating to low earners |
| Highest wage earner may be a supervisor or manager or specialist worker or more highly trained worker | E1 one comment relating to high earners |
# Question 1
## (i)
Positive skewness | B1 | 1
## (ii)
Inter-quartile range = 10.3 – 8.0 = 2.3
Lower limit $8.0 - 1.5 \times 2.3 = 4.55$ | M1
Upper limit $10.3 + 1.5 \times 2.3 = 13.75$ | M1
Lowest value is 7 so no outliers at lower end | A1
Highest value is 17.6 so at least one outlier at upper end | A1
**Total: 5**
## (iii)
Any suitable answers
E.g. minimum wage means no very low values | E1 one comment relating to low earners
Highest wage earner may be a supervisor or manager or specialist worker or more highly trained worker | E1 one comment relating to high earners
**Total: 2**
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**TOTAL: 8**1 A business analyst collects data about the distribution of hourly wages, in $\pounds$, of shop-floor workers at a factory. These data are illustrated in the box and whisker plot.\\
\includegraphics[max width=\textwidth, alt={}, center]{56f1bd5c-4b45-4e36-a324-e7e0edbb5bdd-1_206_1420_505_397}\\
(i) Name the type of skewness of the distribution.\\
(ii) Find the interquartile range and hence show that there are no outliers at the lower end of the distribution, but there is at least one outlier at the upper end.\\
(iii) Suggest possible reasons why this may be the case.
\hfill \mbox{\textit{OCR MEI S1 Q1 [8]}}