OCR MEI S1 2010 June — Question 1 8 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2010
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeOutliers from box plot or summary statistics
DifficultyModerate -0.8 This question requires reading values from a box plot, calculating IQR, and applying the standard outlier rule (1.5×IQR beyond quartiles). These are routine procedural tasks with minimal problem-solving. Part (iii) requires contextual interpretation but is not mathematically demanding. Significantly easier than average A-level questions.
Spec2.02f Measures of average and spread2.02h Recognize outliers

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]{091d6f43-ad01-4849-9f3c-3e58349aa169-2_204_1422_484_363}
  1. Name the type of skewness of the distribution.
  2. 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.
  3. Suggest possible reasons why this may be the case.

AnswerMarks Guidance
(i) Positive skewnessB1 1 mark
(ii) Inter-quartile range = \(10.3 - 8.0 = 2.3\)B1
Lower limit \(8.0 - 1.5 \times 2.3 = 4.55\)
AnswerMarks
Upper limit \(10.3 + 1.5 \times 2.3 = 13.75\)M1 for \(8.0 - 1.5 \times 2.3\), M1 for \(10.3 + 1.5 \times 2.3\)
Lowest value is 7 so no outliers at lower end
AnswerMarks Guidance
Highest value is 17.6 so at least one outlier at upper end.A1, A1 5 marks
(iii) Any suitable answers
AnswerMarks Guidance
Eg minimum wage means no very low valuesE1 one comment relating to low earners
Highest wage earner may be a supervisor or manager or specialist worker or more highly trained workerE1 one comment relating to high earners 2 marks
TOTAL: 8 marks
**(i)** Positive skewness | B1 | 1 mark

**(ii)** Inter-quartile range = $10.3 - 8.0 = 2.3$ | B1 |

Lower limit $8.0 - 1.5 \times 2.3 = 4.55$
Upper limit $10.3 + 1.5 \times 2.3 = 13.75$ | M1 for $8.0 - 1.5 \times 2.3$, M1 for $10.3 + 1.5 \times 2.3$ |

Lowest value is 7 so no outliers at lower end
Highest value is 17.6 so at least one outlier at upper end. | A1, A1 | 5 marks

**(iii)** Any suitable answers
Eg 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 | 2 marks

**TOTAL: 8 marks**

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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]{091d6f43-ad01-4849-9f3c-3e58349aa169-2_204_1422_484_363}\\
(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 2010 Q1 [8]}}