OCR MEI S1 2007 January — Question 1 7 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2007
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCompare multiple measures numerically
DifficultyModerate -0.8 This is a straightforward descriptive statistics question requiring only basic calculations (mean, median, midrange) and standard commentary about outliers affecting the mean. No problem-solving or novel insight needed—purely routine application of definitions with minimal computational complexity.
Spec2.02f Measures of average and spread

1 The total annual emissions of carbon dioxide, \(x\) tonnes per person, for 13 European countries are given below. $$\begin{array} { c c c c c c c c c c c c c } 6.2 & 6.7 & 6.8 & 8.1 & 8.1 & 8.5 & 8.6 & 9.0 & 9.9 & 10.1 & 11.0 & 11.8 & 22.8 \end{array}$$
  1. Find the mean, median and midrange of these data.
  2. Comment on how useful each of these is as a measure of central tendency for these data, giving a brief reason for each of your answers.

Question 1:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
Mean \(= 127.6/13 = 9.8\)M1, A1 CAO M1 for \(127.6/13\) soi
Median \(= 8.6\)B1 CAO
Midrange \(= 14.5\)B1 CAO
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
Mean slightly inflated due to the outlierB1
Median good since it is not affected by the outlierB1
Midrange poor as it is highly inflated due to the outlierB1
# Question 1:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 127.6/13 = 9.8$ | M1, A1 CAO | M1 for $127.6/13$ soi |
| Median $= 8.6$ | B1 CAO | |
| Midrange $= 14.5$ | B1 CAO | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean slightly inflated due to the outlier | B1 | |
| Median good since it is not affected by the outlier | B1 | |
| Midrange poor as it is highly inflated due to the outlier | B1 | |

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1 The total annual emissions of carbon dioxide, $x$ tonnes per person, for 13 European countries are given below.

$$\begin{array} { c c c c c c c c c c c c c } 
6.2 & 6.7 & 6.8 & 8.1 & 8.1 & 8.5 & 8.6 & 9.0 & 9.9 & 10.1 & 11.0 & 11.8 & 22.8
\end{array}$$

(i) Find the mean, median and midrange of these data.\\
(ii) Comment on how useful each of these is as a measure of central tendency for these data, giving a brief reason for each of your answers.

\hfill \mbox{\textit{OCR MEI S1 2007 Q1 [7]}}