OCR MEI S1 2010 January — Question 7 19 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2010
SessionJanuary
Marks19
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeEstimate mean and standard deviation from histogram
DifficultyModerate -0.8 This is a standard S1 histogram question testing routine skills: reading frequency density, calculating mean/SD from grouped data, identifying outliers using the 2SD rule, and drawing cumulative frequency. All parts follow textbook procedures with no problem-solving or novel insight required, making it easier than average but not trivial due to the multi-step calculations involved.
Spec2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers
A pear grower collects a random sample of 120 pears from his orchard. The histogram below shows the lengths, in mm, of these pears. \includegraphics{figure_7}
  1. Calculate the number of pears which are between 90 and 100 mm long. [2]
  2. Calculate an estimate of the mean length of the pears. Explain why your answer is only an estimate. [4]
  3. Calculate an estimate of the standard deviation. [3]
  4. Use your answers to parts (ii) and (iii) to investigate whether there are any outliers. [4]
  5. Name the type of skewness of the distribution. [1]
  6. Illustrate the data using a cumulative frequency diagram. [5]
(i)
AnswerMarks Guidance
\(10 \times 2 = 20\)M1 for \(10 \times 2\), A1 CAO [2]
(ii)
Mean \(= \frac{10 \times 65 + 35 \times 75 + 55 \times 85 + 20 \times 95}{120} = \frac{9850}{120} = 82.08\)
AnswerMarks Guidance
It is an estimate because the data are grouped.M1 for midpoints, M1 for double pairs, A1 CAO, E1 indep [4]
(iii)
\(10 \times 65^2 + 35 \times 75^2 + 55 \times 85^2 + 20 \times 95^2 (= 817000)\)
\(S_{xx} = 817000 - \frac{9850^2}{120} (= 8479.17)\)
AnswerMarks Guidance
\(s = \sqrt{\frac{8479.17}{119}} = 8.44\)M1 for \(\sum fx^2\), M1 for valid attempt at \(S_{xx}\), A1 CAO [3]
(iv)
\(\bar{x} - 2s = 82.08 - 2 \times 8.44 = 65.2\)
\(\bar{x} + 2s = 82.08 + 2 \times 8.44 = 98.96\)
AnswerMarks Guidance
So there are probably some outliers.M1 FT for \(\bar{x} - 2s\), M1 FT for \(\bar{x} + 2s\), A1 for both, E1 dep on A1 [4]
(v)
AnswerMarks Guidance
Negative.E1 [1]
(vi)
AnswerMarks Guidance
Upper bound60 70
Cumulative frequency0 10
[Cumulative frequency curve shown with appropriate scales, labels 'Length and CF', points plotted, and joining points]
AnswerMarks Guidance
C1 for cumulative frequencies, S1 for scales, L1 for labels 'Length and CF', P1 for points, J1 for joining points (dep on P1), All dep on attempt at cumulative frequency[5]
TOTAL [19] 
## (i)
$10 \times 2 = 20$ | M1 for $10 \times 2$, A1 CAO | [2]

## (ii)
Mean $= \frac{10 \times 65 + 35 \times 75 + 55 \times 85 + 20 \times 95}{120} = \frac{9850}{120} = 82.08$

It is an estimate because the data are grouped. | M1 for midpoints, M1 for double pairs, A1 CAO, E1 indep | [4]

## (iii)
$10 \times 65^2 + 35 \times 75^2 + 55 \times 85^2 + 20 \times 95^2 (= 817000)$

$S_{xx} = 817000 - \frac{9850^2}{120} (= 8479.17)$

$s = \sqrt{\frac{8479.17}{119}} = 8.44$ | M1 for $\sum fx^2$, M1 for valid attempt at $S_{xx}$, A1 CAO | [3]

## (iv)
$\bar{x} - 2s = 82.08 - 2 \times 8.44 = 65.2$
$\bar{x} + 2s = 82.08 + 2 \times 8.44 = 98.96$

So there are probably some outliers. | M1 FT for $\bar{x} - 2s$, M1 FT for $\bar{x} + 2s$, A1 for both, E1 dep on A1 | [4]

## (v)
Negative. | E1 | [1]

## (vi)
| Upper bound | 60 | 70 | 80 | 90 | 100 |
|---|---|---|---|---|---|
| Cumulative frequency | 0 | 10 | 45 | 100 | 120 |

[Cumulative frequency curve shown with appropriate scales, labels 'Length and CF', points plotted, and joining points]

| C1 for cumulative frequencies, S1 for scales, L1 for labels 'Length and CF', P1 for points, J1 for joining points (dep on P1), All dep on attempt at cumulative frequency | [5]

**TOTAL** | | [19]

---
A pear grower collects a random sample of 120 pears from his orchard. The histogram below shows the lengths, in mm, of these pears.

\includegraphics{figure_7}

\begin{enumerate}[label=(\roman*)]
\item Calculate the number of pears which are between 90 and 100 mm long. [2]

\item Calculate an estimate of the mean length of the pears. Explain why your answer is only an estimate. [4]

\item Calculate an estimate of the standard deviation. [3]

\item Use your answers to parts (ii) and (iii) to investigate whether there are any outliers. [4]

\item Name the type of skewness of the distribution. [1]

\item Illustrate the data using a cumulative frequency diagram. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S1 2010 Q7 [19]}}
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