OCR MEI S1 — Question 5 22 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks22
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeEstimate mean and standard deviation from histogram
DifficultyModerate -0.3 This is a standard S1 histogram question requiring routine calculations (reading frequency density, finding mean/SD from grouped data, identifying outliers using 2SD rule, sketching cumulative frequency). While multi-part with several marks, each component uses well-practiced techniques with no novel problem-solving required, making it slightly easier than average.
Spec2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers
5 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[max width=\textwidth, alt={}, center]{99c502aa-2c9f-461d-9dc0-ed55e3df32a2-3_815_1628_505_304}
  1. Calculate the number of pears which are between 90 and 100 mm long.
  2. Calculate an estimate of the mean length of the pears. Explain why your answer is only an estimate.
  3. Calculate an estimate of the standard deviation.
  4. Use your answers to parts (ii) and (iii) to investigate whether there are any outliers.
  5. Name the type of skewness of the distribution.
  6. Illustrate the data using a cumulative frequency diagram.
Question 5:
Part (i)
AnswerMarks Guidance
AnswerMark Guidance
\(10 \times 2 = 20\)M1 For \(10 \times 2\)
20A1 CAO
Part (ii)
AnswerMarks Guidance
AnswerMark Guidance
Mean \(= \frac{10\times65 + 35\times75 + 55\times85 + 20\times95}{120} = \frac{9850}{120} = 82.08\)M1 M1 M1 for midpoints; M1 for double pairs
82.08A1 CAO
It is an estimate because the data are groupedE1 Independent
Part (iii)
AnswerMarks Guidance
AnswerMark Guidance
\(10\times65^2 + 35\times75^2 + 55\times85^2 + 20\times95^2 = 817000\)M1 For \(\Sigma fx^2\)
\(S_{xx} = 817000 - \frac{9850^2}{120} = 8479.17\)M1 Valid attempt at \(S_{xx}\)
\(s = \sqrt{\frac{8479.17}{119}} = 8.44\)A1 CAO
Part (iv)
AnswerMarks Guidance
AnswerMark Guidance
\(\bar{x} - 2s = 82.08 - 2\times8.44 = 65.2\)M1 FT for \(\bar{x} - 2s\)
\(\bar{x} + 2s = 82.08 + 2\times8.44 = 98.96\)M1 FT for \(\bar{x} + 2s\)
Both bounds correctA1 For both
So there are probably some outliersE1 dep on A1
Part (v)
AnswerMarks Guidance
AnswerMark Guidance
NegativeB1
Part (vi)
AnswerMarks Guidance
AnswerMark Guidance
Cumulative frequencies: 0, 10, 45, 100, 120C1 For cumulative frequencies
Correct scales and labels ('Length and CF')S1, L1
Points plotted correctlyP1
Points joined correctlyJ1 dep on P1; all dep on attempt at cumulative frequency 
# Question 5:

## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $10 \times 2 = 20$ | M1 | For $10 \times 2$ |
| 20 | A1 | CAO |

## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= \frac{10\times65 + 35\times75 + 55\times85 + 20\times95}{120} = \frac{9850}{120} = 82.08$ | M1 M1 | M1 for midpoints; M1 for double pairs |
| 82.08 | A1 | CAO |
| It is an estimate because the data are grouped | E1 | Independent |

## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $10\times65^2 + 35\times75^2 + 55\times85^2 + 20\times95^2 = 817000$ | M1 | For $\Sigma fx^2$ |
| $S_{xx} = 817000 - \frac{9850^2}{120} = 8479.17$ | M1 | Valid attempt at $S_{xx}$ |
| $s = \sqrt{\frac{8479.17}{119}} = 8.44$ | A1 | CAO |

## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} - 2s = 82.08 - 2\times8.44 = 65.2$ | M1 | FT for $\bar{x} - 2s$ |
| $\bar{x} + 2s = 82.08 + 2\times8.44 = 98.96$ | M1 | FT for $\bar{x} + 2s$ |
| Both bounds correct | A1 | For both |
| So there are probably some outliers | E1 | dep on A1 |

## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| Negative | B1 | |

## Part (vi)
| Answer | Mark | Guidance |
|--------|------|----------|
| Cumulative frequencies: 0, 10, 45, 100, 120 | C1 | For cumulative frequencies |
| Correct scales and labels ('Length and CF') | S1, L1 | |
| Points plotted correctly | P1 | |
| Points joined correctly | J1 | dep on P1; all dep on attempt at cumulative frequency |

---
5 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[max width=\textwidth, alt={}, center]{99c502aa-2c9f-461d-9dc0-ed55e3df32a2-3_815_1628_505_304}\\
(i) Calculate the number of pears which are between 90 and 100 mm long.\\
(ii) Calculate an estimate of the mean length of the pears. Explain why your answer is only an estimate.\\
(iii) Calculate an estimate of the standard deviation.\\
(iv) Use your answers to parts (ii) and (iii) to investigate whether there are any outliers.\\
(v) Name the type of skewness of the distribution.\\
(vi) Illustrate the data using a cumulative frequency diagram.

\hfill \mbox{\textit{OCR MEI S1  Q5 [22]}}
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