| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate mean and standard deviation from histogram |
| Difficulty | Moderate -0.3 This is a standard S1 histogram question covering routine techniques: reading frequency density, calculating mean/SD from grouped data, identifying outliers using the 2SD rule, and drawing cumulative frequency. While multi-part with several marks, each component is textbook-standard with no novel problem-solving required. Slightly easier than average due to being purely procedural. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Answer | Marks | Guidance |
|---|---|---|
| \(10 \times 2 = 20\) | M1 for \(10 \times 2\); A1 CAO | [2 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = \frac{10\times65 + 35\times75 + 55\times85 + 20\times95}{120} = \frac{9850}{120} = 82.08\) | M1 for midpoints; M1 for double pairs; A1 CAO | It is an estimate because the data are grouped |
| Answer | Marks | 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 for valid attempt at \(S_{xx}\) | |
| \(s = \sqrt{\frac{8479.17}{119}} = 8.44\) | A1 CAO | [3 marks] |
| Answer | Marks | 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 \(\bar{x} + 2s\); A1 for both | |
| So there are probably some outliers | E1 dep on A1 | [4 marks] |
| Answer | Marks |
|---|---|
| Negative | [1 mark] |
| Answer | Marks | Guidance |
|---|---|---|
| Upper bound | 60 | 70 |
| Cumulative frequency | 0 | 10 |
| 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 |
# Question 3:
## Part (i)
$10 \times 2 = 20$ | M1 for $10 \times 2$; A1 CAO | **[2 marks]**
## Part (ii)
$\text{Mean} = \frac{10\times65 + 35\times75 + 55\times85 + 20\times95}{120} = \frac{9850}{120} = 82.08$ | M1 for midpoints; M1 for double pairs; A1 CAO | It is an estimate because the data are grouped | E1 indep | **[4 marks]**
## Part (iii)
$10\times65^2 + 35\times75^2 + 55\times85^2 + 20\times95^2 = 817000$ | M1 for $\Sigma fx^2$ |
$S_{xx} = 817000 - \frac{9850^2}{120} = 8479.17$ | M1 for valid attempt at $S_{xx}$ |
$s = \sqrt{\frac{8479.17}{119}} = 8.44$ | A1 CAO | **[3 marks]**
## Part (iv)
$\bar{x} - 2s = 82.08 - 2\times8.44 = 65.2$ | M1 FT for $\bar{x} - 2s$ |
$\bar{x} + 2s = 82.08 + 2\times8.44 = 98.96$ | M1 FT for $\bar{x} + 2s$; A1 for both |
So there are probably some outliers | E1 dep on A1 | **[4 marks]**
## Part (v)
Negative | **[1 mark]**
## Part (vi)
| Upper bound | 60 | 70 | 80 | 90 | 100 |
|---|---|---|---|---|---|
| Cumulative frequency | 0 | 10 | 45 | 100 | 120 |
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 marks]**
---3 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]{56f1bd5c-4b45-4e36-a324-e7e0edbb5bdd-2_825_1634_467_295}\\
(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 Q3 [19]}}