| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Identify appropriate measure with outliers |
| Difficulty | Easy -1.2 This is a straightforward AS-level statistics question requiring basic sampling methodology, calculation of mean and standard deviation (likely with calculator), and application of the standard outlier rule (1.5×IQR). All components are routine recall and standard procedures with no problem-solving or novel insight required. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Allocate numbers 001 to 200 to the trees | B1 | 1.2 |
| Choose 10 (3 digit) random numbers | B1 | 2.4 — e.g. use calculator to get 10 different random numbers |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 27.61\ \text{kg}\) | B1 | 1.1 — BC |
| SD \(= 4.04\ \text{kg}\) (3sf) | B1 | 1.1 — BC |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Upper limit \(= 27.61 + 2 \times 4.04 = 35.69\) | M1 | 1.1 — For mean \(+ 2 \times\) sd OR \(UQ + 1.5\ IQR = 28.3 + 1.5 \times 3.2 = 33.1\) |
| So the value of 38.1 is an outlier | A1 | 1.1 |
| This value should be investigated to check if it is genuine. If so, it should not be removed from the data | B1 | 2.2b — Or e.g. if the value is not representative of the other 199 trees because e.g. this tree is a different type it should be ignored |
| [3] |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Allocate numbers 001 to 200 to the trees | B1 | 1.2 |
| Choose 10 (3 digit) random numbers | B1 | 2.4 — e.g. use calculator to get 10 different random numbers |
| **[2]** | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 27.61\ \text{kg}$ | B1 | 1.1 — BC |
| SD $= 4.04\ \text{kg}$ (3sf) | B1 | 1.1 — BC |
| **[2]** | | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Upper limit $= 27.61 + 2 \times 4.04 = 35.69$ | M1 | 1.1 — For mean $+ 2 \times$ sd OR $UQ + 1.5\ IQR = 28.3 + 1.5 \times 3.2 = 33.1$ |
| So the value of 38.1 is an outlier | A1 | 1.1 |
| This value should be investigated to check if it is genuine. If so, it should not be removed from the data | B1 | 2.2b — Or e.g. if the value is not representative of the other 199 trees because e.g. this tree is a different type it should be ignored |
| **[3]** | | |
---7 A farmer has 200 apple trees. She is investigating the masses of the crops of apples from individual trees. She decides to select a sample of these trees and find the mass of the crop for each tree.
\begin{enumerate}[label=(\alph*)]
\item Explain how she can select a random sample of 10 different trees from the 200 trees.
The masses of the crops from the 10 trees, measured in kg, are recorded as follows.\\
$\begin{array} { l l l l l l l l l l } 23.5 & 27.4 & 26.2 & 29.0 & 25.1 & 27.4 & 26.2 & 28.3 & 38.1 & 24.9 \end{array}$
\item For these data find
\begin{itemize}
\item the mean,
\item the sample standard deviation.
\item Show that there is one outlier at the upper end of the data. How should the farmer decide whether to use this outlier in any further analysis of the data?
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 Q7 [7]}}