| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Standard combined mean and SD |
| Difficulty | Easy -1.2 This is a straightforward statistics question requiring basic interpretation of summary statistics (comparing means and SDs), calculating whether a value is an outlier using the standard 2SD rule, and providing contextual explanations. All parts are routine AS-level tasks with no complex calculations or novel problem-solving required. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Africa | Europe | Oceania | |
| \(n\) | 55 | 49 | 21 |
| mean | 29.3 | 10.0 | 17.8 |
| sd | 8.43 | 1.94 | 4.50 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(29.3 > 10\) o.e. so birth rates on average higher in Africa than in Europe | B1 | 2.4 - Must refer to means e.g. 'averages' is B0 so 'mean of Africa > mean of Europe so birth rates (on average) higher in Africa'. Comparison can be indirect/implicit. Accept reverse statements with \(10 < 29.3\) |
| \(8.43 > 1.94\) so birth rates more variable (larger dispersion) in Africa than in Europe. Condone 'more distributed' or 'more spread out' BOD for 'more variable' | B1 | 2.4 - Must refer to SDs. Accept reverse statements with \(1.94 < 8.43\). In both parts, must refer to specific statistical data given - means/SDs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(29.3 - 2 \times 8.43 (= 12.44)\) | M1 | 1.1 - Calculation of \(\mu - 2\sigma\) only. Ignore attempts at upper bound. '12.44' not needed for this mark |
| \(9.86 < 12.44\) so 9.86 is outlier o.e. | A1 | 2.3 - May see e.g. so 9.86 is more than 2 standard deviations away from the mean. Must give a concluding statement and must also compare 9.86 with correct value. Comparison can be indirect e.g. 9.86 outside acceptable range etc. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| It should not be discarded as it's genuine data | B1 | 2.4 - Must have 'No/should not' and reason. 'No as Mauritius does has a very low birth rate' is B1. 'The data for Mauritius is accurate data' o.e. is B1. 'No as there are some countries in the LDS with very low birth rates' is B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Not all of the data was available | B1 | 2.2a - 'No data for the other countries' – BOD B1. Any reference to countries being small or 'data may not be available' is B0 - needs to convey the fact that it wasn't in the LDS |
## Question 10:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $29.3 > 10$ o.e. so **birth rates on average higher** in Africa than in Europe | B1 | 2.4 - Must refer to means e.g. 'averages' is B0 so 'mean of Africa > mean of Europe so birth rates (on average) higher in Africa'. Comparison can be indirect/implicit. Accept reverse statements with $10 < 29.3$ |
| $8.43 > 1.94$ so **birth rates more variable** (larger dispersion) in Africa than in Europe. Condone 'more distributed' or 'more spread out' BOD for 'more variable' | B1 | 2.4 - Must refer to SDs. Accept reverse statements with $1.94 < 8.43$. In both parts, must refer to specific statistical data given - means/SDs |
**Total: [2]**
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $29.3 - 2 \times 8.43 (= 12.44)$ | M1 | 1.1 - Calculation of $\mu - 2\sigma$ only. Ignore attempts at upper bound. '12.44' not needed for this mark |
| $9.86 < 12.44$ so **9.86 is outlier** o.e. | A1 | 2.3 - May see e.g. so 9.86 is more than 2 standard deviations away from the mean. Must give a **concluding statement** and must also **compare 9.86 with correct value**. Comparison can be indirect e.g. 9.86 outside acceptable range etc. |
**Total: [2]**
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| It should **not** be discarded as it's genuine data | B1 | 2.4 - Must have 'No/should not' **and** reason. 'No as Mauritius does has a very low birth rate' is B1. 'The data for Mauritius is accurate data' o.e. is B1. 'No as there are some countries in the LDS with very low birth rates' is B1 |
**Total: [1]**
### Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not all of the data was available | B1 | 2.2a - 'No data for the other countries' – BOD B1. Any reference to countries being small or 'data may not be available' is B0 - needs to convey the fact that it wasn't in the LDS |
**Total: [1]**
---10 The pre-release material contains information about the birth rate per 1000 people in different countries of the world. These countries have been classified into different regions.
The table shows some data for three of these regions: the mean and standard deviation (sd) of the birth rate per 1000, and the number of countries for which data was used, n.
\section*{Birth rate per 1000 by region}
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Africa & Europe & Oceania \\
\hline
$n$ & 55 & 49 & 21 \\
\hline
mean & 29.3 & 10.0 & 17.8 \\
\hline
sd & 8.43 & 1.94 & 4.50 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the information in the table to compare and contrast the birth rate per 1000 in Africa with the birth rate per 1000 in Europe.
\item The birth rate per 1000 in Mauritius, which is in Africa, is recorded as 9.86. Use the information in the table to show that this value is an outlier.
\item Use your knowledge of the pre-release material to explain whether the value for Mauritius should be discarded.
\item The pre-release material identifies 27 countries in Oceania. Suggest a reason why only 21 values were used to calculate the mean and standard deviation.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2024 Q10 [6]}}