| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Find median and quartiles from stem-and-leaf diagram |
| Difficulty | Easy -1.8 This is a straightforward stem-and-leaf diagram question requiring only basic data reading skills and standard quartile calculations. The data are already ordered, making median/quartile identification mechanical (counting positions), and the outlier test is a direct formula application. No problem-solving or conceptual insight needed—pure routine recall. |
| Spec | 2.01a Population and sample: terminology2.02f Measures of average and spread2.02h Recognize outliers |
| 1 | 9 | |||||||||||
| 2 | 6 | |||||||||||
| 3 | 8 | 9 | ||||||||||
| 4 | 0 | 1 | 2 | 2 | 3 | 5 | 6 | 7 | 9 | 9 | ||
| 5 | 1 | 2 | 2 | 2 | 3 | 4 | 5 | 5 | 7 | 8 | 9 | 9 |
| 6 | 0 | 1 | 1 | 3 | 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Population because all the available data are used | E1 | Must include 'population' and a correct justification. Accept 'population as it is data from every single day the phone was used' scores |
| Answer | Marks | Guidance |
|---|---|---|
| Negative skew | B1 | 'Negative' is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_3 = 58\) or \(Q_1 = 42\) identified | B1 | |
| \(IQR = 16\) | B1 | '\(IQR = 16\)' implies B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(42 - 1.5\times16 = 18\) | M1 FT(c) | For calculating \(Q_1 - 1.5\times IQR\) for their values |
| Smallest value is 19 which is not an outlier, so no outliers in lower tail | A1 | Comparison of lower bound with 19 and conclusion e.g. '\(18 < 19\) or \(19 > 18\)' so no. '18 so no as all values \(> 18\)' is A1. '18 so no etc' is A0 as they need to compare to smallest value |
# Question 6:
## Part (a)
Population because all the available data are used | E1 | Must include 'population' and a correct justification. Accept 'population as it is data from every single day the phone was used' scores
## Part (b)
Negative skew | B1 | 'Negative' is B0
## Part (c)
$Q_3 = 58$ or $Q_1 = 42$ identified | B1 |
$IQR = 16$ | B1 | '$IQR = 16$' implies B1B1
## Part (d)
$42 - 1.5\times16 = 18$ | M1 FT(c) | For calculating $Q_1 - 1.5\times IQR$ for their values
Smallest value is 19 which is not an outlier, so no outliers in lower tail | A1 | Comparison of lower bound with 19 and conclusion e.g. '$18 < 19$ or $19 > 18$' so no. '18 so no as all values $> 18$' is A1. '18 so no etc' is A0 as they need to compare to smallest value
---6 An app on my new smartphone records the number of times in a day I use the phone. The data for each day since I bought the phone are shown in the stem and leaf diagram.
\begin{center}
\begin{tabular}{ l | l l l l l l l l l l l l l l }
1 & 9 & & & & & & & & & & & \\
2 & 6 & & & & & & & & & & & \\
3 & 8 & 9 & & & & & & & & & & \\
4 & 0 & 1 & 2 & 2 & 3 & 5 & 6 & 7 & 9 & 9 & & \\
5 & 1 & 2 & 2 & 2 & 3 & 4 & 5 & 5 & 7 & 8 & 9 & 9 \\
6 & 0 & 1 & 1 & 3 & 9 & & & & & & & \\
\end{tabular}
\end{center}
Key: 3|1 means 31
\begin{enumerate}[label=(\alph*)]
\item Explain whether these data are a sample or a population.
\item Describe the shape of the distribution.
\item Determine the interquartile range.
\item Use your answer to part (c) to determine whether there are any outliers in the lower tail.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2023 Q6 [6]}}