| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Compare multiple measures numerically |
| Difficulty | Moderate -0.3 This is a straightforward statistics question requiring calculation of mean from grouped data (using midpoints), finding the median from a frequency table, and explaining why median is preferred when data is skewed. All techniques are standard AS-level procedures with no novel problem-solving required, though part (c) requires some statistical understanding beyond pure calculation. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Number of days absent | 0 | 1 | 2 to 4 | 5 to 10 | 11 to 20 | 21 to 30 | More than 30 |
| Number of students | 7 | 12 | 9 | 1 | 0 | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\Sigma fx}{\Sigma f} \left(=\frac{72}{30}\right)\) | M1 | attempted. May be implied by working seen or eg \(\geq 2\) terms in numerator correct; Dep \(\div \Sigma f\) (NOT \(=7\)); Allow end points or incorrect midpoints |
| \(= 2.4\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(1\) | B1 [1] | Ignore method except \(0,0,1,1,7,9,12\) or \(7,12,9,1,0,1,0\): B0 |
| Answer | Marks | Guidance |
|---|---|---|
| The median is less influenced than the mean by the one student in the \(21\)-\(30\) class | B1 [1] | or equivalent statement. Must clearly refer to the actual outlier in \(21\)-\(30\) class; Not "It's the same as the mode". Ignore "skew". |
## Question 9:
### Part (a):
$\frac{\Sigma fx}{\Sigma f} \left(=\frac{72}{30}\right)$ | **M1** | attempted. May be implied by working seen or eg $\geq 2$ terms in numerator correct; Dep $\div \Sigma f$ (NOT $=7$); Allow end points or incorrect midpoints
$= 2.4$ | **A1** [2] |
### Part (b):
$1$ | **B1** [1] | Ignore method except $0,0,1,1,7,9,12$ or $7,12,9,1,0,1,0$: B0
### Part (c):
The median is less influenced than the mean by the one student in the $21$-$30$ class | **B1** [1] | or equivalent statement. Must clearly refer to the **actual** outlier in $21$-$30$ class; Not "It's the same as the mode". Ignore "skew".
---9 The table shows information about the number of days absent last year by students in class 2A at a certain school.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
Number of days absent & 0 & 1 & 2 to 4 & 5 to 10 & 11 to 20 & 21 to 30 & More than 30 \\
\hline
Number of students & 7 & 12 & 9 & 1 & 0 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate an estimate of the mean for these data.
\item Find the median of these data.
The headteacher is writing a report on the numbers of absences at her school. She wishes to include a figure for the average number of absences in class 2A. A governor suggests that she should quote the mean. The class teacher suggests that she should quote the median, because it is lower than the mean.
\item Give another reason for using the median rather than the mean for the average number of absences in class 2A.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q9 [4]}}