| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Construct stem-and-leaf then find median and quartiles |
| Difficulty | Easy -1.8 This is a straightforward data handling question requiring only basic skills: ordering 12 values, constructing a stem-and-leaf diagram with given stems, finding the median by position, and making a simple comparison between mean and median. No calculation complexity or statistical insight required beyond GCSE level. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread |
| Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
| 8 | 15 | 25 | 29 | 31 | 31 | 34 | 36 | 34 | 26 | 15 | 8 |
2 The mean daily maximum temperatures at a research station over a 12 -month period, measured to the nearest degree Celsius, are given below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Jan & Feb & Mar & Apr & May & Jun & Jul & Aug & Sep & Oct & Nov & Dec \\
\hline
8 & 15 & 25 & 29 & 31 & 31 & 34 & 36 & 34 & 26 & 15 & 8 \\
\hline
\end{tabular}
\end{center}
(i) Construct a sorted stem and leaf diagram to represent these data, taking stem values of $0,10 , \ldots$.\\
(ii) Write down the median of these data.\\
(iii) The mean of these data is 24.3. Would the mean or the median be a better measure of central tendency of the data? Briefly explain your answer.
\hfill \mbox{\textit{OCR MEI S1 Q2 [7]}}