| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 12 |
| 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 -1.3 This is a straightforward S1 question requiring basic statistical skills: reading values from a stem-and-leaf diagram (positions 20, 21 for median; 10, 11 and 30, 31 for quartiles), calculating outlier boundaries using a given formula, and drawing a standard box plot. All steps are routine recall and mechanical calculation with no problem-solving or conceptual challenge. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.02i Select/critique data presentation |
| Angle | ( \(6 \mid 4\) means \(64 ^ { \circ }\) ) | Totals |
| 4 | 1 | (1) |
| 4 | (0) | |
| 5 | 024 | (3) |
| 5 | 589 | (3) |
| 6 | 11334 | (5) |
| 6 | 55789 | (5) |
| 7 | 011233444 | (9) |
| 7 | 5667799 | (7) |
| 8 | 01134 | (5) |
| 8 | 56 | (2) |
| Answer | Marks |
|---|---|
| \(Q_1 = 63°\) | A1 |
| \(Q_2 = \frac{71+72}{2} = 71.5°\) | M1 A1 |
| \(Q_3 = 77°\) | A1 |
| Answer | Marks |
|---|---|
| \(Q_3 - Q_1 = 77 - 63 = 14\) | M1 |
| limits are \(63 - (1.5 \times 14) = 42\) and \(77 + (1.5 \times 14) = 98\) | M1 |
| \(\therefore 41\) is an outlier | A1 |
| Answer | Marks |
|---|---|
| [Boxplot diagram with outlier at 41, whiskers extending to approximately 42 and 98, box from 63 to 77 with median line at 71.5] | B3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\)ve skew. e.g. people know 90° so less likely to draw much larger than 75° | B1 B1 | (12) |
**(a)**
$Q_1 = 63°$ | A1 |
$Q_2 = \frac{71+72}{2} = 71.5°$ | M1 A1 |
$Q_3 = 77°$ | A1 |
**(b)**
$Q_3 - Q_1 = 77 - 63 = 14$ | M1 |
limits are $63 - (1.5 \times 14) = 42$ and $77 + (1.5 \times 14) = 98$ | M1 |
$\therefore 41$ is an outlier | A1 |
**(c)**
[Boxplot diagram with outlier at 41, whiskers extending to approximately 42 and 98, box from 63 to 77 with median line at 71.5] | B3 |
**(d)**
$-$ve skew. e.g. people know 90° so less likely to draw much larger than 75° | B1 B1 | (12)
---5. For a project, a student asked 40 people to draw two straight lines with what they thought was an angle of $75 ^ { \circ }$ between them, using just a ruler and a pencil. She then measured the size of the angles that had been drawn and her data are summarised in this stem and leaf diagram.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Angle & ( $6 \mid 4$ means $64 ^ { \circ }$ ) & Totals \\
\hline
4 & 1 & (1) \\
\hline
4 & & (0) \\
\hline
5 & 024 & (3) \\
\hline
5 & 589 & (3) \\
\hline
6 & 11334 & (5) \\
\hline
6 & 55789 & (5) \\
\hline
7 & 011233444 & (9) \\
\hline
7 & 5667799 & (7) \\
\hline
8 & 01134 & (5) \\
\hline
8 & 56 & (2) \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the median and quartiles of these data.
Given that any values outside of the limits $\mathrm { Q } _ { 1 } - 1.5 \left( \mathrm { Q } _ { 3 } - \mathrm { Q } _ { 1 } \right)$ and $\mathrm { Q } _ { 3 } + 1.5 \left( \mathrm { Q } _ { 3 } - \mathrm { Q } _ { 1 } \right)$ are to be regarded as outliers,
\item determine if there are any outliers in these data,
\item draw a box plot representing these data on graph paper,
\item describe the skewness of the distribution and suggest a reason for it.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q5 [12]}}