| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| 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 data handling skills: reading a stem-and-leaf diagram, finding median/quartiles using position formulas, applying the standard outlier definition, and drawing a box plot. All steps are routine recall with no problem-solving or conceptual challenge beyond standard textbook exercises. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Age | Key: \(4 \mid 3\) is an age of 43 | |||||||||||||||||||||||
| 4 | 3 | \(( 1 )\) | ||||||||||||||||||||||
| 5 | 4 | |||||||||||||||||||||||
| 6 | 2 | 3 | 5 | 6 | 8 | 8 | 8 | 9 | 9 | \(( 1 )\) | ||||||||||||||
| 7 | 1 | 1 | 3 | 4 | 4 | 6 | 6 | 6 | 8 | 8 | 9 | \(( 9 )\) | ||||||||||||
| 8 | 0 | 0 | 2 | 7 | 8 | 8 | 9 | \(( 11 )\) | ||||||||||||||||
| 9 | 3 | 7 | ||||||||||||||||||||||
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{Median} =]\ \mathbf{74}\) | B1 (1 mark) | For 74 |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_1 = 68\), \(Q_3 = 80\) | M1 | M1 for attempt at both, at least one correct; may appear in calculation e.g. \(80 - A\) (where \(60 < A < 80\)) or \(B - 68\) (where \(68 < B < 90\)) |
| \([\text{IQR} = 80 - 68 =]\ \mathbf{12}\) | A1 (2 marks) | A1 for 12 |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_1 - 1.5 \times (\text{IQR}) = \text{"68"} - 1.5 \times \text{"12"} [= 50]\) | M1 | For correct attempt at at least one limit; can ft their quartiles and IQR |
| Answer | Marks | Guidance |
|---|---|---|
| Outliers are \(< \mathbf{50}\) or \(> \mathbf{98}\) | A1ft | 1st A1ft for correct attempts at both limits with at least one correct limit, or correct ft using their quartiles and IQR; sight of 50 and 98 scores M1A1 |
| So there is just one outlier at \(\mathbf{43}\) | A1 (3 marks) | 2nd A1 for identifying only one outlier at 43 (e.g. may say "43 < 50"); must be stated in (c); just stating outlier is 43 without sight of limits is M0A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Box plot with two whiskers, one at each end | M1 | For drawing a box with only two whiskers one at each end |
| \(Q_1, Q_2, Q_3\) correctly drawn box | A1ft | 1st A1ft for \(Q_1, Q_2\) and \(Q_3\) as correctly drawn box (or ft for \(Q_1 < Q_2 < Q_3\)) |
| Upper whisker ending at 97 and lower whisker ending at 54 or 50 and only one outlier shown at 43; allow \(\pm 0.5\) of a square for accuracy | A1 (3 marks) | NB A fully correct box plot can score full marks in (d) even if other parts are missing or incorrect |
## Question 2:
### Part (a)
$[\text{Median} =]\ \mathbf{74}$ | B1 (1 mark) | For 74
### Part (b)
$Q_1 = 68$, $Q_3 = 80$ | M1 | M1 for attempt at both, at least one correct; may appear in calculation e.g. $80 - A$ (where $60 < A < 80$) or $B - 68$ (where $68 < B < 90$)
$[\text{IQR} = 80 - 68 =]\ \mathbf{12}$ | A1 (2 marks) | A1 for 12
### Part (c)
$Q_1 - 1.5 \times (\text{IQR}) = \text{"68"} - 1.5 \times \text{"12"} [= 50]$ | M1 | For correct attempt at at least one limit; can ft their quartiles and IQR
or $Q_3 + 1.5 \times (\text{IQR}) = \text{"80"} + 1.5 \times \text{"12"} [= 98]$
Outliers are $< \mathbf{50}$ or $> \mathbf{98}$ | A1ft | 1st A1ft for correct attempts at **both** limits with at least one correct limit, or correct ft using their quartiles and IQR; sight of 50 and 98 scores M1A1
So there is just one outlier at $\mathbf{43}$ | A1 (3 marks) | 2nd A1 for identifying only one outlier at 43 (e.g. may say "43 < 50"); must be stated in (c); just stating outlier is 43 without sight of limits is M0A0A0
### Part (d)
Box plot with two whiskers, one at each end | M1 | For drawing a box with only two whiskers one at each end
$Q_1, Q_2, Q_3$ correctly drawn box | A1ft | 1st A1ft for $Q_1, Q_2$ and $Q_3$ as correctly drawn box (or ft for $Q_1 < Q_2 < Q_3$)
Upper whisker ending at 97 **and** lower whisker ending at 54 or 50 **and** only one outlier shown at 43; allow $\pm 0.5$ of a square for accuracy | A1 (3 marks) | **NB** A fully correct box plot can score full marks in (d) even if other parts are missing or incorrect2. The stem and leaf diagram below shows the ages (in years) of the residents in a care home.
\begin{center}
\begin{tabular}{ c | c c c c c c c c c c c c }
\multicolumn{14}{c}{Age} & & & \multicolumn{9}{c}{Key: $4 \mid 3$ is an age of 43} \\
\hline
4 & 3 & & & & & & & & & & & $( 1 )$ & & & & & & & & & & & & \\
5 & 4 & & & & & & & & & & & & & & & & & & & & & & & \\
6 & 2 & 3 & 5 & 6 & 8 & 8 & 8 & 9 & 9 & & & $( 1 )$ & & & & & & & & & & & & \\
7 & 1 & 1 & 3 & 4 & 4 & 6 & 6 & 6 & 8 & 8 & 9 & $( 9 )$ & & & & & & & & & & & & \\
8 & 0 & 0 & 2 & 7 & 8 & 8 & 9 & & & & & $( 11 )$ & & & & & & & & & & & & \\
9 & 3 & 7 & & & & & & & & & & & & & & & & & & & & & & \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the median age of the residents.
\item Find the interquartile range (IQR) of the ages of the residents.
An outlier is defined as a value that is either\\
more than $1.5 \times ( \mathrm { IQR } )$ below the lower quartile or more than $1.5 \times ( \mathrm { IQR } )$ above the upper quartile.
\item Determine any outliers in these data. Show clearly any calculations that you use.
\item On the grid on page 5, draw a box plot to summarise these data.\\
Ages
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2021 Q2 [9]}}