| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate range and interquartile range |
| Difficulty | Easy -1.2 This is a straightforward S1 question testing basic statistical measures (range, IQR, median via linear interpolation) and outlier detection. Parts (a) and (b) require simple reading from a box plot and subtraction. The linear interpolation and quartile calculations are standard textbook procedures with no conceptual challenges. This is easier than average A-level content as it's purely procedural recall with no problem-solving or novel insight required. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.02i Select/critique data presentation2.02j Clean data: missing data, errors |
| Angle, \(\boldsymbol { a }\), (degrees) | Number of students |
| \(55 \leqslant a < 60\) | 6 |
| \(60 \leqslant a < 65\) | 15 |
| \(65 \leqslant a < 70\) | 13 |
| \(70 \leqslant a < 75\) | 11 |
| \(75 \leqslant a < 80\) | 8 |
| \(80 \leqslant a < 85\) | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Range} = 48 - 9 = \mathbf{39}\) | B1 (1 mark) | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{IQR} = 25 - 12 = \mathbf{13}\) | B1 (1 mark) | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Median} = 65 + \dfrac{[9]}{13} \times 5 = \dfrac{890}{13} \approx \mathbf{68.5°}\) | M1 A1 (2 marks) | M1 for attempt (should have 65 or 70, 13 and 5). Allow working down: \(70 - \dfrac{[4]}{13} \times 5\). Condone: \(65 + \dfrac{[9.5]}{13} \times 5 = 68.7\). A1 awrt 68.5 (condone 68.7 if \((n+1)\) used). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Lower Quartile} = 60 + \dfrac{9}{15} \times 5 = \mathbf{63}\) \((*)\) | M1 A1cso (2 marks) | M1 for correct expression for lower quartile (condone 9.25 if \((n+1)\) used). Watch out for working down e.g. \(65 - \dfrac{6}{15} \times 5\) (M1) but \(\dfrac{60+65}{2} = 62.5 = 63\) is M0. A1 for correct solution with no incorrect working seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(63 - 1.5 \times (75 - 63) = 45\) and \(75 + 1.5 \times (75 - 63) = 93\); no data above 93 and no data below 45, so there are no outliers. | M1 A1, A1 (3 marks) | M1 for either correct calculation (may be implied by one correct limit). A1 for either 45 or 93. A1 for both 45 and 93 and conclusion. |
| Answer | Marks | Guidance |
|---|---|---|
| Box plot drawn with whiskers, median \(63 < Q_2 < 75\), quartiles at 63 and 75, whiskers to 55 and 84. | M1 A1ft (2 marks) | M1 for a box with 1 whisker drawn on each side (line must be shown). A1ft their median \(63 < Q_2 < 75\), but quartiles (63 and 75), 55 and 84 must be correct. Use 0.5 sq. accuracy; condone median on 68 or 69 if 68.5 seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Median for the \(70°\) angle is closer to \(70°\) than the \(20°\) median is to \(20°\). | B1 | \(1^{\text{st}}\) B1 for correct comparison of their medians (\(63 < (c) < 75\)) to true value. |
| The range/IQR for the \(70°\) angle box plot is smaller/shorter. | B1 | \(2^{\text{nd}}\) B1 for correct comparison of their range or IQR ("spread" is B0). Allow saying IQRs of 12 and 13 are similar. Ignore mention of "skewness" or "outliers". |
| Therefore, students were more accurate at drawing the \(70°\) angle. | dB1 (3 marks) | \(3^{\text{rd}}\) dB1 dependent upon at least one previous B1 being scored, for choosing \(70°\). |
# Question 1:
## Part (a)
| $\text{Range} = 48 - 9 = \mathbf{39}$ | B1 (1 mark) | — |
## Part (b)
| $\text{IQR} = 25 - 12 = \mathbf{13}$ | B1 (1 mark) | — |
## Part (c)
| $\text{Median} = 65 + \dfrac{[9]}{13} \times 5 = \dfrac{890}{13} \approx \mathbf{68.5°}$ | M1 A1 (2 marks) | M1 for attempt (should have 65 or 70, 13 and 5). Allow working down: $70 - \dfrac{[4]}{13} \times 5$. Condone: $65 + \dfrac{[9.5]}{13} \times 5 = 68.7$. A1 awrt 68.5 (condone 68.7 if $(n+1)$ used). |
## Part (d)
| $\text{Lower Quartile} = 60 + \dfrac{9}{15} \times 5 = \mathbf{63}$ $(*)$ | M1 A1cso (2 marks) | M1 for correct expression for lower quartile (condone 9.25 if $(n+1)$ used). Watch out for working down e.g. $65 - \dfrac{6}{15} \times 5$ (M1) but $\dfrac{60+65}{2} = 62.5 = 63$ is M0. A1 for correct solution with no incorrect working seen. |
## Part (e)(i)
| $63 - 1.5 \times (75 - 63) = 45$ and $75 + 1.5 \times (75 - 63) = 93$; no data above 93 and no data below 45, so there are no outliers. | M1 A1, A1 (3 marks) | M1 for either correct calculation (may be implied by one correct limit). A1 for either 45 or 93. A1 for both 45 and 93 and conclusion. |
## Part (e)(ii)
| Box plot drawn with whiskers, median $63 < Q_2 < 75$, quartiles at 63 and 75, whiskers to 55 and 84. | M1 A1ft (2 marks) | M1 for a box with 1 whisker drawn on each side (line must be shown). A1ft their median $63 < Q_2 < 75$, but quartiles (63 and 75), 55 and 84 must be correct. Use 0.5 sq. accuracy; condone median on 68 or 69 if 68.5 seen. |
## Part (f)
| Median for the $70°$ angle is closer to $70°$ than the $20°$ median is to $20°$. | B1 | $1^{\text{st}}$ B1 for correct comparison of their **medians** ($63 < (c) < 75$) to true value. |
| The range/IQR for the $70°$ angle box plot is smaller/shorter. | B1 | $2^{\text{nd}}$ B1 for correct comparison of their **range** or **IQR** ("spread" is B0). Allow saying IQRs of 12 and 13 are similar. Ignore mention of "skewness" or "outliers". |
| Therefore, students were more accurate at drawing the $70°$ angle. | dB1 (3 marks) | $3^{\text{rd}}$ dB1 dependent upon at least one previous B1 being scored, for choosing $70°$. |
**Total: 14 marks**\begin{enumerate}
\item Each of 60 students was asked to draw a $20 ^ { \circ }$ angle without using a protractor. The size of each angle drawn was measured. The results are summarised in the box plot below.\\
\includegraphics[max width=\textwidth, alt={}, center]{9626e3ce-35d6-41b5-a0bd-1185f38b9e36-02_371_1040_340_461}\\
(a) Find the range for these data.\\
(b) Find the interquartile range for these data.
\end{enumerate}
The students were then asked to draw a $70 ^ { \circ }$ angle.\\
The results are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Angle, $\boldsymbol { a }$, (degrees) & Number of students \\
\hline
$55 \leqslant a < 60$ & 6 \\
\hline
$60 \leqslant a < 65$ & 15 \\
\hline
$65 \leqslant a < 70$ & 13 \\
\hline
$70 \leqslant a < 75$ & 11 \\
\hline
$75 \leqslant a < 80$ & 8 \\
\hline
$80 \leqslant a < 85$ & 7 \\
\hline
\end{tabular}
\end{center}
(c) Use linear interpolation to estimate the size of the median angle drawn. Give your answer to 1 decimal place.\\
(d) Show that the lower quartile is $63 ^ { \circ }$
For these data, the upper quartile is $75 ^ { \circ }$, the minimum is $55 ^ { \circ }$ and the maximum is $84 ^ { \circ }$ An outlier is an observation that falls either more than $1.5 \times$ (interquartile range) above the upper quartile or more than $1.5 \times$ (interquartile range) below the lower quartile.\\
(e) (i) Show that there are no outliers for these data.\\
(ii) Draw a box plot for these data on the grid on page 3.\\
(f) State which angle the students were more accurate at drawing. Give reasons for your answer.\\
(3)
\includegraphics[max width=\textwidth, alt={}, center]{9626e3ce-35d6-41b5-a0bd-1185f38b9e36-03_378_1059_2067_447}\\
\hfill \mbox{\textit{Edexcel S1 2015 Q1 [14]}}