| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | October |
| 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 statistics question testing basic data representation skills. Parts (a) and (b) require simple reading from a box plot (max - min, Q3 - Q1). Parts (c) and (d) involve standard linear interpolation from grouped data using cumulative frequencies. Part (e) applies the 1.5×IQR outlier rule mechanically. Part (f) requires basic comparison of two distributions. All techniques are routine textbook exercises with no problem-solving insight required, making this easier than average for A-level. |
| Spec | 2.02f Measures of average and spread2.02h Recognize outliers |
| Error ( \(\boldsymbol { e }\) cm) | Number of apprentices |
| \(- 40 < e \leqslant - 16\) | 2 |
| \(- 16 < e \leqslant - 8\) | 18 |
| \(- 8 < e \leqslant 0\) | 33 |
| \(0 < e \leqslant 8\) | 14 |
| \(8 < e \leqslant 16\) | 10 |
| \(16 < e \leqslant 40\) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | [Range] = 63 | B1 |
| (b) | [IQR] = 18 | B1 |
| (c) | \([Q_2 = ] (-8) + \frac{20}{33} \times 8\) or \((0) - \frac{13}{33} \times 8\) [NB \((n+1)\) will have 20.5 or 12.5] | M1 |
| \(= -3.1515...\) awrt \(-3.15\) | A1 | A1 for awrt \(-3.15\) (allow use of \(n+1\) leading to \(-3.03\)) Accept \(-\frac{104}{33}\) if box plot is OK or 3sf value is quoted in (f) |
| (d) | \([Q_3 = ]\) mid-point of [0, 8] group so therefore = 4 | B1cso |
| (e)(i) | IQR = \(4 - (-8) = 12\) so upper limit is \(4 + 1.5 \times 12 = 22\) lower limit is \(-8 - 1.5 \times 12 = -26\) So the outliers are 23 and 28 | M1, A1, A1, M1 |
| (e)(ii) | [Box plot with correct features] | A1, A1 |
| (f) | Interquartile range is smaller (12 compared to 18) or range is smaller (53 v 63) Median is closer to zero (\(-3.15\) is closer than 5) So they have improved | B1, B1, dB1 |
| **(a)** | [Range] = 63 | B1 | |
|---|---|---|---|
| **(b)** | [IQR] = 18 | B1 | |
| **(c)** | $[Q_2 = ] (-8) + \frac{20}{33} \times 8$ or $(0) - \frac{13}{33} \times 8$ [NB $(n+1)$ will have 20.5 or 12.5] | M1 | M1 for a correct fraction and $\times 8$ (ignore end point) |
| | $= -3.1515...$ awrt $-3.15$ | A1 | A1 for awrt $-3.15$ (allow use of $n+1$ leading to $-3.03$) Accept $-\frac{104}{33}$ if box plot is OK or 3sf value is quoted in (f) |
| **(d)** | $[Q_3 = ]$ mid-point of [0, 8] group so therefore = 4 | B1cso | B1cso for a clear argument with no incorrect working seen. Allow $4.14...$ from $7.25$ for $(n+1)$ case |
| **(e)(i)** | IQR = $4 - (-8) = 12$ so upper limit is $4 + 1.5 \times 12 = 22$ lower limit is $-8 - 1.5 \times 12 = -26$ So the outliers are **23 and 28** | M1, A1, A1, M1 | M1 for at least one correct calculation e.g. $4 + 1.5(8 - (-4))$ (implied by one correct limit) 1st A1 for one correct limit 2nd A1 for both correct limits and the two correct outliers identified |
| **(e)(ii)** | [Box plot with correct features] | A1, A1 | M1 for a box with 2 whiskers (one at each end) 1st A1 for $-8$ and $4$ and ft $Q_2$ between them and lower whisker ending at $-25$ no outliers 2nd A1 for upper whisker ending at 18 or 22 and 2 outliers marked at 23 and 28 |
| **(f)** | Interquartile range is smaller (12 compared to 18) or range is smaller (53 v 63) Median is closer to zero ($-3.15$ is closer than 5) So they **have improved** | B1, B1, dB1 | 1st B1 for a statement about range or IQR saying that 2nd estimates are better Allow range or IQR "has decreased" or "is smaller" o.e. 2nd B1 for a statement about medians saying that 2nd one is closer to zero Don't allow "decreased" or "smaller" unless clearly using |median| or say e.g. $3.15 < 5$ 3rd dB1 dep on at least one other B1 for concluding that they have improved based on change in median or range/IQR. Must clearly state "improved" not just "yes" |
**[Total 14]**
---\begin{enumerate}
\item At the start of a course, an instructor asked a group of 80 apprentices to estimate the length of a piece of pipe. The error (true length - estimated length) was recorded in centimetres. The results are summarised in the box plot below.\\
\includegraphics[max width=\textwidth, alt={}, center]{77ae01cd-2b58-48ab-889f-272e27ecf99d-02_291_1445_397_246}\\
(a) Find the range for these data.\\
(b) Find the interquartile range for these data.
\end{enumerate}
One month later, the instructor asked the 80 apprentices to estimate the length of a different piece of pipe and recorded their errors. The results are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Error ( $\boldsymbol { e }$ cm) & Number of apprentices \\
\hline
$- 40 < e \leqslant - 16$ & 2 \\
\hline
$- 16 < e \leqslant - 8$ & 18 \\
\hline
$- 8 < e \leqslant 0$ & 33 \\
\hline
$0 < e \leqslant 8$ & 14 \\
\hline
$8 < e \leqslant 16$ & 10 \\
\hline
$16 < e \leqslant 40$ & 3 \\
\hline
\end{tabular}
\end{center}
(c) Use linear interpolation to estimate the median error for these data.\\
(d) Show that the upper quartile for these data, to the nearest centimetre, is 4 .
For these data, the lower quartile is - 8 and the five worst errors were $- 25 , - 21,18,23,28$ An outlier is a value 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 only 2 outliers for these data.\\
(ii) Draw a box plot for these data on the grid on page 3.\\
(f) State, giving reasons, whether or not the apprentices' ability to estimate the length of a piece of pipe has improved over the first month of the course.
\includegraphics[max width=\textwidth, alt={}, center]{77ae01cd-2b58-48ab-889f-272e27ecf99d-03_412_1520_2222_173}
\hfill \mbox{\textit{Edexcel S1 2017 Q1 [14]}}