| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Identify outliers using IQR rule |
| Difficulty | Easy -1.2 This is a straightforward S1 question requiring routine application of standard procedures: ordering data, finding quartiles/median/IQR, applying the given outlier rule, drawing a boxplot, and calculating the mean. All steps are mechanical with no problem-solving or conceptual insight required—significantly easier than average A-level questions. |
| 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 |
| 15, | 14, | 16, | 15, | 17, | 16, | 15, | 14, | 15, | 16, |
| 17, | 16, | 15, | 14, | 16, | 17, | 15, | 25, | 18, | 16 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Q_2 = \frac{16+16}{2} = 16\) | M1A1 | Method for median |
| \(Q_1 = 15\) | B1 | |
| \(Q_3 = 16.5\) | B1 | |
| \(\text{IQR} = 1.5\) | B1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.5 \times \text{IQR} = 1.5 \times 1.5 = 2.25\) | M1A1 | |
| \(Q_1 - 1.5 \times \text{IQR} = 12.75 \Rightarrow\) no outliers below \(Q_1\) | A1 | |
| \(Q_3 + 1.5 \times \text{IQR} = 18.75 \Rightarrow 25\) is an outlier | A1 | |
| Boxplot, label scale | M1 | |
| 14, 15, 16, 16.5, 18.75 (18) | A1 | |
| Outlier marked | A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = \frac{322}{20} = 16.1\) | M1A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Almost symmetrical/slight negative skew | B1 | |
| Mean \((16.1) \approx\) Median \((16)\) and \(Q_3 - Q_2\ (0.5) \approx Q_2 - Q_1\ (1.0)\) | B1 | (2 marks) |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Q_2 = \frac{16+16}{2} = 16$ | M1A1 | Method for median |
| $Q_1 = 15$ | B1 | |
| $Q_3 = 16.5$ | B1 | |
| $\text{IQR} = 1.5$ | B1 | **(5 marks)** |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.5 \times \text{IQR} = 1.5 \times 1.5 = 2.25$ | M1A1 | |
| $Q_1 - 1.5 \times \text{IQR} = 12.75 \Rightarrow$ no outliers below $Q_1$ | A1 | |
| $Q_3 + 1.5 \times \text{IQR} = 18.75 \Rightarrow 25$ is an outlier | A1 | |
| Boxplot, label scale | M1 | |
| 14, 15, 16, 16.5, 18.75 (18) | A1 | |
| Outlier marked | A1 | **(7 marks)** |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = \frac{322}{20} = 16.1$ | M1A1 | **(2 marks)** |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Almost symmetrical/slight negative skew | B1 | |
| Mean $(16.1) \approx$ Median $(16)$ and $Q_3 - Q_2\ (0.5) \approx Q_2 - Q_1\ (1.0)$ | B1 | **(2 marks)** |
---4. A restaurant owner is concerned about the amount of time customers have to wait before being served. He collects data on the waiting times, to the nearest minute, of 20 customers. These data are listed below.
\begin{center}
\begin{tabular}{ l l l l l l l l l l }
15, & 14, & 16, & 15, & 17, & 16, & 15, & 14, & 15, & 16, \\
17, & 16, & 15, & 14, & 16, & 17, & 15, & 25, & 18, & 16 \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the median and inter-quartile range of the waiting times.
An outlier is an observation that falls either $1.5 \times$ (inter-quartile range) above the upper quartile or $1.5 \times$ (inter-quartile range) below the lower quartile.
\item Draw a boxplot to represent these data, clearly indicating any outliers.
\item Find the mean of these data.
\item Comment on the skewness of these data. Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2003 Q4 [16]}}