| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Outliers from box plot or summary statistics |
| Difficulty | Easy -1.2 This is a straightforward S1 question requiring routine application of the outlier formula (1.5×IQR), drawing a standard box plot, and making basic interpretations from given summary statistics. All steps are mechanical with no problem-solving or novel insight required. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Amount received in sales (£1000s) | |
| Two lowest values | 3,4 |
| Lower quartile | 7 |
| Median | 12 |
| Upper quartile | 14 |
| Two highest values | 20,25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(14 + 1.5 \times (14-7) = 24.5\) | M1 | For at least one correct calculation seen |
| \(7 - 1.5 \times (14-7) = -3.5\) | A1 | For 24.5 and \(-3.5\) seen; may be read off graph |
| Box plot with upper and lower whiskers, at least 2 correct values | M1 | Condone no median marked |
| Values 3, 7, 12, 14 and 20 or 24.5 in appropriate places | A1ft | Apply ft for whiskers compatible with outlier limits; apply \(\pm 0.5\) square accuracy |
| Outlier at 25 marked | B1 | For only one outlier appropriately marked at 25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Since \(Q_3 - Q_2 < Q_2 - Q_1\) | B1 | Statement or equivalent in words; use of inequality does not require differences to be calculated |
| Negatively skew | dB1 | Dependent on suitable reason above; "correlation" is B0; "positive skew" with whisker argument can score B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Not true | B1 | For rejecting the company's claim |
| Since the lower quartile is 7000 and therefore 75% above 7000 not 10000, or 10 is inside the box | dB1 | Appropriate supporting reason; dependent on rejecting company's claim |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $14 + 1.5 \times (14-7) = 24.5$ | M1 | For at least one correct calculation seen |
| $7 - 1.5 \times (14-7) = -3.5$ | A1 | For 24.5 and $-3.5$ seen; may be read off graph |
| Box plot with upper and lower whiskers, at least 2 correct values | M1 | Condone no median marked |
| Values 3, 7, 12, 14 and 20 or 24.5 in appropriate places | A1ft | Apply ft for whiskers compatible with outlier limits; apply $\pm 0.5$ square accuracy |
| Outlier at 25 marked | B1 | For only one outlier appropriately marked at 25 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Since $Q_3 - Q_2 < Q_2 - Q_1$ | B1 | Statement or equivalent in words; use of inequality does not require differences to be calculated |
| Negatively skew | dB1 | Dependent on suitable reason above; "correlation" is B0; "positive skew" with whisker argument can score B1B1 |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Not true | B1 | For rejecting the company's claim |
| Since the lower quartile is 7000 and therefore 75% above 7000 not 10000, or 10 is inside the box | dB1 | Appropriate supporting reason; dependent on rejecting company's claim |3. Over a long period of time a small company recorded the amount it received in sales per month. The results are summarised below.
\begin{center}
\begin{tabular}{ | c | c | }
\hline
& Amount received in sales (£1000s) \\
\hline
Two lowest values & 3,4 \\
\hline
Lower quartile & 7 \\
\hline
Median & 12 \\
\hline
Upper quartile & 14 \\
\hline
Two highest values & 20,25 \\
\hline
\end{tabular}
\end{center}
An outlier is an observation that falls\\
either $1.5 \times$ interquartile range above the upper quartile or $1.5 \times$ interquartile range below the lower quartile.
\begin{enumerate}[label=(\alph*)]
\item On the graph paper below, draw a box plot to represent these data, indicating clearly any outliers.\\
(5)\\
\includegraphics[max width=\textwidth, alt={}, center]{c78ec7b6-dd06-4de1-94c2-052a5577dd10-05_933_1226_1283_367}
\item State the skewness of the distribution of the amount of sales received. Justify your answer.
\item The company claims that for $75 \%$ of the months, the amount received per month is greater than $\pounds 10000$. Comment on this claim, giving a reason for your answer.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2011 Q3 [9]}}