| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Coding to simplify calculation |
| Difficulty | Moderate -0.3 This is a standard S1 coding question requiring routine application of formulas for mean and standard deviation with grouped data. Part (a) involves straightforward calculation of coded values and their sum, part (b) applies standard decoding formulas (given Σfy² already), and part (c) requires basic interpretation of skewed data. While multi-step, it follows textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Value of goods stolen (£) | Number of weeks |
| 0-199 | 31 |
| 200-399 | 6 |
| 400-599 | 3 |
| 600-799 | 4 |
| 800-999 | 5 |
| 1000-1999 | 2 |
| 2000-2999 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y\) values = \(-3, -2, -1, 0, 1, 4, 9\) | M1 | |
| \(\sum fy = (3 \times 31) + (2 \times 6) + \ldots = 86\) | M1 A1 | |
| (b) \(\sum f = 52; \quad \bar{y} = \frac{-86}{52} = -1.6538\) | M1 | |
| \(x = (200 \times -1.6538) + 699.5 = 368.73 = £369\) (nearest £) | M1 A1 | |
| std. dev. of \(y = \sqrt{\frac{424}{52} - (-1.6538)^2} = 2.3278\) | M1 | |
| std. dev. of \(x = 200 \times 2.3278 = 465.56 = £466\) (nearest £) | M1 A1 | |
| (c) e.g. mean is raised by a few very large values, most weeks a lot less is stolen; median is more typical but would suggest that the amount stolen is much less of a problem than it really is | B3 | (12 marks) |
(a) $y$ values = $-3, -2, -1, 0, 1, 4, 9$ | M1 |
$\sum fy = (3 \times 31) + (2 \times 6) + \ldots = 86$ | M1 A1 |
(b) $\sum f = 52; \quad \bar{y} = \frac{-86}{52} = -1.6538$ | M1 |
$x = (200 \times -1.6538) + 699.5 = 368.73 = £369$ (nearest £) | M1 A1 |
std. dev. of $y = \sqrt{\frac{424}{52} - (-1.6538)^2} = 2.3278$ | M1 |
std. dev. of $x = 200 \times 2.3278 = 465.56 = £466$ (nearest £) | M1 A1 |
(c) e.g. mean is raised by a few very large values, most weeks a lot less is stolen; median is more typical but would suggest that the amount stolen is much less of a problem than it really is | B3 | (12 marks)
---5. An antiques shop recorded the value of items stolen to the nearest pound during each week for a year giving the data in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Value of goods stolen (£) & Number of weeks \\
\hline
0-199 & 31 \\
\hline
200-399 & 6 \\
\hline
400-599 & 3 \\
\hline
600-799 & 4 \\
\hline
800-999 & 5 \\
\hline
1000-1999 & 2 \\
\hline
2000-2999 & 1 \\
\hline
\end{tabular}
\end{center}
Letting $x$ represent the mid-point of each group and using the coding $y = \frac { x - 699.5 } { 200 }$,
\begin{enumerate}[label=(\alph*)]
\item find $\sum$ fy.
\item estimate to the nearest pound the mean and standard deviation of the value of the goods stolen each week using your value for $\sum f y$ and $\sum f y ^ { 2 } = 424$.\\
(6 marks)\\
The median for these data is $\pounds 82$.
\item Explain why the manager of the shop might be reluctant to use either the mean or the median in summarising these data.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q5 [12]}}