| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Find median and quartiles from stem-and-leaf diagram |
| Difficulty | Easy -1.3 This is a routine statistics question testing standard procedures: reading a stem-and-leaf diagram, finding quartiles using position formulas, calculating mean/SD from given sums, and applying linear transformations. All techniques are straightforward recall with no problem-solving or insight required, making it easier than average. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Mark | Totals | ||
| 3 | 699 | 9 | \(( 3 )\) |
| 4 | 012223 | 4 | \(( 6 )\) |
| 4 | 566668 | \(( 5 )\) | |
| 5 | 0233344 | \(( 6 )\) | |
| 5 | 5566779 | \(( 6 )\) | |
| 6 | 00000013444 | \(( 9 )\) | |
| 6 | 5566789 | \(( 6 )\) | |
| 7 | 12333 | \(( 4 )\) |
| Key | (3|6 means 36) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(60\) | B1 | B1 60 only |
| (1 mark) | ||
| (b) \(Q_1 = 46\) | B1 | Award each B1 for correct answer only in this order. |
| \(Q_2 = 56\) | B1 | |
| \(Q_3 = 64\) | B1 | |
| (3 marks) | ||
| (c) mean \(= 55.48\ldots\) or \(\frac{2497}{45}\) | B1 | awrt 55.5 |
| \(\text{sd} = \sqrt{\frac{143369}{45} - \left(\frac{2497}{45}\right)^2}\) | M1 | |
| \(= 10.342\ldots\) (s = 10.459...) | A1 | anything which rounds to 10.3 (or s = 10.5) |
| (3 marks) | ||
| (d) Mean \(<\) median \(<\) mode or \(Q_2 - Q_1 > Q_3 - Q_2\) with or without their numbers or median closer to upper quartile (than lower quartile) or (mean-median)/sd \(<0\); | B1 | B1 any correct comparison of a pair of mean, median and mode using their values. |
| negative skew; | B1 dep | B1 for 'negative skew' or allow (almost) symmetrical dependent upon correct reason. |
| (2 marks) | ||
| (e) mean \(= (55 - 5) \times 0.9\) | M1 | |
| \(= 45\) | A1 | |
| sd \(= 10 \times 0.9\) | M1 | |
| \(= 9\) | A1 | |
| (4 marks) |
**(a)** $60$ | B1 | B1 60 only
| | (1 mark) |
**(b)** $Q_1 = 46$ | B1 | Award each B1 for correct answer only in this order.
| | |
| $Q_2 = 56$ | B1 | |
| | |
| $Q_3 = 64$ | B1 | |
| | (3 marks) |
**(c)** mean $= 55.48\ldots$ or $\frac{2497}{45}$ | B1 | awrt 55.5
| | |
| $\text{sd} = \sqrt{\frac{143369}{45} - \left(\frac{2497}{45}\right)^2}$ | M1 | |
| | |
| $= 10.342\ldots$ (s = 10.459...) | A1 | anything which rounds to 10.3 (or s = 10.5)
| | (3 marks) |
**(d)** Mean $<$ median $<$ mode or $Q_2 - Q_1 > Q_3 - Q_2$ with or without their numbers or median closer to upper quartile (than lower quartile) or (mean-median)/sd $<0$; | B1 | B1 any correct comparison of a pair of mean, median and mode using their values.
| | |
| negative skew; | B1 dep | B1 for 'negative skew' or allow (almost) symmetrical dependent upon correct reason.
| | (2 marks) |
**(e)** mean $= (55 - 5) \times 0.9$ | M1 | |
| | |
| $= 45$ | A1 | |
| | |
| sd $= 10 \times 0.9$ | M1 | |
| | |
| $= 9$ | A1 | |
| | (4 marks) |
**Total: 13 marks**
---\begin{enumerate}
\item The marks, $x$, of 45 students randomly selected from those students who sat a mathematics examination are shown in the stem and leaf diagram below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | l l l }
\multicolumn{1}{l}{Mark} & \multicolumn{1}{l}{Totals} & & \\
\hline
3 & 699 & 9 & $( 3 )$ \\
4 & 012223 & 4 & $( 6 )$ \\
4 & 566668 & $( 5 )$ & \\
5 & 0233344 & $( 6 )$ & \\
5 & 5566779 & $( 6 )$ & \\
6 & 00000013444 & $( 9 )$ & \\
6 & 5566789 & $( 6 )$ & \\
7 & 12333 & $( 4 )$ & \\
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{ | l | l | }
\hline
Key & (3|6 means 36) \\
\hline
\end{tabular}
\end{center}
(a) Write down the modal mark of these students.\\
(b) Find the values of the lower quartile, the median and the upper quartile.
For these students $\sum x = 2497$ and $\sum x ^ { 2 } = 143369$\\
(c) Find the mean and the standard deviation of the marks of these students.\\
(d) Describe the skewness of the marks of these students, giving a reason for your answer.
The mean and standard deviation of the marks of all the students who sat the examination were 55 and 10 respectively. The examiners decided that the total mark of each student should be scaled by subtracting 5 marks and then reducing the mark by a further $10 \%$.\\
(e) Find the mean and standard deviation of the scaled marks of all the students.
\hfill \mbox{\textit{Edexcel S1 2012 Q4 [13]}}