| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Interpret or analyse given back-to-back stem-and-leaf |
| Difficulty | Moderate -0.8 This is a straightforward S1 question requiring basic stem-and-leaf reading skills and standard quartile/outlier calculations. Finding quartiles from ordered data and applying the 1.5×IQR outlier rule are routine procedures covered early in statistics courses, requiring minimal problem-solving beyond careful counting and arithmetic. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.02i Select/critique data presentation2.02j Clean data: missing data, errors |
| Totals | Greenslax | Penville | Totals | ||||||||||||||||
| (2) | 8 | 7 | 2 | 5 | 5 | 6 | 7 | 8 | 8 | 9 | (7) | ||||||||
| (3) | 9 | 8 | 7 | 3 | 1 | 1 | 1 | 2 | 3 | 4 | 4 | 5 | 6 | 9 | (11) | ||||
| (4) | 4 | 4 | 4 | 0 | 4 | 0 | 1 | 2 | 4 | 7 | (5) | ||||||||
| (5) | 6 | 6 | 5 | 2 | 2 | 5 | 0 | 0 | 5 | 5 | 5 | (5) | |||||||
| (7) | 8 | 6 | 5 | 4 | 2 | 1 | 1 | 6 | 2 | 5 | 6 | 6 | (4) | ||||||
| (8) | 8 | 6 | 6 | 4 | 3 | 1 | 1 | 7 | 0 | 5 | (2) | ||||||||
| (5) | 9 | 8 | 4 | 3 | 2 | 8 | (0) | ||||||||||||
| (1) | 4 | 9 | 9 | (1) | |||||||||||||||
| Greenslax | Penville | |
| Lower quartile, \(Q _ { 1 }\) | \(a\) | 31 |
| Median, \(Q _ { 2 }\) | 64 | 39 |
| Upper quartile, \(Q _ { 3 }\) | \(b\) | 55 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 44\) | B1 | These answers may be in or near the table |
| \(b = 76\) | B1 | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(55 + 1.5(55-31) = 91\) [and \(31 - 1.5(55-31) = -5\)] | M1 | For sight of \(55 + 1.5(55-31)\) or 91 seen (possibly implied by RH whisker of box plot). May be implied by a fully correct box plot |
| Box with whiskers drawn | B1 | 1st B1: box with whiskers (condone missing median) |
| Values 25, 31, 39, 55; RH whisker to end at 75 or 91 | B1 | 2nd B1: accuracy must be within 0.5 of a square; lower quartile at 30 or 32 is OK. Two RH whiskers is B0 |
| Outlier plotted at 99 only | A1 | Allow cross to be vertically displaced. If RH whisker goes to 99, 2nd B0 and A0 even if outlier identified (require horizontal "gap" between RH whisker and outlier) |
| (4) | A fully correct box plot scores 4/4. If not fully correct apply scheme and need evidence for M1. If two box plots are seen ignore the one for Greenslax. If not on graph paper M1 max for (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Greenslax: \([Q_2 - Q_1 = 20,\ Q_3 - Q_2 = 12\) or \((Q_2 - Q_1) > (Q_3 - Q_2)] \Rightarrow\) \(-\)ve (skew) | B1 | 1st B1: Greenslax \(-\)ve skew. We must be able to tell which is which but labels may be implied by their values but not simply from \(Q_3 - Q_2 > Q_2 - Q_1\). If there is just one unlabelled comment assume Penville |
| Penville: \([Q_2 - Q_1 = 8,\ Q_3 - Q_2 = 16\) or \((Q_3 - Q_2) > (Q_2 - Q_1)] \Rightarrow\) \(+\)ve (skew) | B1 | 2nd B1: Penville \(+\)ve skew. Don't insist on seeing "skew" so just \(-\)ve and \(+\)ve will do. Treat "correlation" as ISW |
| Justification that is consistent | ddB1 | 3rd ddB1: dependent on 1st and 2nd B marks being scored. Justification for both based on: quartiles, median relative to quartiles, or "tail". If only values for \(Q_3 - Q_2\) etc are given they should be correct ft for Greenslax and correct for Penville. If values for Greenslax imply \(+\)ve skew then 1st B0 and 3rd B0 |
| (3) | ||
| Total 9 |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 44$ | B1 | These answers may be in or near the table |
| $b = 76$ | B1 | |
| | **(2)** | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $55 + 1.5(55-31) = 91$ [and $31 - 1.5(55-31) = -5$] | M1 | For sight of $55 + 1.5(55-31)$ or 91 seen (possibly implied by RH whisker of box plot). May be implied by a fully correct box plot |
| Box with whiskers drawn | B1 | 1st B1: box with whiskers (condone missing median) |
| Values 25, 31, 39, 55; RH whisker to end at 75 or 91 | B1 | 2nd B1: accuracy must be within 0.5 of a square; lower quartile at 30 or 32 is OK. Two RH whiskers is B0 |
| Outlier plotted at 99 only | A1 | Allow cross to be vertically displaced. If RH whisker goes to 99, 2nd B0 and A0 even if outlier identified (require horizontal "gap" between RH whisker and outlier) |
| | **(4)** | A fully correct box plot scores 4/4. If **not** fully correct apply scheme and need evidence for M1. If two box plots are seen ignore the one for Greenslax. If not on graph paper M1 max for (b) |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Greenslax: $[Q_2 - Q_1 = 20,\ Q_3 - Q_2 = 12$ or $(Q_2 - Q_1) > (Q_3 - Q_2)] \Rightarrow$ $-$ve (skew) | B1 | 1st B1: Greenslax $-$ve skew. We must be able to tell which is which but labels may be implied by their values but not simply from $Q_3 - Q_2 > Q_2 - Q_1$. If there is just one unlabelled comment assume Penville |
| Penville: $[Q_2 - Q_1 = 8,\ Q_3 - Q_2 = 16$ or $(Q_3 - Q_2) > (Q_2 - Q_1)] \Rightarrow$ $+$ve (skew) | B1 | 2nd B1: Penville $+$ve skew. Don't insist on seeing "skew" so just $-$ve and $+$ve will do. Treat "correlation" as ISW |
| Justification that is consistent | ddB1 | 3rd ddB1: dependent on 1st and 2nd B marks being scored. Justification for **both** based on: quartiles, median relative to quartiles, or "tail". If only values for $Q_3 - Q_2$ etc are given they should be correct ft for Greenslax and correct for Penville. If values for Greenslax imply $+$ve skew then 1st B0 and 3rd B0 |
| | **(3)** | |
| | **Total 9** | |\begin{enumerate}
\item A random sample of 35 homeowners was taken from each of the villages Greenslax and Penville and their ages were recorded. The results are summarised in the back-to-back stem and leaf diagram below.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
Totals & \multicolumn{8}{|c|}{Greenslax} & \multicolumn{3}{|c|}{} & \multicolumn{4}{|r|}{Penville} & & \multicolumn{3}{|r|}{Totals} \\
\hline
(2) & & & & & & 8 & 7 & 2 & 5 & 5 & 6 & 7 & 8 & 8 & 9 & & & & (7) \\
\hline
(3) & & & & & 9 & 8 & 7 & 3 & 1 & 1 & 1 & 2 & 3 & 4 & 4 & 5 & 6 & 9 & (11) \\
\hline
(4) & & & & 4 & 4 & 4 & 0 & 4 & 0 & 1 & 2 & 4 & 7 & & & & & & (5) \\
\hline
(5) & & & 6 & 6 & 5 & 2 & 2 & 5 & 0 & 0 & 5 & 5 & 5 & & & & & & (5) \\
\hline
(7) & 8 & 6 & 5 & 4 & 2 & 1 & 1 & 6 & 2 & 5 & 6 & 6 & & & & & & & (4) \\
\hline
(8) & 8 & 6 & 6 & 4 & 3 & 1 & 1 & 7 & 0 & 5 & & & & & & & & & (2) \\
\hline
(5) & & & 9 & 8 & 4 & 3 & 2 & 8 & & & & & & & & & & & (0) \\
\hline
(1) & & & & & & & 4 & 9 & 9 & & & & & & & & & & (1) \\
\hline
\end{tabular}
\end{center}
Key: 7 | 3 | 1 means 37 years for Greenslax and 31 years for Penville\\
Some of the quartiles for these two distributions are given in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& Greenslax & Penville \\
\hline
Lower quartile, $Q _ { 1 }$ & $a$ & 31 \\
\hline
Median, $Q _ { 2 }$ & 64 & 39 \\
\hline
Upper quartile, $Q _ { 3 }$ & $b$ & 55 \\
\hline
\end{tabular}
\end{center}
(a) Find the value of $a$ and the value of $b$.
An outlier is a value that falls either
$$\begin{aligned}
& \text { more than } 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right) \text { above } Q _ { 3 } \\
& \text { or more than } 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right) \text { below } Q _ { 1 }
\end{aligned}$$
(b) On the graph paper opposite draw a box plot to represent the data from Penville. Show clearly any outliers.\\
(c) State the skewness of each distribution. Justify your answers.\\
\includegraphics[max width=\textwidth, alt={}, center]{8270bcae-494c-4248-8229-a72e9e84eab0-03_930_1237_1800_367}\\
\hfill \mbox{\textit{Edexcel S1 2014 Q1 [9]}}