| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Convert between different coding systems |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard coding transformations and correlation. Part (a) requires solving two simultaneous equations from the coding formula (routine algebra), part (b) is direct substitution into the correlation formula, part (c) tests understanding that linear coding doesn't affect correlation, and part (d) requires basic interpretation. All techniques are standard S1 material with no novel problem-solving required. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.08a Pearson correlation: calculate pmcc5.08b Linear coding: effect on pmcc |
| \(p\) | 1840 | 1848 | 1830 | 1824 | 1819 | 1834 | 1850 |
| \(q\) | 4.0 | 4.8 | 3.0 | 2.4 | 1.9 | 3.4 | 5.0 |
| House | Price \(( \pounds )\) | Size \(\left( \mathrm { m } ^ { 2 } \right)\) |
| \(H\) | 156400 | 85 |
| \(J\) | 172900 | 95 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1840-a}{b}=4.0\) and \(\frac{1848-a}{b}=4.8\) | M1 | Setting up two suitable equations which could lead to \(a\) and \(b\) |
| \(a=\mathbf{1800}\), \(b=\mathbf{10}\) | A1 | \(a=10\) and \(b=1800\) is A0; correct answer only is 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r=\frac{-2.17}{\sqrt{1.02\times8.22}}=-0.749417343...\) awrt \(-\mathbf{0.749}\) | M1A1 | M1 for correct expression (condone missing \(-\)); A1 for awrt \(-0.749\); \((-0.75\) or awrt \(0.749\) with no working scores M1 A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-0.749\) | B1ft | ft for \(-0.749\) or ft their answer to (b) to at least 2sf; must be in range \(-1<'(b)'<1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| House H: \(156400/85=[£1840/\text{m}^2\) or \(q=4]\) | M1 | M1 for calculating price/square metre for both \(H\) and \(J\); can be implied by sight of 1840 and 1820 |
| House J: \(172900/95=[£1820/\text{m}^2\) or \(q=2]\) | Allow "\(H\) is £20/square metre more than \(J\)" | |
| Since \((r=-0.749)\) there is negative correlation, or: the higher the price per square metre, the lower the distance from the train station | dM1 | Dependent on 1st M1; statement that correlation is negative or contextualised interpretation |
| Therefore House H is likely to be closer | A1 | Dependent on both Ms; no ft if \(r>0\) |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1840-a}{b}=4.0$ and $\frac{1848-a}{b}=4.8$ | M1 | Setting up two suitable equations which could lead to $a$ and $b$ |
| $a=\mathbf{1800}$, $b=\mathbf{10}$ | A1 | $a=10$ and $b=1800$ is A0; correct answer only is 2/2 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r=\frac{-2.17}{\sqrt{1.02\times8.22}}=-0.749417343...$ awrt $-\mathbf{0.749}$ | M1A1 | M1 for correct expression (condone missing $-$); A1 for awrt $-0.749$; $(-0.75$ or awrt $0.749$ with no working scores M1 A0) |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-0.749$ | B1ft | ft for $-0.749$ or ft their answer to (b) to at least 2sf; must be in range $-1<'(b)'<1$ |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| House H: $156400/85=[£1840/\text{m}^2$ or $q=4]$ | M1 | M1 for calculating price/square metre for both $H$ and $J$; can be implied by sight of 1840 and 1820 |
| House J: $172900/95=[£1820/\text{m}^2$ or $q=2]$ | | Allow "$H$ is £20/square metre more than $J$" |
| Since $(r=-0.749)$ there is negative correlation, or: the higher the price per square metre, the lower the distance from the train station | dM1 | Dependent on 1st M1; statement that correlation is negative or contextualised interpretation |
| Therefore House H is likely to be closer | A1 | Dependent on both Ms; no ft if $r>0$ |
---2. An estate agent recorded the price per square metre, $p \pounds / \mathrm { m } ^ { 2 }$, for 7 two-bedroom houses. He then coded the data using the coding $q = \frac { p - a } { b }$, where $a$ and $b$ are positive constants. His results are shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$p$ & 1840 & 1848 & 1830 & 1824 & 1819 & 1834 & 1850 \\
\hline
$q$ & 4.0 & 4.8 & 3.0 & 2.4 & 1.9 & 3.4 & 5.0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$
The estate agent also recorded the distance, $d \mathrm {~km}$, of each house from the nearest train station. The results are summarised below.
$$\mathrm { S } _ { d d } = 1.02 \quad \mathrm {~S} _ { q q } = 8.22 \quad \mathrm {~S} _ { d q } = - 2.17$$
\item Calculate the product moment correlation coefficient between $d$ and $q$
\item Write down the value of the product moment correlation coefficient between $d$ and $p$
The estate agent records the price and size of 2 additional two-bedroom houses, $H$ and $J$.
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
House & Price $( \pounds )$ & Size $\left( \mathrm { m } ^ { 2 } \right)$ \\
\hline
$H$ & 156400 & 85 \\
\hline
$J$ & 172900 & 95 \\
\hline
\end{tabular}
\end{center}
\item Suggest which house is most likely to be closer to a train station. Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2015 Q2 [8]}}