| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Effect of data transformation on correlation |
| Difficulty | Moderate -0.8 This question tests standard results about the effect of linear transformations on mean, standard deviation, and correlation. Part (a) requires applying the formulas for how linear transformations affect summary statistics (routine A-level content), while part (b) tests knowledge that correlation is invariant under linear transformation. These are bookwork results that require recall and direct application rather than problem-solving or derivation. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.08a Pearson correlation: calculate pmcc5.08c Pearson: measure of straight-line fit |
| Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |
| Maximum (u) | 39 | 40 | 48 | 61 | 71 | 81 | 85 | 83 | 77 | 67 | 54 | 41 |
| Minimum (v) | 26 | 27 | 34 | 44 | 53 | 63 | 68 | 66 | 60 | 51 | 41 | 30 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= \frac{39+40+48+61+71+81+85+83+77+67+54+41}{12} = \frac{747}{12} = 62.25 \approx 62.3\) | B1 | Accept 62.2 or 62.3 |
| Standard deviation \(= \sqrt{\frac{\sum u^2}{12} - \bar{u}^2}\) where \(\sum u^2 = 49814\) | M1 | Correct method for s.d. |
| \(s.d. = \sqrt{\frac{49814}{12} - 62.25^2} = \sqrt{4151.17 - 3875.0625} = \sqrt{276.1} \approx 16.6\) | A1 | Accept 16.6 or 16.7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using \(c = \frac{5}{9}(f-32)\), new mean \(= \frac{5}{9}(62.25 - 32) = \frac{5}{9}(30.25) = 16.8\) °C | M1 | Applying linear transformation to mean |
| A1 | Accept 16.8 or 16.9 | |
| New s.d. \(= \frac{5}{9} \times 16.6... = 9.2\) °C | M1 | Multiplying s.d. by \(\frac{5}{9}\) only |
| A1 | Accept 9.2 or 9.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r_{xy} = 0.997\) | B1 | Correct value stated |
| Since \(x\) and \(y\) are linear transformations of \(u\) and \(v\) respectively (of the form \(x = au + b\), \(y = av + b\) with \(a > 0\)), the PMCC is unchanged | B1 | Must give reason relating to linear transformation |
# Question 1:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= \frac{39+40+48+61+71+81+85+83+77+67+54+41}{12} = \frac{747}{12} = 62.25 \approx 62.3$ | B1 | Accept 62.2 or 62.3 |
| Standard deviation $= \sqrt{\frac{\sum u^2}{12} - \bar{u}^2}$ where $\sum u^2 = 49814$ | M1 | Correct method for s.d. |
| $s.d. = \sqrt{\frac{49814}{12} - 62.25^2} = \sqrt{4151.17 - 3875.0625} = \sqrt{276.1} \approx 16.6$ | A1 | Accept 16.6 or 16.7 |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Using $c = \frac{5}{9}(f-32)$, new mean $= \frac{5}{9}(62.25 - 32) = \frac{5}{9}(30.25) = 16.8$ °C | M1 | Applying linear transformation to mean |
| A1 | Accept 16.8 or 16.9 |
| New s.d. $= \frac{5}{9} \times 16.6... = 9.2$ °C | M1 | Multiplying s.d. by $\frac{5}{9}$ only |
| A1 | Accept 9.2 or 9.3 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $r_{xy} = 0.997$ | B1 | Correct value stated |
| Since $x$ and $y$ are linear transformations of $u$ and $v$ respectively (of the form $x = au + b$, $y = av + b$ with $a > 0$), the PMCC is unchanged | B1 | Must give reason relating to linear transformation |
---1 The average maximum monthly temperatures, $u$ degrees Fahrenheit, and the average minimum monthly temperatures, $v$ degrees Fahrenheit, in New York City are as follows.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
& Jan & Feb & Mar & Apr & May & Jun & Jul & Aug & Sep & Oct & Nov & Dec \\
\hline
Maximum (u) & 39 & 40 & 48 & 61 & 71 & 81 & 85 & 83 & 77 & 67 & 54 & 41 \\
\hline
Minimum (v) & 26 & 27 & 34 & 44 & 53 & 63 & 68 & 66 & 60 & 51 & 41 & 30 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Calculate, to one decimal place, the mean and the standard deviation of the 12 values of the average maximum monthly temperature.
\item For comparative purposes with a UK city, it was necessary to convert the temperatures from degrees Fahrenheit ( ${ } ^ { \circ } \mathrm { F }$ ) to degrees Celsius ( ${ } ^ { \circ } \mathrm { C }$ ). The formula used to convert $f ^ { \circ } \mathrm { F }$ to $c ^ { \circ } \mathrm { C }$ is:
$$c = \frac { 5 } { 9 } ( f - 32 )$$
Use this formula and your answers in part (a)(i) to calculate, in ${ } ^ { \circ } \mathbf { C }$, the mean and the standard deviation of the 12 values of the average maximum monthly temperature.\\
(3 marks)
\end{enumerate}\item The value of the product moment correlation coefficient, $r _ { u v }$, between the above 12 values of $u$ and $v$ is 0.997 , correct to three decimal places.
State, giving a reason, the corresponding value of $r _ { x y }$, where $x$ and $y$ are the exact equivalent temperatures in ${ } ^ { \circ } \mathrm { C }$ of $u$ and $v$ respectively.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2013 Q1 [7]}}