| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Calculate regression line equation |
| Difficulty | Moderate -0.3 This is a standard S1 statistics question testing routine calculations: mean, standard deviation, correlation coefficient, and understanding linear transformations of data. All parts follow textbook procedures with no problem-solving required. The transformation part (c-d) tests conceptual understanding but is well-rehearsed material. Slightly easier than average due to being purely procedural with given formulas. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02c Linear coding: effects on mean and variance5.08a Pearson correlation: calculate pmcc |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = 71.83\ldots\) | B1 | awrt 71.8, allow \(\frac{431}{6}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Standard deviation} = \sqrt{\frac{62802}{12}-\left(\frac{862}{12}\right)^2}\) or variance \(= \frac{62802}{12}-\left(\frac{862}{12}\right)^2\) | M1 | Correct method to find s.d. or variance, ft their mean for M1 only; also allow \(\text{s.d.} = \sqrt{\frac{S_{xx}}{n}}\) |
| \(\sqrt{73.47\ldots} = 8.571\ldots\), \(8.57\) (to 3 s.f.) | A1* | Must see at least one simplification of working and the given answer 8.57; \(\sqrt{\frac{62802}{12}-71.8^2}\) scores M1A0; need at least 71.833 used |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xx} = 62802 - \frac{862^2}{12}\left[=\frac{2645}{3}=881.66\ldots\right]\) | M1 | Correct method to find \(S_{xx}\) (implied by awrt 882) |
| \(r = \frac{512.67}{\sqrt{413.67 \times 881.66\ldots}}\) | M1 | Correct method to find PMCC using their value of \(S_{xx}\) |
| \(= 0.8489\ldots\) awrt 0.849 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = \frac{5}{9}\times(71.8' - 32)\) | M1 | Correct method to find mean ft their mean in part (a) |
| \(= 22.11\ldots\) awrt 22.1 | A1ft | awrt 22.1 ft their mean in part (a) |
| \(\text{Standard deviation} = \frac{5}{9}\times 8.57\ldots\) | M1 | Correct method to find s.d.; do not isw if any further calculation is done after multiplying by \(\frac{5}{9}\) |
| \(= 4.76\ldots\) awrt 4.76 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 0.8489\ldots\) / same (as for \(x\) and \(y\)) | M1 | \(r =\) their part (b) provided \(-1 \leq r \leq 1\); allow 2 s.f. on the ft |
| \(r\) not affected by (linear) coding | A1 | Any correct reasoning; allow e.g. 'addition/subtraction and multiplication/division does not affect \(r\)' |
## Question 2:
### Part (a)(i):
$\text{Mean} = 71.83\ldots$ | B1 | awrt 71.8, allow $\frac{431}{6}$
### Part (a)(ii):
$\text{Standard deviation} = \sqrt{\frac{62802}{12}-\left(\frac{862}{12}\right)^2}$ or variance $= \frac{62802}{12}-\left(\frac{862}{12}\right)^2$ | M1 | Correct method to find s.d. or variance, ft their mean for M1 only; also allow $\text{s.d.} = \sqrt{\frac{S_{xx}}{n}}$
$\sqrt{73.47\ldots} = 8.571\ldots$, $8.57$ (to 3 s.f.) | A1* | Must see at least one simplification of working and the given answer 8.57; $\sqrt{\frac{62802}{12}-71.8^2}$ scores M1A0; need at least 71.833 used
### Part (b):
$S_{xx} = 62802 - \frac{862^2}{12}\left[=\frac{2645}{3}=881.66\ldots\right]$ | M1 | Correct method to find $S_{xx}$ (implied by awrt 882)
$r = \frac{512.67}{\sqrt{413.67 \times 881.66\ldots}}$ | M1 | Correct method to find PMCC using their value of $S_{xx}$
$= 0.8489\ldots$ awrt 0.849 | A1 |
### Part (c):
$\text{Mean} = \frac{5}{9}\times(71.8' - 32)$ | M1 | Correct method to find mean ft their mean in part (a)
$= 22.11\ldots$ awrt 22.1 | A1ft | awrt 22.1 ft their mean in part (a)
$\text{Standard deviation} = \frac{5}{9}\times 8.57\ldots$ | M1 | Correct method to find s.d.; do not isw if any further calculation is done after multiplying by $\frac{5}{9}$
$= 4.76\ldots$ awrt 4.76 | A1 |
### Part (d):
$r = 0.8489\ldots$ / same (as for $x$ and $y$) | M1 | $r =$ their part (b) provided $-1 \leq r \leq 1$; allow 2 s.f. on the ft
$r$ not affected by (linear) coding | A1 | Any correct reasoning; allow e.g. 'addition/subtraction and multiplication/division does not affect $r$'
---\begin{enumerate}
\item The average minimum monthly temperature, $x$ degrees Fahrenheit ( ${ } ^ { \circ } \mathrm { F }$ ), and the average maximum monthly temperature, $y$ degrees Fahrenheit ( ${ } ^ { \circ } \mathrm { F }$ ), in Kolkata were recorded for 12 months.
\end{enumerate}
Some of the summary statistics are given below.
$$\sum x = 862 \quad \sum x ^ { 2 } = 62802 \quad \mathrm {~S} _ { y y } = 413.67 \quad S _ { x y } = 512.67 \quad n = 12$$
(a) (i) Calculate the mean of the 12 values of the average minimum\\
monthly temperature.\\
(ii) Show that the standard deviation of the 12 values of the average minimum monthly temperature is $8.57 ^ { \circ } \mathrm { F }$ to 3 significant figures.\\
(b) Calculate the product moment correlation coefficient between $x$ and $y$
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 was
$$c = \frac { 5 } { 9 } ( f - 32 )$$
where $f$ is the temperature in ${ } ^ { \circ } \mathrm { F }$ and $c$ is the temperature in ${ } ^ { \circ } \mathrm { C }$\\
(c) Use this formula and the values from part (a) to calculate, in ${ } ^ { \circ } \mathrm { C }$, the mean and the standard deviation of the 12 values of the average minimum monthly temperature in Kolkata.\\
Give your answers to 3 significant figures.
Given that
\begin{itemize}
\item $u$ is the equivalent temperature in ${ } ^ { \circ } \mathrm { C }$ of $x$
\item $\quad v$ is the equivalent temperature in ${ } ^ { \circ } \mathrm { C }$ of $y$\\
(d) state, giving a reason, the product moment correlation coefficient between $u$ and $v$
\end{itemize}
\hfill \mbox{\textit{Edexcel S1 2024 Q2 [12]}}