| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Find missing data values |
| Difficulty | Standard +0.8 This Further Statistics 2 question requires understanding of coded data transformations, applying variance formulas through coding relationships (S_xx = 100·S_ss), and interpreting correlation. While methodical, it demands careful algebraic manipulation of summary statistics and understanding that correlation is invariant under linear coding—more sophisticated than standard A-level but accessible with FS2 knowledge. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08c Pearson: measure of straight-line fit |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xx} = \sum(10s)^2 - \frac{(\sum 10s)^2}{10}\) | M1 | Attempting to use \(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{10}\) with \(x = 10s\) |
| \(2658.9 = 100\sum(s)^2 - \frac{100(\sum s)^2}{10}\) | M1 | Substituting in 2658.9 and dealing with the 10 correctly |
| \(2658.9 = 100 S_{ss}\) | ||
| \(S_{ss} = 26.589 *\) | A1*cso | A complete solution with no errors leading to 26.589 only |
| Answer | Marks | Guidance |
|---|---|---|
| \(64 = \sum_{1}^{10}(d_i - 9)\) | M1 | Realising that either \(64 = \sum_{1}^{10}(d_i - 9)\) or \(64 = 10\sum_{1}^{10}d_i - 900\) o.e. must be used. May be implied by seeing 96.4 |
| \(64 = 10\sum_{1}^{10} d_i - 900\) | ||
| \(\sum_{1}^{10} d_i = 96.4\) | A1 | 96.4 only |
| \(S_{dd} = 1081.74 - \frac{(\text{"96.4"})^2}{10}\) | M1 | |
| \(= 152.444\) | ||
| \(r = 0.935\) | A1ft | awrt 0.935 ft "their 96.4" |
| Answer | Marks | Guidance |
|---|---|---|
| Linear correlation is significant but scatter diagram suggests a non-linear relationship between the level of serum magnesium, and the level of the disease protein | B1 | A correct comment comparing their value of \(r\) and the scatter diagram in context |
**(a)**
| $S_{xx} = \sum(10s)^2 - \frac{(\sum 10s)^2}{10}$ | M1 | Attempting to use $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{10}$ with $x = 10s$ |
| $2658.9 = 100\sum(s)^2 - \frac{100(\sum s)^2}{10}$ | M1 | Substituting in 2658.9 and dealing with the 10 correctly |
| $2658.9 = 100 S_{ss}$ | | |
| $S_{ss} = 26.589 *$ | A1*cso | A complete solution with no errors leading to 26.589 only |
**(b)**
| $64 = \sum_{1}^{10}(d_i - 9)$ | M1 | Realising that either $64 = \sum_{1}^{10}(d_i - 9)$ or $64 = 10\sum_{1}^{10}d_i - 900$ o.e. must be used. May be implied by seeing 96.4 |
| $64 = 10\sum_{1}^{10} d_i - 900$ | | |
| $\sum_{1}^{10} d_i = 96.4$ | A1 | 96.4 only |
| $S_{dd} = 1081.74 - \frac{(\text{"96.4"})^2}{10}$ | M1 | |
| $= 152.444$ | | |
| $r = 0.935$ | A1ft | awrt 0.935 ft "their 96.4" |
**(c)**
| Linear correlation is significant but scatter diagram suggests a non-linear relationship between the level of serum magnesium, and the level of the disease protein | B1 | A correct comment comparing their value of $r$ and the scatter diagram in context |\begin{enumerate}
\item Over a period of time, researchers took 10 blood samples from one patient with a blood disease. For each sample, they measured the levels of serum magnesium, $s \mathrm { mg } / \mathrm { dl }$, in the blood and the corresponding level of the disease protein, $d \mathrm { mg } / \mathrm { dl }$. One of the researchers coded the data for each sample using $x = 10 s$ and $y = 10 ( d - 9 )$ but spilt ink over his work.
\end{enumerate}
The following summary statistics and unfinished scatter diagram are the only remaining information.
$$\sum d ^ { 2 } = 1081.74 \quad \mathrm {~S} _ { d s } = 59.524$$
and
$$\sum y = 64 \quad \mathrm {~S} _ { x x } = 2658.9$$
$d \mathrm { mg } / \mathrm { dl }$\\
\includegraphics[max width=\textwidth, alt={}, center]{e777c787-0d39-4d84-a0f9-fc4a6712184f-22_983_1534_840_303}\\
(a) Use the formula for $\mathrm { S } _ { x x }$ to show that $\mathrm { S } _ { s s } = 26.589$\\
(b) Find the value of the product moment correlation coefficient between $s$ and $d$.\\
(c) With reference to the unfinished scatter diagram, comment on your result in part (b).\\
\hfill \mbox{\textit{Edexcel FS2 Q7 [8]}}