| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Relate two regression lines |
| Difficulty | Standard +0.3 This is a straightforward application of standard regression formulas with given summary statistics. Parts (i) and (ii) are routine recall and calculation, while part (iii) requires knowing the relationship between regression coefficients (b' = r·σ_x/σ_y), which is a standard A-level further maths result. All steps are mechanical with no problem-solving insight needed. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09c Calculate regression line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Approx. straight line with negative gradient | B1 | Valid comment on scatter diagram (A.E.F.) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EITHER: \(b = r\sqrt{S_{yy}/S_{xx}} = r\sqrt{\{(8245 - 240^2/20)/(2125 - 200^2/20)\}}\) | M2 | Find gradient \(b\) directly using \(r\) |
| \(= -0.992\sqrt{5365/125}\) | A1 | |
| OR: \(S_{xy} = r\sqrt{S_{xx}\cdot S_{yy}} = -812.37\); \(b = S_{xy}/S_{xx} = [-812.37/125]\) | (M1 A1), (M1) | Find \(S_{xy}\) and \(b\) |
| \([= -6.499] = -6.50\) | A1 | Evaluate \(b\) |
| \(y = b(x - 10) + 12 = 77.0 - 6.50x\) | M1, A1 | Find equation of regression line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b' = -0.992^2/6.499 = -0.151\) | M1 A1 | Find \(b'\) using \(r^2 = bb'\): [or \(S_{xy}/S_{yy}\)] |
| Part total: 2 marks | Question total: 9 marks |
## Question 9(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Approx. straight line with negative gradient | B1 | Valid comment on scatter diagram (A.E.F.) |
**Part total: 1 mark**
## Question 9(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER:** $b = r\sqrt{S_{yy}/S_{xx}} = r\sqrt{\{(8245 - 240^2/20)/(2125 - 200^2/20)\}}$ | M2 | Find gradient $b$ directly using $r$ |
| $= -0.992\sqrt{5365/125}$ | A1 | |
| **OR:** $S_{xy} = r\sqrt{S_{xx}\cdot S_{yy}} = -812.37$; $b = S_{xy}/S_{xx} = [-812.37/125]$ | (M1 A1), (M1) | Find $S_{xy}$ and $b$ |
| $[= -6.499] = -6.50$ | A1 | Evaluate $b$ |
| $y = b(x - 10) + 12 = 77.0 - 6.50x$ | M1, A1 | Find equation of regression line |
**Part total: 6 marks**
## Question 9(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b' = -0.992^2/6.499 = -0.151$ | M1 A1 | Find $b'$ using $r^2 = bb'$: [or $S_{xy}/S_{yy}$] |
**Part total: 2 marks | Question total: 9 marks**
---9 A set of 20 pairs of bivariate data $( x , y )$ is summarised by
$$\Sigma x = 200 , \quad \Sigma x ^ { 2 } = 2125 , \quad \Sigma y = 240 , \quad \Sigma y ^ { 2 } = 8245 .$$
The product moment correlation coefficient is - 0.992 .\\
(i) What does the value of the product moment correlation coefficient indicate about a scatter diagram of the data points?\\
(ii) Find the equation of the regression line of $y$ on $x$.\\
(iii) The equation of the regression line of $x$ on $y$ is $x = a ^ { \prime } + b ^ { \prime } y$. Find the value of $b ^ { \prime }$.
\hfill \mbox{\textit{CAIE FP2 2010 Q9 [9]}}