| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Find missing data values |
| Difficulty | Challenging +1.2 This is a multi-part statistics question requiring understanding of regression line properties (both lines pass through the mean point), calculation of correlation from regression slopes, and hypothesis testing. While it involves several steps and concepts, the techniques are standard for Further Maths statistics with no novel insight required—students follow established procedures for finding means, using the relationship r² = b_yx × b_xy, and performing a correlation test. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09c Calculate regression line |
| \(x\) | 1 | 5 | 7 | 9 |
| \(y\) | 5 | 6 | 7 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_{\text{disc}} = \frac{1}{2}(\frac{1}{4}ma^2) + \frac{1}{2}m\{(3a/2)^2 + (2a)^2\}\) | M1 A1 | Verify MI of either disc about axis \(l\) at \(O\) |
| \(= ma^2/8 + 25ma^2/8 = 13ma^2/4\) | A1 | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_{OC} = (4/3)\,ma^2\) | B1 | Find or state MI of rod \(OC\) about \(l\) |
| \(I_{AB} = \frac{1}{3}\cdot 2m\,(3a/2)^2 + 2m\,(2a)^2 = (19/2)\,ma^2\) | M1 A1 | Find MI of rod \(AB\) about \(l\) |
| \(I = (2\times 13/4 + 4/3 + 19/2)\,ma^2 = (52/3)\,ma^2\) | A1 | A.G. Verify MI of object about \(l\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\omega_0 = \sqrt{(2ag)}/2a\) or \(\sqrt{(g/2a)}\) | B1 | Find and use initial angular speed |
| \(\frac{1}{2}I\,\omega_0^2 = \frac{1}{2}(52/3)\,ma^2 \times (g/2a) = (13/3)\,mga\) | M1 A1 | Find initial rotational KE |
| \((3mg\times 2a + mga)(1-\cos\theta)\) or \((4mg\times 7a/4)(1-\cos\theta) = 7mga(1-\cos\theta)\) | M1 A1 | Find gain in P.E. at instantaneous rest |
| \(1 - \cos\theta = (13/3)\,mga/7mga = 13/21,\ \cos\theta = 8/21\) | A.G. M1 A1 | Verify \(\cos\theta\) by equating KE and PE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: \(\bar{x}=5,\ \bar{y}=6\) by solving simultaneous equations | M1, A1; A1 | Find \(\bar{x},\ \bar{y}\) |
| \(22 + x_5 = 5\times 5,\ x_5=3\); \(25+y_5=5\times 6,\ y_5=5\) | A1, A1 | Hence find missing values \(x_5,\ y_5\) |
| OR: \(25+y_5 = 5\times 4.5 + 0.3(22+x_5)\); \(22+x_5=3(25+y_5)-5\times 13\) | (M1 A1)(A1) | Formulate simultaneous equations for \(x_5,\ y_5\) |
| \(y_5 = 0.3\,x_5 + 4.1,\ x_5 = 3\,y_5-12\); \(x_5=3,\ y_5=5\) | (A1 A1) | Hence find missing values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = \sqrt{(0.3\times 3)}\) or \(12/\sqrt{(40\times 4)} = 0.949\) | M1 A1 | Find correlation coefficient \(r\); A0 for \(-0.949\) or \(\pm 0.949\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum x = 25+20=45,\ \sum x^2=165+100=265\) | B1 | Find corresponding summations for combined data (B1 needs all 5 correct) |
| \(\sum y=30+17=47,\ \sum y^2=184+69=253,\ \sum xy=162+75=237\) | ||
| \(S_{xy}=237-45\times 47/10=25.5\) | ||
| \(S_{xx}=265-45^2/10=62.5\) | ||
| \(S_{yy}=253-47^2/10=32.1\) | ||
| \(r' = S_{xy}/\sqrt{(S_{xx}S_{yy})} = 25.5/44.79 = 0.569\) | M1 *A1 | Find correlation coefficient \(r'\) |
| \(H_0: \rho=0,\ H_1: \rho\neq 0\) | B1 | State both hypotheses (B0 for \(r\ldots\)) |
| \(r_{10,\ 5\%} = 0.632\) | *B1 | State or use correct tabular two-tail \(r\)-value |
| Accept \(H_0\) if \( | r' | <\) tabular value (AEF) |
| \(0.569 < 0.632\) so popln. pmcc not different from 0 | DA1 | Correct conclusion (AEF, dep *A1, *B1) |
# Question 10(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{\text{disc}} = \frac{1}{2}(\frac{1}{4}ma^2) + \frac{1}{2}m\{(3a/2)^2 + (2a)^2\}$ | M1 A1 | Verify MI of either disc about axis $l$ at $O$ |
| $= ma^2/8 + 25ma^2/8 = 13ma^2/4$ | A1 | A.G. |
---
# Question 10(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{OC} = (4/3)\,ma^2$ | B1 | Find or state MI of rod $OC$ about $l$ |
| $I_{AB} = \frac{1}{3}\cdot 2m\,(3a/2)^2 + 2m\,(2a)^2 = (19/2)\,ma^2$ | M1 A1 | Find MI of rod $AB$ about $l$ |
| $I = (2\times 13/4 + 4/3 + 19/2)\,ma^2 = (52/3)\,ma^2$ | A1 | A.G. Verify MI of object about $l$ |
---
# Question 10(a)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\omega_0 = \sqrt{(2ag)}/2a$ or $\sqrt{(g/2a)}$ | B1 | Find and use initial angular speed |
| $\frac{1}{2}I\,\omega_0^2 = \frac{1}{2}(52/3)\,ma^2 \times (g/2a) = (13/3)\,mga$ | M1 A1 | Find initial rotational KE |
| $(3mg\times 2a + mga)(1-\cos\theta)$ or $(4mg\times 7a/4)(1-\cos\theta) = 7mga(1-\cos\theta)$ | M1 A1 | Find gain in P.E. at instantaneous rest |
| $1 - \cos\theta = (13/3)\,mga/7mga = 13/21,\ \cos\theta = 8/21$ | A.G. M1 A1 | Verify $\cos\theta$ by equating KE and PE |
---
# Question 10(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| EITHER: $\bar{x}=5,\ \bar{y}=6$ by solving simultaneous equations | M1, A1; A1 | Find $\bar{x},\ \bar{y}$ |
| $22 + x_5 = 5\times 5,\ x_5=3$; $25+y_5=5\times 6,\ y_5=5$ | A1, A1 | Hence find missing values $x_5,\ y_5$ |
| OR: $25+y_5 = 5\times 4.5 + 0.3(22+x_5)$; $22+x_5=3(25+y_5)-5\times 13$ | (M1 A1)(A1) | Formulate simultaneous equations for $x_5,\ y_5$ |
| $y_5 = 0.3\,x_5 + 4.1,\ x_5 = 3\,y_5-12$; $x_5=3,\ y_5=5$ | (A1 A1) | Hence find missing values |
---
# Question 10(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = \sqrt{(0.3\times 3)}$ or $12/\sqrt{(40\times 4)} = 0.949$ | M1 A1 | Find correlation coefficient $r$; A0 for $-0.949$ or $\pm 0.949$ |
---
# Question 10(b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum x = 25+20=45,\ \sum x^2=165+100=265$ | B1 | Find corresponding summations for combined data (B1 needs all 5 correct) |
| $\sum y=30+17=47,\ \sum y^2=184+69=253,\ \sum xy=162+75=237$ | | |
| $S_{xy}=237-45\times 47/10=25.5$ | | |
| $S_{xx}=265-45^2/10=62.5$ | | |
| $S_{yy}=253-47^2/10=32.1$ | | |
| $r' = S_{xy}/\sqrt{(S_{xx}S_{yy})} = 25.5/44.79 = 0.569$ | M1 *A1 | Find correlation coefficient $r'$ |
| $H_0: \rho=0,\ H_1: \rho\neq 0$ | B1 | State both hypotheses (B0 for $r\ldots$) |
| $r_{10,\ 5\%} = 0.632$ | *B1 | State or use correct tabular two-tail $r$-value |
| Accept $H_0$ if $|r'| <$ tabular value (AEF) | M1 | |
| $0.569 < 0.632$ so popln. pmcc not different from 0 | DA1 | Correct conclusion (AEF, dep *A1, *B1) |For a random sample, $A$, of 5 pairs of values of $x$ and $y$, the equations of the regression lines of $y$ on $x$ and $x$ on $y$ are respectively $y = 4.5 + 0.3 x$ and $x = 3 y - 13$. Four of the five pairs of data are given in the following table.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | }
\hline
$x$ & 1 & 5 & 7 & 9 \\
\hline
$y$ & 5 & 6 & 7 & 7 \\
\hline
\end{tabular}
\end{center}
Find\\
(i) the fifth pair of values of $x$ and $y$,\\
(ii) the value of the product moment correlation coefficient.
A second random sample, $B$, of 5 pairs of values of $x$ and $y$ is summarised as follows.
$$\Sigma x = 20 \quad \Sigma x ^ { 2 } = 100 \quad \Sigma y = 17 \quad \Sigma y ^ { 2 } = 69 \quad \Sigma x y = 75$$
The two samples, $A$ and $B$, are combined to form a single random sample of size 10 .\\
(iii) Use this combined sample to test, at the $5 \%$ significance level, whether the population product moment correlation coefficient is different from zero.
\hfill \mbox{\textit{CAIE FP2 2016 Q10 OR}}