| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate x on y regression line |
| Difficulty | Standard +0.3 This question requires calculating the x-on-y regression line (slightly less routine than y-on-x) and performing a paired t-test on differences. Both are standard A-level Further Maths procedures with all summary statistics provided, requiring straightforward formula application rather than problem-solving insight. The two-part structure and context make it slightly above average difficulty but well within typical FM statistics questions. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Child | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) | \(I\) | \(J\) |
| \(x\) | 5.2 | 4.1 | 3.7 | 5.4 | 7.6 | 6.1 | 3.2 | 4.0 | 3.5 | 8.0 |
| \(y\) | 6.2 | 4.8 | 5.0 | 5.6 | 7.7 | 7.0 | 4.0 | 4.5 | 3.6 | 8.5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xy} = 313.28 - 50.8 \times 56.9/10 = 24.228\) | ||
| \(S_{xx} = 284.16 - 50.8^2/10 = 26.096\) | ||
| \(b_1 = S_{xy}/S_{xx} = 0.928\) | M1 A1 | |
| \(y = 56.9/10 + b_1(7 - 50.8/10)\) | M1 A1 | |
| \(= 5.69 + 0.928(7 - 5.08)\) | ||
| \([= 0.928 \times 7 + 0.976]\) | ||
| \(y = 7.47\) (allow 7.48 or 7.5) | A1 | PA −1 so max 4/5 for 0.93, giving \(y = 7.48\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{yy} = 347.59 - 56.9^2/10 = 23.829\) | ||
| \(b_2 = S_{xy}/S_{yy} = 1.017\) | (M1) | |
| \(7 = 50.8/10 + b_2(y - 56.9/10)\) | (M1 A1) | |
| \(= 1.017y - 0.707\) | ||
| \(y = 7.58\) (allow 7.6) | (A1) | can earn at most 4/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Differences: \(1\ 0.7\ 1.3\ 0.2\ 0.1\ 0.9\ 0.8\ 0.5\ 0.1\ 0.5\) | ||
| \(\bar{d} = 6.1/10 = 0.61\) | M1 A1 | |
| \(s^2 = (5.19 - 6.1^2/10)/9 = 0.1632\) or \(0.404^2\) | B1 | allow biased: \(0.1469\) or \(0.3833^2\) |
| \(H_0: \mu_y - \mu_x = 0.4,\ H_1: \mu_y - \mu_x > 0.4\) | B1 | State hypotheses (AEF; B0 for \(\bar{x}\)) |
| \(t = (\bar{d} - 0.4)/(s/\sqrt{10}) = 1.64\) | M1 A1 | |
| \(t_{9,\ 0.95} = 1.83\ [3]\) | B1 | |
| Accept \(H_0\): No improvement of more than 0.4 | B1\(\checkmark\) | Consistent conclusion (AEF, ft on both \(t\)-values) |
| Wrong test can earn only B1 for hypotheses and B1 for conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_{\text{disc}} = \frac{1}{2}(2ma^2) = ma^2\) | B1 | |
| \(I_{\text{ring}} = 2m(3a)^2 = 18ma^2\) | B1 | |
| \(I_{\text{rod}} = \frac{1}{3}(3m/2)a^2 + (3m/2)(2a)^2 = \frac{13}{2}ma^2\) | B1 | EITHER method |
| OR: \(\frac{1}{3}(9m/2)(3a)^2 - \frac{1}{3}(3m/2)a^2 = 13ma^2\) | (B1) | |
| \(I_O = ma^2 + 18ma^2 + 4(\frac{13}{2})ma^2 = 45ma^2\) | B1 | AG |
| \(I_O' = \frac{1}{2}I_O\) | M1 | |
| \(I_A = I_O' + 10m(3a)^2 = \frac{225}{2}ma^2\) | M1 A1 | |
| \(I_A' = I_A + 3m(6a)^2 = \frac{441}{2}ma^2\) | M1 | |
| \(\omega_0 = u/3a\) | B1 | |
| Gain in P.E. at instantaneous rest: \((10mg \times 3a + 3mg \times 6a)(1 - \sin\theta)\) | M1 A1 | |
| \(= 45mga/2 + 27mga/2\) | ||
| or \(48mga(1 - \sin\theta) = 36mga\) | A1 | |
| \(\frac{1}{2}I_A'\omega_0^2 = \frac{1}{2}(441) \times 36ag\ \ [441 = 21^2]\) | M1 | |
| \(u^2 = (36/441) \times 36ag\) | ||
| \(u = \frac{12}{7}\sqrt{ag}\) or \(1.71\sqrt{ag}\) | A1 | |
| SR: Taking \(AC\) at \(\sin^{-1}(\frac{1}{4})\) to vertical: P.E. \(= 48mga(1 - \cos\sin^{-1}(\frac{1}{4})) = 1.524mga\) | (A0) | |
| \(u = 0.35[3]\sqrt{ag}\) | (A1) | max 6/7; SR: Overlooking added particle can earn M0 B1 M1 A0 A0 M1 A0 (max 3/7) |
| Answer | Marks | Guidance |
|---|---|---|
| Geometric distribution | B1 | |
| \(H_0\): Distribution fits data or \(p = 0.6\) | B1 | (AEF; B0 for "It is a good fit") |
| Expected values using \(200pq^{x-1}\) with \(p=0.6,\ q=0.4\): \(120\ \ 48\ \ 19.2\ \ 7.68\ \ 3.072\ \ 1.2288\ \ 0.8192\) | M1 | ignore incorrect final value; e.g. 0.4915 can earn max 5/8 |
| Combine last 3 cells since exp. value \(< 5\): \(O:\ \ldots\ 6\); \(E:\ \ldots\ 5.12\) | B1 | |
| \(\chi^2 = 0.3 + 0.5208 + 0.4083 + 2.8519 + 0.15125 = 4.23\ [2]\) | M1 A1 | |
| 5 cells: \(\chi^2_{4,\ 0.95} = 9.488\) | B1\(\checkmark\) | |
| 6 cells: \(\chi^2_{5,\ 0.95} = 11.07\) | ||
| 7 cells: \(\chi^2_{6,\ 0.95} = 12.59\) | ||
| 4 cells: \(\chi^2_{3,\ 0.95} = 7.815\) | if 4 cells combined with 0.4915 | |
| Accept \(H_0\) if \(\chi^2 <\) tabular value | M1 | |
| \(4.23 < 9.49\) so distribution fits or \(p = 0.6\) | A1 | (Allow A1 for "It is a good fit") |
| Answer | Marks |
|---|---|
| Find \(p\) of at least one 6 on 5 throws of one die: \(p = 1 - 0.75^5 = 0.7627\) (4 s.f.) | M1 A1 |
| Find prob. of at least one 6 on exactly 4 of 10 dice: \({}^{10}C_4 \times p^4(1-p)^6\) | M1; M1 |
| \(= 210 \times 6.043 \times 10^{-5}\) | |
| \(= 0.0126\) or \(0.0127\) | A1 |
# Question 10:
## Part 1 - Regression line of y on x
$S_{xy} = 313.28 - 50.8 \times 56.9/10 = 24.228$ | | |
$S_{xx} = 284.16 - 50.8^2/10 = 26.096$ | | |
$b_1 = S_{xy}/S_{xx} = 0.928$ | M1 A1 | |
$y = 56.9/10 + b_1(7 - 50.8/10)$ | M1 A1 | |
$= 5.69 + 0.928(7 - 5.08)$ | | |
$[= 0.928 \times 7 + 0.976]$ | | |
$y = 7.47$ (allow 7.48 or 7.5) | A1 | PA −1 so max 4/5 for 0.93, giving $y = 7.48$ |
## Part 1 SR - Regression line of x on y
$S_{yy} = 347.59 - 56.9^2/10 = 23.829$ | | |
$b_2 = S_{xy}/S_{yy} = 1.017$ | (M1) | |
$7 = 50.8/10 + b_2(y - 56.9/10)$ | (M1 A1) | |
$= 1.017y - 0.707$ | | |
$y = 7.58$ (allow 7.6) | (A1) | can earn at most 4/5 |
## Part 2 - Hypothesis Test
Differences: $1\ 0.7\ 1.3\ 0.2\ 0.1\ 0.9\ 0.8\ 0.5\ 0.1\ 0.5$ | | |
$\bar{d} = 6.1/10 = 0.61$ | M1 A1 | |
$s^2 = (5.19 - 6.1^2/10)/9 = 0.1632$ or $0.404^2$ | B1 | allow biased: $0.1469$ or $0.3833^2$ |
$H_0: \mu_y - \mu_x = 0.4,\ H_1: \mu_y - \mu_x > 0.4$ | B1 | State hypotheses (AEF; B0 for $\bar{x}$) |
$t = (\bar{d} - 0.4)/(s/\sqrt{10}) = 1.64$ | M1 A1 | |
$t_{9,\ 0.95} = 1.83\ [3]$ | B1 | |
Accept $H_0$: No improvement of more than 0.4 | B1$\checkmark$ | Consistent conclusion (AEF, ft on both $t$-values) |
| | Wrong test can earn only B1 for hypotheses and B1 for conclusion |
---
# Question 11A:
## Moments of Inertia
$I_{\text{disc}} = \frac{1}{2}(2ma^2) = ma^2$ | B1 | |
$I_{\text{ring}} = 2m(3a)^2 = 18ma^2$ | B1 | |
$I_{\text{rod}} = \frac{1}{3}(3m/2)a^2 + (3m/2)(2a)^2 = \frac{13}{2}ma^2$ | B1 | EITHER method |
OR: $\frac{1}{3}(9m/2)(3a)^2 - \frac{1}{3}(3m/2)a^2 = 13ma^2$ | (B1) | |
$I_O = ma^2 + 18ma^2 + 4(\frac{13}{2})ma^2 = 45ma^2$ | B1 | AG |
$I_O' = \frac{1}{2}I_O$ | M1 | |
$I_A = I_O' + 10m(3a)^2 = \frac{225}{2}ma^2$ | M1 A1 | |
$I_A' = I_A + 3m(6a)^2 = \frac{441}{2}ma^2$ | M1 | |
$\omega_0 = u/3a$ | B1 | |
Gain in P.E. at instantaneous rest: $(10mg \times 3a + 3mg \times 6a)(1 - \sin\theta)$ | M1 A1 | |
$= 45mga/2 + 27mga/2$ | | |
or $48mga(1 - \sin\theta) = 36mga$ | A1 | |
$\frac{1}{2}I_A'\omega_0^2 = \frac{1}{2}(441) \times 36ag\ \ [441 = 21^2]$ | M1 | |
$u^2 = (36/441) \times 36ag$ | | |
$u = \frac{12}{7}\sqrt{ag}$ or $1.71\sqrt{ag}$ | A1 | |
SR: Taking $AC$ at $\sin^{-1}(\frac{1}{4})$ to vertical: P.E. $= 48mga(1 - \cos\sin^{-1}(\frac{1}{4})) = 1.524mga$ | (A0) | |
$u = 0.35[3]\sqrt{ag}$ | (A1) | max 6/7; SR: Overlooking added particle can earn M0 B1 M1 A0 A0 M1 A0 (max 3/7) |
---
# Question 11B:
## Part 1 - Chi-squared test
Geometric distribution | B1 | |
$H_0$: Distribution fits data or $p = 0.6$ | B1 | (AEF; B0 for "It is a good fit") |
Expected values using $200pq^{x-1}$ with $p=0.6,\ q=0.4$: $120\ \ 48\ \ 19.2\ \ 7.68\ \ 3.072\ \ 1.2288\ \ 0.8192$ | M1 | ignore incorrect final value; e.g. 0.4915 can earn max 5/8 |
Combine last 3 cells since exp. value $< 5$: $O:\ \ldots\ 6$; $E:\ \ldots\ 5.12$ | B1 | |
$\chi^2 = 0.3 + 0.5208 + 0.4083 + 2.8519 + 0.15125 = 4.23\ [2]$ | M1 A1 | |
5 cells: $\chi^2_{4,\ 0.95} = 9.488$ | B1$\checkmark$ | |
6 cells: $\chi^2_{5,\ 0.95} = 11.07$ | | |
7 cells: $\chi^2_{6,\ 0.95} = 12.59$ | | |
4 cells: $\chi^2_{3,\ 0.95} = 7.815$ | | if 4 cells combined with 0.4915 |
Accept $H_0$ if $\chi^2 <$ tabular value | M1 | |
$4.23 < 9.49$ so distribution fits or $p = 0.6$ | A1 | (Allow A1 for "It is a good fit") |
## Part 2 - Probability
Find $p$ of at least one 6 on 5 throws of one die: $p = 1 - 0.75^5 = 0.7627$ (4 s.f.) | M1 A1 | |
Find prob. of at least one 6 on exactly 4 of 10 dice: ${}^{10}C_4 \times p^4(1-p)^6$ | M1; M1 | |
$= 210 \times 6.043 \times 10^{-5}$ | | |
$= 0.0126$ or $0.0127$ | A1 | |10 Young children at a primary school are learning to throw a ball as far as they can. The distance thrown at the beginning of the school year and the distance thrown at the end of the same school year are recorded for each child. The distance thrown, in metres, at the beginning of the year is denoted by $x$; the distance thrown, in metres, at the end of the year is denoted by $y$. For a random sample of 10 children, the results are shown in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
Child & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $J$ \\
\hline
$x$ & 5.2 & 4.1 & 3.7 & 5.4 & 7.6 & 6.1 & 3.2 & 4.0 & 3.5 & 8.0 \\
\hline
$y$ & 6.2 & 4.8 & 5.0 & 5.6 & 7.7 & 7.0 & 4.0 & 4.5 & 3.6 & 8.5 \\
\hline
\end{tabular}
\end{center}
$$\left[ \Sigma x = 50.8 , \quad \Sigma x ^ { 2 } = 284.16 , \quad \Sigma y = 56.9 , \quad \Sigma y ^ { 2 } = 347.59 , \quad \Sigma x y = 313.28 . \right]$$
A particular child threw the ball a distance of 7.0 metres at the beginning of the year, but he could not throw at the end of the year because he had broken his arm. By finding the equation of an appropriate regression line, estimate the distance this child would have thrown at the end of the year.
The teacher suspects that, on average, the distance thrown by a child increases between the two throws by more than 0.4 metres. Stating suitable hypotheses and assuming a normal distribution, test the teacher's suspicion at the $5 \%$ significance level.
\hfill \mbox{\textit{CAIE FP2 2015 Q10 [13]}}