# Question 10:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(x) = \frac{1}{2}(x-1)$ | B1 | Find $F(x)$ for $1 \leq x \leq 3$ |
| $G(y) = P(Y < y) = P(X^3 < y) = P\!\left(X < y^{\frac{1}{3}}\right) = F\!\left(y^{\frac{1}{3}}\right) = \frac{1}{2}\left(y^{\frac{1}{3}}-1\right)$; $1 \leq y \leq 27$ | M1 A1; B1 | Find $G(y)$ from $Y=X^3$; result may be stated |
| $0\ (y<1)$ and $1\ (y>27)$ | B1 | State $G(y)$ for other values of $y$ |
| $g(y) = \frac{y^{-\frac{2}{3}}}{6}$ or $\frac{1}{6y^{\frac{2}{3}}}$ for $1 \leq y \leq 27$ | B1$\checkmark$ | Find $g(y)$; follow through on $G(y)$ |
| Sketch of $g(y)$ for $1 \leq y \leq 27$ with $g(y)=0$ on either side | B1, B1 | |
| $E(Y) = \int y\,g(y)\,dy = \int\frac{y^{\frac{1}{3}}}{6}\,dy = \left[\frac{y^{\frac{4}{3}}}{8}\right]_1^{27} = 10$ | M1 A1 | Find mean of $Y$; no need to find median $= 8$ |
| $G(10) - G(8)$ or $\left\|G(10) - \frac{1}{2}\right\| = \frac{1}{2}\!\left(10^{\frac{1}{3}}-8^{\frac{1}{3}}\right)$ or $\left\|\frac{1}{2}(10^{\frac{1}{3}}-1)-\frac{1}{2}\right\| = 0.077\ [2]$ | M1 A1 | Find probability $Y$ lies between median and mean; 2 s.f. sufficient |

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# Question 11a:

## Part (i) — EITHER method
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_X = \frac{1}{2}mr^2$ | M1 A1 | State or find MI of $X$ about $AB$ by $\perp$ axes |
| $I_Y = \frac{1}{2}mr^2 + mr^2 = \frac{3mr^2}{2}$ | M1 A1 | State or find MI of $Y$ (or $Z$) about $AB$ |
| $I_W = \frac{1}{2}(3m)R^2 = \frac{1}{2}(3m)r^2\!\left(1+\frac{2}{\sqrt{3}}\right)^2 = \frac{1}{2}(7+4\sqrt{3})mr^2$ | M1 A1 | State or find MI of $W$ about $AB$ by $\perp$ axes |
| $I = \left(\frac{1}{2}+2\times\frac{3}{2}+\frac{7}{2}+2\sqrt{3}\right)mr^2 = (7+2\sqrt{3})mr^2$ | M1 A1 | Find MI of object about $AB$; A.G. |

## Part (i) — OR method
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_X = mr^2 + m\!\left(\frac{2r}{\sqrt{3}}\right)^2 = \frac{7mr^2}{3}$ | M1 A1 | State or find MI of $X$, $Y$ or $Z$ about centre $O$ |
| $I_W = 3mR^2 = 3mr^2\!\left(1+\frac{2}{\sqrt{3}}\right)^2 = (7+4\sqrt{3})mr^2$ | M1 A1 | State or find MI of $W$ about $O$ |
| $I_O = 3I_X + I_W = (14+4\sqrt{3})mr^2$ | M1 A1 | Find MI of object about $O$ |
| $I = \frac{1}{2}I_O = (7+2\sqrt{3})mr^2$ | M1 A1 | Find MI about $AB$ by $\perp$ axes; A.G. |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I' = I + 9mR^2 = I + 3(7+4\sqrt{3})mr^2 = 14(2+\sqrt{3})mr^2$ | M1 A1 | Find new MI of object plus particle about $AB$ |
| $\frac{1}{2}I'\omega^2 = 9mg \times R\sin 60°$ | M1 A1 | Find equation for angular speed $\omega$ using energy |
| $\omega^2 = \frac{9mgR\sqrt{3}}{I'}$; $\omega = \sqrt{\frac{9g}{14r}}$ or $0.802\sqrt{\frac{g}{r}}$ | M1 A1 | Substitute and simplify; A.E.F. |

# Question 11b:

## Part 1 - Confidence Interval (5 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $s_A^2 = \dfrac{\left(1422.34 - \dfrac{106^2}{8}\right)}{7}$ | M1 A1 | Allow use of biased: $\sigma_{A,8}^2 = 2.23$ or $1.493^2$ |
| $= \dfrac{446}{175}$ or $2.549$ or $1.596^2$ | | |
| $\dfrac{106}{8} \pm t\sqrt{\left(\dfrac{s_X^2}{8}\right)}$ | M1 | Find confidence interval |
| $t_{7,0.975} = 2.36\ [5]$ | A1 | State or use correct tabular value of $t$ |
| $13.25 \pm 1.335$ or $[11.9,\ 14.6]$ | A1 | Evaluate C.I. correct to 1 d.p. | **5** |

## Part 2 - Hypothesis Test (7 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| Distribution of $B$ is Normal with same population variance | B1 | State suitable assumptions (A.E.F.) |
| $H_0: \mu_A = \mu_B$, $H_1: \mu_A > \mu_B$ | B1 | State hypotheses (B0 for $\bar{a}\ ...$) |
| $s_B^2 = \dfrac{\left(971.53 - \dfrac{75.9^2}{6}\right)}{5}$ | M1 | Estimate (or imply) $B$'s population variance; allow biased: $\sigma_{B,6}^2 = 1.899$ or $1.378^2$ |
| $= 2.279$ or $1.510^2$ | | |
| $s^2 = \dfrac{7s_A^2 + 5s_B^2}{12} = \dfrac{17.84 + 11.395}{12} = 2.436$ or $1.561^2$ | M1 | Find pooled estimate of common variance $s^2$ |
| $t = \dfrac{(13.25 - 12.65)}{s\sqrt{(8^{-1} + 6^{-1})}} = \dfrac{0.6}{0.8430} = 0.712$ | M1 A1 | Calculate value of $t$ (to 2 d.p., either sign) |
| $t_{12,0.95} = 1.782$ | B1 | State or use correct tabular $t$-value (to 2 d.p.); or compare $0.6$ with $1.782\ s\sqrt{(8^{-1}+6^{-1})} = 1.50$ |
| Accept $H_0$; mean lengths are the same | B1$\checkmark$ | Consistent conclusion (A.E.F., $\checkmark$ on two $t$ values) | **7** |

## Part 3 - Confidence Interval for Difference (2 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $13.25 - 12.65 \pm t\,s\sqrt{(8^{-1}+6^{-1})}$ | M1 | Find confidence interval for the difference |
| $0.6 \pm 1.84$ or $[-1.24,\ 2.44]$ | A1 | Evaluate C.I. with $t_{12,0.975} = 2.179$, to 2 d.p. | **2** |

**Total: [14]**