## Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[-ae^{-(x-2)}\right]_2^\infty = a = 1$ | B1 | **AG** Find $a$ by equating $\int_2^\infty f(x)\,dx$ to 1 |
| $F(x) = \int f(x)\,dx = -ae^{-(x-2)} + c$ | | Find or state distribution function $F(x)$ for $x \geq 2$ |
| $F(2) = 0$ so $F(x) = 1 - e^{-(x-2)}$ | M1 A1 | |
| $F(x) = 0$ for $x < 2$ | A1 | State $F(x)$ for $x < 2$ |
| $1 - e^{-(m-2)} = \frac{1}{2}$, $e^{m-2} = 2$ | | Find median value $m$ from $F(m) = \frac{1}{2}$ |
| $m = 2 + \ln 2$ or $2 - \ln\frac{1}{2}$ or $2.69$ | M1 A1 | |

**Part marks: 1, 5**

### Question 9(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $G(y) = P(Y < y) = P(e^X < y)$ | | Find or state $G(y)$ from $Y = e^X$ for $x \geq 2$ |
| $= P(X < \ln y) = F(\ln y)$ | | |
| $= 1 - e^{-(\ln y - 2)} = 1 - e^2/y$ | M1 A1 | |
| $g(y) = e^2/y^2$ | A1 | Find $g(y)$ by differentiation (allow $e^2 = 7.39$) |
| for $y \geq e^2$ [$g(y) = 0$ for $y < e^2$] | A1 | State corresponding range of $y$ |

**Part marks: 4**

### Question 9(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(Y > 10) = 1 - G(10)$ | | Find $P(Y > 10)$; M0 for $G(10)$ or $1 - g(10)$ |
| $= e^2/10$ or $0.739$ | M1 A1 | |

**Part marks: 2 | Total: 12**

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## Question 10A:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv_1^2 = \frac{1}{2}mu^2 - \frac{1}{2}mga$ | B1 | Use conservation of energy when $P$ is level with $A$ |
| $T_1 = mv_1^2/\frac{1}{2}a$ | B1 | Use $F = ma$ radially |
| $T_1 = (2m/a)(u^2 - ag)$ | B1 | **AG** Eliminate $v_1^2$ |

**Part marks: 3**

### Question 10A(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| State or imply that string loses contact when vertical | M1 | |
| $\frac{1}{2}mv_2^2 = \frac{1}{2}mu^2 - mga$ | B1 | Use conservation of energy |
| $T_2 = mv_2^2/\frac{1}{2}a - mg$ | B1 | Use $F = ma$ radially just before this point |
| $T_3 = mv_2^2/(3a/2) - mg$ | B1 | Use $F = ma$ radially just after this point |
| EITHER: $mv_2^2/\frac{1}{2}a - mg = 5\{mv_2^2/(3a/2) - mg\}$ | | Find $v_2^2$ and hence $u^2$ from $T_2/T_3 = 5$ |
| $v_2^2 = 3ag$, $u^2 = 3ag + 2ag = 5ag$ | M1 A1 | **AG** |
| OR: Eliminate $v_2^2$ in $T_2/T_3 = 5$: $(m/a)(2u^2 - 5ag) = 5(m/a)(2u^2/3 - 7ag/3)$ | | |
| $4u^2/3 = 20ag/3$, $u^2 = 5ag$ | M1 A1 | **AG** |

**Part marks: 6**

## Question (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find $\cos\theta$, where $\theta = \angle AOP$, when $P$ is level with $A$: $\cos\theta = \frac{a}{3a/2} = \frac{2}{3}$ | B1 | |
| Use conservation of energy: $\frac{1}{2}mv_4^2 = \frac{1}{2}mu^2 - \frac{1}{2}mga$ | B1 | |
| $or\ \frac{1}{2}mu^2 - mg\{(3a/2)\cos\theta - \frac{1}{2}a\}$ | | |
| $or\ \frac{1}{2}mv_2^2 + \frac{1}{2}mga$ | | |
| $or\ \frac{1}{2}mv_2^2 + mg(3a/2)(1-\cos\theta)$ | B1 | |
| $[v_4^2 = 4ag]$ | | |
| Use $F = ma$ radially: $T_4 = \frac{mv_4^2}{(3a/2)} - mg\cos\theta$ | B1 | |
| Eliminate $v_4^2$ to find $T_4$: $T_4 = 4mag \times \frac{2}{3a} - \frac{2mg}{3}$ | | |
| $= \frac{8mg}{3} - \frac{2mg}{3}$ | | |
| $= 2mg$ | M1 A1 | **5 marks** |

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## Question 10B:

| Answer/Working | Mark | Guidance |
|---|---|---|
| State hypotheses: $H_0: \mu_A = \mu_B$, $H_1: \mu_A < \mu_B$ | B1 | B0 for $\bar{t}_A$ ... |
| Estimate both population variances (to 3 d.p.): $s_A^2 = (215.18 - \frac{102^2}{50})/49$ | M1 A1 | |
| $and\ s_B^2 = (282.3 - \frac{129^2}{60})/59$ | | |
| (allow biased: $0.1420\ or\ 0.3768^2$) $s_A^2 = 0.1449\ or\ 71/490$ | | |
| $or\ 0.3807^2$ | | |
| $and\ 0.0825\ or\ 0.2872^2$) $and\ s_B^2 = 0.0839\ or\ 99/1180$ | A1 | |
| $or\ 0.2897^2$ | | |
| **EITHER:** Estimate combined variance: $s^2 = \frac{s_A^2}{50} + \frac{s_B^2}{60}$ | M1 | |
| (to 3 s.f.) $= 0.004296\ or\ 0.06555^2$ | A1 | |
| Calculate value of $z$ (or $-z$): $z = (\bar{t}_B - \bar{t}_A)/s$ | | |
| (to 3 s.f.) $= (2.15 - 2.04)/s = 1.68$ | M1 A1 | |
| **OR:** Pooled estimate of common variance: $s^2 = (49s_A^2 + 59s_B^2)/108$ | (M1 A1) | note $s_A^2$ and $s_B^2$ not needed explicitly |
| $or\ (215.18 - \frac{102^2}{50} + 282.3 - \frac{129^2}{60})/108$ | | |
| (to 3 s.f.) $= 0.1116\ or\ 0.334^2$ | | |
| Calculate value of $z$ (or $-z$): $z = (\bar{t}_B - \bar{t}_A)/s\ \sqrt{1/50 + 1/60}$ | (M1 A1) | |
| (to 3 s.f.) $= 0.11/0.06396 = 1.72$ | | |
| State or use correct tabular $z$ value (to 3 s.f.): $z_{0.95} = 1.645$ (allow 1.658) | *B1 | |
| (or can compare $\bar{t}_B - \bar{t}_A = 0.11$ with $0.105$) | | |
| Correct conclusion (AEF, ft on $z$, dep *B1): $z >$ tabular value, so club $A$ takes less time | B1 | **10 marks** |
| Calculate value of $z$ (either sign): $z = (0.11 - 0.05)/s = 0.06/s$ | M1 A1 | |
| (to 3 s.f.) $= 0.9154\ [or\ 0.9381]$ | | |
| Find $\Phi(z)$ and set of possible values of $\alpha$: $\Phi(z) = 0.820\ [or\ 0.826]$ | | M0 for $\alpha \sim 82$ or $0.82$ (to 1 d.p.) |
| $\alpha \geq (or >) 18.0\ [or\ 17.4]$ | M1 A1 | **4 marks** |
| **SR** Allow B1 (max 3/4) for $\alpha < 18.0\ [or\ 17.4]$ | | **Total: 14** |