## Question 10:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = ae^{-kx}$; $k = \ln 2$ | M1; A1 | Replace $2^x$ by $e^{kx}$ to find $k$ |
| $[-(a/k)e^{-kx}]_0^\infty = 1$, $a = k$ or $\ln 2$ | M1 A1 | Show $a = k$ by e.g. $\int_0^\infty f(x) = 1$ |
| $E(X) = 1/\ln 2$ or $1.44$ | B1 | State value of $E(X)$; **Part total: 4 then 1** |
| $G(y) = P(Y \leq y) = P(X \leq k^{-1}\ln y)$ | M1 A1 | Find distribution fn $G$ of $Y$ |
| $= F(k^{-1}\ln y) = (a/k)(1 - e^{-\ln y})$ | M1 A1 | |
| $= 1 - 1/y$ (CAO) | A1 | |
| $g(y) = 1/y^2$ (CAO) | M1 A1 | Find probability density function $g$ of $Y$ |
| $y \geq 1$ | B1 | State interval for either $G$ or $g$; **Part total: 8** |
| | | **Total: 13** |

# Question 11a:

**EITHER approach:**

Observe or deduce when $R_B$ is maximised: $R_{B,max}$ occurs when dog at $B$ | M1 |

Moments for ladder about $A$: $6a R_{B,max} = 4aW + 8a \times \frac{1}{4}W$ | M1 |

**OR approach:**

Moments when dog is $x$ hor. from wall: $6a R_B = 4aW + (8a-x)\frac{1}{4}W$ | (M1) |

Deduce limit on $R_B$: $R_{B,max}$ occurs when dog at $B$ | (M1) |

Find max. value of $R_B$: $R_{B,max} = W$ | A1 |

Resolve horizontally for ladder $AB$: $F_A = R_B$ | B1 |

Resolve vertically for ladder $AB$: $R_A = W + \frac{1}{4}W\ [= \frac{5W}{4}]$ | B1 |

Find bound on $\mu$ from $F_A \leq \mu R_A$: $\mu \geq F_A / R_A \geq R_{B,max} / (\frac{5W}{4})$ | M1 |

$\mu \geq \frac{4}{5}$ **A.G.** | A1 | **7 marks**

Find friction $F_{cube}$ along $DE$ by hor. resolution: $F_{cube} = F_A\ or\ R_B$ | B1 |

Find reaction $R_{cube}$ from floor by vert. resolution: $R_{cube} = W + \frac{1}{4}W + kW$ | B1 |

Show that $F_{cube} \leq \mu R_{cube}$: $\mu R_{cube} \geq W + \frac{4kW}{5} \geq W$ | |

$= R_{B,max} \geq F_{cube}$ | M1 A1 | **4 marks**

Find moments about $D$ opposing effect of $R_{cube}$: $2akW + \frac{a5W}{4} - 4a F_A$ | M1 |

Find smallest value of $k$ for which moments $\geq 0$: $\frac{(4W - 5W/4)}{2W} = \frac{11}{8}$ | M1 A1 | **3 marks** | **[14]**

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# Question 11b:

State hypotheses: $H_0: \mu_2 = \mu_1,\quad H_1: \mu_2 > \mu_1$ | B1 |

Calculate $\sum(x_i - \bar{x})^2$ (M1 for either): $8.24,\ 4.62[4]$ | M1 |

*or* estimate variances: $0.168\ or\ 0.165,\ 0.0784\ or\ 0.0771$ | M1 A1 A1 |

Find $s^2$ (M0 if inconsistent denominators used): $s^2 = \frac{0.168}{50} + \frac{0.0784}{60}$

$or\ \frac{0.165}{49} + \frac{0.0771}{59}$

$[= 0.00467]$ | *M1 |

Find value of $z$ (dep *M1): $z = \frac{(1803.6/60 - 1492.0/50)}{s}$ | M1 |

$= \frac{(30.06 - 29.84)}{0.0683} = 3.22$ | *A1 |

**S.R.** Using pooled estimate of variance: $s^2 = \frac{(8.24 + 4.62)}{108} = 0.119$ | (M0) |

$z = \frac{0.22}{s\sqrt{(1/50 + 1/60)}} = 3.33$ | (B1) |

Find tabular value (to 2 dp): $\Phi^{-1}(0.98) = 2.05[4]$ | *B1 |

Compare values for conclusion (A1 dep *A1, *B1): $\mu_2 > \mu_1$ (A.E.F., M1$\sqrt{}$ on values) | M1$\sqrt{}$ A1 | **10 marks**

Find limiting value of $z$ (to 2 dp): $z = \frac{(0.22 - 0.1)}{s} = 1.756$ | M1 A1 |

Find $\Phi(z)$ and hence values of $\alpha$ (to 1 dp): $\Phi(z) = 0.9604,\ \alpha \geq 3.9\ or\ 4.0$ | M1 A1 |

$s^2 = 0.119$ gives: $z = 1.816,\ \alpha \geq 3.47$ | (M1 M1) | **4 marks** | **[14]**