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2017 Paper 3 Q11
D: 1700.0 B: 1484.0

A railway truck, initially at rest, can move forwards without friction on a long straight \mbox{horizontal} track. On the truck, \(n\) guns are mounted parallel to the track and facing backwards, where \(n>1\). Each of the guns is loaded with a single projectile of mass \(m\). The mass of the truck and guns (but not including the projectiles) is \(M\). When a gun is fired, the projectile leaves its muzzle horizontally with a speed \(v-V\) relative to the ground, where~\(V\) is the speed of the truck immediately before the gun is fired.

  1. All \(n\) guns are fired simultaneously. Find the speed, \(u\), with which the truck moves, and show that the kinetic energy, \(K\), which is gained by the system (truck, guns and projectiles) is given by \[ K= \tfrac{1}{2}nmv^2\left(1 +\frac{nm}{M} \right) . \]
  2. Instead, the guns are fired one at a time. Let \(u_r\) be the speed of the truck when \(r\) guns have been fired, so that \(u_0= 0\). Show that, for \(1\le r \le n\,\), \[ u_r - u_{r-1} = \frac{mv}{M+(n-r)m} \tag{\(*\)} \] and hence that \(u_n < u\,\).
  3. Let \(K_r\) be the total kinetic energy of the system when \(r\) guns have been fired (one at a time), so that \(K_0 = 0\). Using \((*)\), show that, for \(1\le r\le n\,\), \[ K_r -K_{r-1} = \tfrac 12 mv^2 + \tfrac12 mv (u_r-u_{r-1}) \] and hence show that \[ K_n = \tfrac{1}{2}nmv^2 +\tfrac{1}{2}mvu_n \,. \] Deduce that \(K_n

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2017 Paper 3 Q12
D: 1700.0 B: 1500.2

The discrete random variables \(X\) and \(Y\) can each take the values \(1\), \(\ldots\,\), \(n\) (where \(n\ge2\)). Their joint probability distribution is given by \[ \P(X=x, \ Y=y) = k(x+y) \,, \] where \(k\) is a constant.

  1. Show that \[ \P(X=x) = \dfrac{n+1+2x}{2n(n+1)}\,. \] Hence determine whether \(X\) and \(Y\) are independent.
  2. Show that the covariance of \(X\) and \(Y\) is negative.

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(X = x) &= \sum_{y=1}^n \mathbb{P}(X=x,Y=y) \\ &&&= \sum_{y=1}^n k(x+y) \\ &&&= nkx + k\frac{n(n+1)}2 \\ \\ && 1 &= \sum_{x=1}^n \mathbb{P}(X=x) \\ &&&= nk\frac{n(n+1)}{2} + kn\frac{n(n+1)}2 \\ &&&= kn^2(n+1) \\ \Rightarrow && k &= \frac{1}{n^2(n+1)} \\ \Rightarrow && \mathbb{P}(X = x) &= \frac{nx}{n^2(n+1)} + \frac{n(n+1)}{2n^2(n+1)} \\ &&&= \frac{n+1+2x}{2n(n+1)} \\ \\ && \mathbb{P}(X=x)\mathbb{P}(Y=y) &= \frac{(n+1)^2+2(n+1)(x+y)+4xy}{4n^2(n+1)^2} \\ &&&\neq \frac{x+y}{n^2(n+1)} \end{align*} Therefore \(X\) and \(Y\) are not independent.
  2. \(\,\) \begin{align*} && \E[X] &= \sum_{x=1}^n x \mathbb{P}(X=x) \\ &&&= \sum_{x=1}^n x \mathbb{P}(X=x)\\ &&&= \sum_{x=1}^n x \frac{n+1+2x}{2n(n+1)} \\ &&&= \frac{1}{2n(n+1)} \left ( (n+1) \sum x + 2\sum x^2\right)\\ &&&= \frac{1}{2n(n+1)} \left ( \frac{n(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{3} \right) \\ &&&= \frac{1}{2} \left ( \frac{n+1}{2} + \frac{2n+1}{3} \right)\\ &&&= \frac{1}{2} \left ( \frac{7n+5}{6} \right)\\ &&&= \frac{7n+5}{12} \\ \\ && \textrm{Cov}(X,Y) &= \mathbb{E}\left[XY\right] - \E[X] \E[Y] \\ &&&= \sum_{x=1}^n \sum_{y=1}^n xy \frac{x+y}{n^2(n+1)} - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \sum \sum (x^2 y+xy^2) - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \left (\sum y \right )\left (\sum x^2\right ) - \E[X]^2 \\ &&&=\frac{(n+1)(2n+1)}{12} - \left ( \frac{7n+5}{12}\right)^2 \\ &&&= \frac1{144} \left (12(2n^2+3n+1) - (49n^2+70n+25) \right)\\ &&&= \frac{1}{144} \left (-25n^2-34n-13 \right) \\ &&& < 0 \end{align*} since \(\Delta = 34^2 - 4 \cdot 25 \cdot 13 = 4(17^2-25 \times 13) = -4 \cdot 36 < 0\)

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2017 Paper 3 Q13
D: 1700.0 B: 1500.0

The random variable \(X\) has mean \(\mu\) and variance \(\sigma^2\), and the function \({\rm V}\) is defined, for \(-\infty < x < \infty\), by \[ {\rm V}(x) = \E \big( (X-x)^2\big) . \] Express \({\rm V}(x)\) in terms of \(x\), \( \mu\) and \(\sigma\). The random variable \(Y\) is defined by \(Y={\rm V}(X)\). Show that \[ \E(Y) = 2 \sigma^2 %\text{ \ \ and \ \ } %\Var(Y) = \E(X-\mu)^4 -\sigma^4 . \tag{\(*\)} \] Now suppose that \(X\) is uniformly distributed on the interval \(0\le x \le1\,\). Find \({\rm V}(x)\,\). Find also the probability density function of \(Y\!\) and use it to verify that \((*)\) holds in this case.

Solution: \begin{align*} {\rm V}(x) &= \E \big( (X-x)^2\big) \\ &= \E \l X^2 - 2xX + x^2\r \\ &= \E [ X^2 ]- 2x\E[X] + x^2 \\ &= \sigma^2+\mu^2 - 2x\mu + x^2 \\ &= \sigma^2 + (\mu - x)^2 \end{align*} \begin{align*} \E[Y] &= \E[\sigma^2 + (\mu - X)^2] \\ &= \sigma^2 + \E[(\mu - X)^2]\\ &= \sigma^2 + \sigma^2 \\ &= 2\sigma^2 \end{align*} If \(X \sim U(0,1)\) then \(V(x) = \frac{1}{12} + (\frac12 - x)^2\). \begin{align*} \P(Y \leq y) &= \P(\frac1{12} + (\frac12 - X)^2 \leq y) \\ &= \P((\frac12 -X)^2 \leq y - \frac1{12}) \\ &= \P(|\frac12 -X| \leq \sqrt{y - \frac1{12}}) \\ &= \begin{cases} 1 & \text{if } y - \frac1{12} > \frac14 \\ 2 \sqrt{y - \frac1{12}} & \text{if } \frac14 > y - \frac1{12} > 0 \\ \end{cases} \\ &= \begin{cases} 1 & \text{if } y> \frac13 \\ \sqrt{4y - \frac1{3}} & \text{if } \frac13 > y > \frac1{12} \\ \end{cases} \end{align*} Therefore $f_Y(y) = \begin{cases} \frac{2}{\sqrt{4y-\frac{1}{3}}} & \text{if } \frac1{12} < y < \frac13 \\ 0 & \text{otherwise} \end{cases}$ \begin{align*} \E[Y] &= \int_{1/12}^{1/3} \frac{2x}{\sqrt{4x-\frac13}} \, dx \\ &= 2\int_{u = 0}^{u=1} \frac{\frac{1}{4}u +\frac1{12}}{\sqrt{u}} \,\frac{1}{4} du \tag{\(u = 4x - \frac13, \frac{du}{dx} = 4\)}\\ &= \frac{1}{2 \cdot 12}\int_{u = 0}^{u=1} 3\sqrt{u} +\frac{1}{\sqrt{u}} \, du \\ &= \frac{1}{2 \cdot 12} \left [2 u^{3/2} + 2u^{1/2} \right ]_0^1 \\ &= \frac{1}{2 \cdot 12} \cdot 4 \\ &= \frac{2}{12} \end{align*} as required

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