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1992 Paper 1 Q14
D: 1500.0 B: 1484.8
central limit theorem Poisson distribution hypothesis testing statistical interpretation normal approximation critical analysis rates and ratios newspaper headlines

The average number of pedestrians killed annually in road accidents in Poldavia during the period 1974-1989 was 1080 and the average number killed annually in commercial flight accidents during the same period was 180. Discuss the following newspaper headlines which appeared in 1991. (The percentage figures in square brackets give a rough indication of the weight of marks attached to each discussion.)

  1. [\(10\%\)] Six Times Safer To Fly Than To Walk. 1974-1989 Figures Prove It.
  2. [\(10\%\)] Our Skies Are Safer. Only 125 People Killed In Air Accidents In 1990.
  3. [\(30\%\)] Road Carnage Increasing. 7 People Killed On Tuesday.
  4. [\(50\%\)] Alarming Rise In Pedestrian Casualties. 1350 Pedestrians Killed In Road Accidents During 1990.

Solution:

  1. We cannot say this, since we do not know how many people were flying or walking each year.
  2. This is difficult to say without knowing the variance. We might expect this to have quite a skewed distribution (one big air crash causes lots of deaths infrequently) so it's impossible to know, although it is substantially lower.
  3. If we have 1080 deaths annually, we should expect ~3 deaths per day. While a day with \(7\) deaths might seem unlikely, over the course of a year it is very likely to occur. (Perhaps the weather was bad). It is also probably a case of selective reporting, we are seeing this data point because it's notable and being reported rather than because it is significant).
  4. This is certainly the most alarming, a ~25% increase is very unlikely without something else going on. (We'd expect it to be ~Po(1080) approximalely N(1080, 1080) but then this is many standard deviations away). However we also know that other factors could drive this (more walking, more people, change in reporting etc)

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1991 Paper 1 Q6
D: 1500.0 B: 1484.8
curve sketching stationary points rational function critical analysis integration discontinuity improper integral chain rule power rule

Criticise each step of the following arguments. You should correct the arguments where necessary and possible, and say (with justification) whether you think the conclusion are true even though the argument is incorrect.

  1. The function \(g\) defined by \[ \mathrm{g}(x)=\frac{2x^{3}+3}{x^{4}+4} \] satisfies \(\mathrm{g}'(x)=0\) only for \(x=0\) or \(x=\pm1.\) Hence the stationary values are given by \(x=0\), \(\mathrm{g}(x)=\frac{3}{4}\) and \(x=\pm1,\) \(\mathrm{g}(x)=1.\) Since \(\frac{3}{4}<1,\) there is a minimum at \(x=0\) and maxima at \(x=\pm1.\) Thus we must have \(\frac{3}{4}\leqslant\mathrm{g}(x)\leqslant1\) for all \(x\).
  2. \({\displaystyle \int(1-x)^{-3}\,\mathrm{d}x=-3(1-x)^{-4}}\quad\) and so \(\quad{\displaystyle \int_{-1}^{3}(1-x)^{-3}\,\mathrm{d}x=0.}\)

Solution:

  1. \begin{align*} && g(x) &= \frac{2x^3+3}{x^4+4} \\ \Rightarrow && g'(x) &= \frac{6x^2(x^4+4) - (2x^3+3)(4x^3)}{(x^4+4)^2} \\ &&&= \frac{-2x^6-12x^3+24x^2}{(x^4+4)} \\ &&&= \frac{-2x^2(x^4+6x-12)}{(x^4+4)} \end{align*} So \(g'(x)\) is not \(0\) for \(x = \pm 1\). We can also note that \(g(-1) = \frac1{5} \neq 1\) Even if the other turning point was \(1\), we would also need to check the behaviour as \(x \to \pm \infty\). We can also note that \(g(-1) = \frac{1}{5} < \frac34\) so the conclusion is also not true.
  2. There are several errors. \[ \int (1-x)^{-3} \d x = \underbrace{\frac{1}{4}}_{\text{correct constant}}(1-x)^{-4} + \underbrace{C}_{\text{constant of integration}} \] We cannot integrate through the asymptote at \(1\). There is a sense in which we could argue \(\displaystyle \int_{-1}^3 (1-x)^{-3} \d x = 0\), specifically using Cauchy principal value \begin{align*} \mathrm {p.v.} \int_{-1}^3 (1-x)^{-3} &=\lim_{\epsilon \to 0} \left [ \int_{-1}^{1-\epsilon} (1-x)^{-3} \d x+ \int_{1+\epsilon}^{3} (1-x)^{-3} \d x\right] \\ &=\lim_{\epsilon \to 0} \left [ \left[ \frac14 (1-x)^{-4}\right]_{-1}^{1-\epsilon}+ \left[ \frac14 (1-x)^{-4}\right]_{1+\epsilon}^3\right] \\ &=\lim_{\epsilon \to 0} \left [ \frac14 \epsilon^{-4}-\frac14 \frac1{2^4} + \frac14 \frac1{2^4} - \frac14 \epsilon^{-4} \right] \\ &= \lim_{\epsilon \to 0} 0 \\ &= 0 \end{align*} However, in many normal ways of treating this integral it would be undefined.

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