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2014 Paper 1 Q13
D: 1500.0 B: 1483.3
continuous probability distributions triangular distribution probability density function piecewise linear function mean median integration cases

A continuous random variable \(X\) has a triangular distribution, which means that it has a probability density function of the form \[ \f(x) = \begin{cases} \g(x) & \text{for \(a< x \le c\)} \\ \h(x) & \text{for \(c\le x < b\)} \\ 0 & \text{otherwise,} \end{cases} \] where \(\g(x)\) is an increasing linear function with \(\g(a)=0\), \(\h(x)\) is a decreasing linear function with \(\h(b) =0\), and \(\g(c)=\h(c)\). Show that \(\g(x) = \dfrac{2(x-a)}{(b-a)(c-a)}\) and find a similar expression for \(\h(x)\).

  1. Show that the mean of the distribution is \(\frac13(a+b+c)\).
  2. Find the median of the distribution in the different cases that arise.

Solution: Since \(\int f(x) \, dx = 1\), and \(f(x)\) is a triangle with base \(b-a\), it must have height \(\frac{2}{b-a}\) in order to have the desired area. Since \(g(a) = 0, g(c) = \frac{2}{b-a}\), \(g(x) = A(x-a)\) and \(\frac{2}{b-a} = A (c-a) \Rightarrow g(x) = \frac{2(x-a)}{(b-a)(c-a)}\) as required. Similarly, \(h(x) = B(x-b)\) and \(\frac{2}{b-a} = B(c-b) \Rightarrow h(x) = \frac{2(b-x)}{(b-a)(b-c)}\) The mean of the distribution will be: \begin{align*} \int_a^b xf(x) \, dx &= \int_a^c xg(x) \, dx + \int_c^b xh(x) \, dx \\ &= \frac{2}{(b-a)(c-a)} \int_a^c x(x-a) dx + \frac{2}{(b-a)(b-c)} \int_c^b x(b-x) \, dx \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \left [ \frac{x^3}{3} - a\frac{x^2}{2} \right ]_a^c + \frac{1}{b-c} \left [ b\frac{x^2}{2} - \frac{x^3}{3} \right ]_c^b\r \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \l \frac{c^3}{3} - a\frac{c^2}{2} - \frac{a^3}{3} + \frac{a^3}{2} \r + \frac{1}{b-c} \l \frac{b^3}{2} - \frac{b^3}{3} - \frac{bc^2}{2} + \frac{c^3}{3} \r \r \\ &= \frac{2}{(b-a)} \l \l \frac{c^2+ac+a^2}{3} - \frac{a(a+c)}{2} \r +\l \frac{b(b+c)}{2} - \frac{b^2+bc+c^2}{3} \r\r \\ &= \frac{2}{(b-a)} \l \frac{2c^2+2ac+2a^2}{6} - \frac{3a^2+3ac}{6} + \frac{3b^2+3bc}{6} - \frac{2b^2+2bc+2c^2}{6} \r \\ &= \frac{2}{(b-a)} \l \frac{-a^2+b^2-ac+bc}{6} \r \\ &= \frac{a+b+c}{3} \\ \end{align*} The median \(M\) satisfies: \begin{align*} && \int_a^M f(x) \, dx &= \frac12 \\ \end{align*} The left hand triangle will have area: \(\frac{c-a}{b-a}\) which will be \(\geq \frac12\) if \(c \geq \frac{a+b}{2}\). In this case we need \begin{align*} && \frac{(M-a)^2}{(b-a)(c-a)} &= \frac12 \\ \Rightarrow && M &= a + \sqrt{\frac12 (b-a)(c-a)} \end{align*} Otherwise, we need: \begin{align*} && \frac{(b-M)^2}{(b-a)(b-c)} &= \frac12 \\ \Rightarrow && M &= b - \sqrt{\frac12 (b-a)(b-c)} \end{align*} These are consistent, if \(c = \frac{b+a}{2}\)

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2011 Paper 1 Q13
D: 1484.0 B: 1471.5
probability continuous probability distribution probability density function integration mean quantile piecewise function arcsin

In this question, you may use without proof the following result: \[ \int \sqrt{4-x^2}\, \d x = 2 \arcsin (\tfrac12 x ) + \tfrac 12 x \sqrt{4-x^2} +c\,. \] A random variable \(X\) has probability density function \(\f\) given by \[ \f(x) = \begin{cases} 2k & -a\le x <0 \\[3mm] k\sqrt{4-x^2} & \phantom{-} 0\le x \le 2 \\[3mm] 0 & \phantom{-}\text{otherwise}, \end{cases} \] where \(k\) and \(a\) are positive constants.

  1. Find, in terms of \(a\), the mean of \(X\).
  2. Let \(d\) be the value of \(X\) such that \(\P(X> d)=\frac1 {10}\,\). Show that \(d < 0\) if \(2a> 9\pi\) and find an expression for \(d\) in terms of \(a\) in this case.
  3. Given that \(d=\sqrt 2\), find \(a\).

Solution: First notice that \begin{align*} && 1 &= \int_{-a}^2 f(x) \d x \\ &&&= 2ka + k\pi \\ \Rightarrow && k &= (\pi + 2a)^{-1} \end{align*}

  1. \(\,\) \begin{align*} && \E[X] &= \int_{-a}^2 x f(x) \d x\\ &&&= \int_{-a}^0 2kx \d x + k\int_0^{2} x\sqrt{4-x^2} \d x\\ &&&= \left [kx^2 \right]_{-a}^0 +k \left [-\frac13(4-x^2)^{\frac32} \right]_0^2 \\ &&&= -ka^2 + \frac83k \\ &&&= \frac{\frac83-a^2}{\pi + 2a} \end{align*}
  2. Consider \(\mathbb{P}(X < 0)\) then \(d < 0 \Leftrightarrow \mathbb{P}(X < 0) > \frac{9}{10}\) \begin{align*} && \frac{9}{10} &< \mathbb{P}(X < 0) \\ &&&= \int_{-a}^0 2k \d x \\ &&&= \frac{2a}{\pi+2a} \\ \Leftrightarrow && 9\pi &< 2a \\ \\ && \frac{9}{10} &= \int_{-a}^d 2k \d x \\ &&&= \frac{2(d+a)}{\pi + 2a} \\ \Rightarrow && 9\pi &= 2a + 20d \\ \Rightarrow && d &= \frac{2a-9\pi}{20} \end{align*}
  3. Suppose \(d=\sqrt 2\) then \begin{align*} && \frac1{10} &= \int_{\sqrt{2}}^2 f(x) \d x \\ &&&= \int_{\sqrt{2}}^2 k\sqrt{4-x^2} \d x \\ &&&= k\left [ 2 \sin^{-1} \tfrac12 x + \tfrac12 x \sqrt{4-x^2}\right]_{\sqrt{2}}^2 \\ &&&= k\left (\pi -\frac{\pi}{2} - 1 \right) \\ \Rightarrow && \pi + 2a &= 5\pi - 10 \\ \Rightarrow && a &= 2\pi-5 \end{align*}

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2001 Paper 2 Q13
D: 1600.0 B: 1517.3
continuous probability distributions median mean survival function order statistics parameter estimation conditional probability Pareto distribution memoryless property

The life times of a large batch of electric light bulbs are independently and identically distributed. The probability that the life time, \(T\) hours, of a given light bulb is greater than \(t\) hours is given by \[ \P(T>t) \; = \; \frac{1}{(1+kt)^\alpha}\;, \] where \(\alpha\) and \(k\) are constants, and \(\alpha >1\). Find the median \(M\) and the mean \(m\) of \(T\) in terms of \(\alpha\) and \(k\). Nine randomly selected bulbs are switched on simultaneously and are left until all have failed. The fifth failure occurs at 1000 hours and the mean life time of all the bulbs is found to be 2400 hours. Show that \(\alpha\approx2\) and find the approximate value of \(k\). Hence estimate the probability that, if a randomly selected bulb is found to last \(M\) hours, it will last a further \(m-M\) hours.

Solution: The median \(M\) is the value such that \begin{align*} && \frac12 &= \mathbb{P}(T > M) \\ &&&= \frac1{(1+kM)^\alpha} \\ \Rightarrow && 2 &= (1+kM)^{\alpha} \\ \Rightarrow && M &= \frac{2^{1/\alpha}-1}{k} \end{align*} The distribution of \(T\) is \(f_T(t) = \frac{k \alpha}{(1+kt)^{\alpha+1}}\) and so \begin{align*} && m &= \int_0^\infty t f_T(t) \d t \\ &&&= \int_0^\infty \frac{tk \alpha}{(1+kt)^{\alpha+1}} \d t \\ &&&= \int_0^\infty \frac{\alpha+tk \alpha-\alpha}{(1+kt)^{\alpha+1}} \d t \\ &&&= \alpha \int_0^\infty (1+kt)^{-\alpha} \d t - \alpha \int_0^\infty (1+kt)^{-(\alpha+1)} \d t \\ &&&= \alpha \left [ -\frac1{k(\alpha-1)}(1+kt)^{-\alpha+1}\right]_0^\infty- \alpha \left [ -\frac1{k\alpha}(1+kt)^{-\alpha}\right]_0^\infty \\ &&&= \frac{\alpha}{k(\alpha-1)} - \frac{1}{k} \\ &&&= \frac{1}{k(\alpha-1)} \end{align*} \begin{align*} && \frac{2^{1/\alpha}-1}{k} &= 1000 \\ && \frac{1}{k(\alpha-1)} &= 2400 \\ \Rightarrow && \frac{\alpha-1}{2^{1/\alpha}-1} &\approx 2.4 \\ && \frac{2-1}{\sqrt2-1} &= \sqrt{2}+1 \approx 2.4 \\ \Rightarrow && \alpha &\approx 2 \\ && k &= \frac{1}{2400} \end{align*} \begin{align*} && \mathbb{P}(T > m | T > M) &= \frac{\mathbb{P}(T > m)}{\mathbb{P}(T > M)} \\ &&&= \frac{2}{(1+km)^{\alpha}} \\ &&&= \frac{2}{(1 + \frac{1}{\alpha-1})^\alpha} \\ &&&\approx \frac{2}{4} =\frac12 \end{align*}

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