19 problems found
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*}
The random variables \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\) are independently and uniformly distributed on the interval \(0 \le x \le 1\). The random variable \(Y\) is defined to be the median of \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\). Given that the probability density function of \(Y\) is \(\g(y)\), where \[ \mathrm{g}(y)=\begin{cases} ky^{n}(1-y)^{n} & \mbox{ if }0\leqslant y\leqslant1\\ 0 & \mbox{ otherwise} \end{cases} \] use the result $$ \int_0^1 {y^{r}}{{(1-y)}^{s}}\,\d y = \frac{r!s!}{(r+s+1)!} $$ to show that \(k={(2n+1)!}/{{(n!)}^2}\), and evaluate \(\E(Y)\) and \({\rm Var}\,(Y)\). Hence show that, for any given positive number \(d\), the inequality $$ {\P\left({\vert {Y - 1/2} \vert} < {d/{\sqrt {n}}} \right)} < {\P\left({\vert {{\bar X} - 1/2} \vert} < {d/{\sqrt {n}}} \right)} $$ holds provided \(n\) is large enough, where \({\bar X}\) is the mean of \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\). [You may assume that \(Y\) and \(\bar X\) are normally distributed for large \(n\).]
A goat \(G\) lies in a square field \(OABC\) of side \(a\). It wanders randomly round its field, so that at any time the probability of its being in any given region is proportional to the area of this region. Write down the probability that its distance, \(R\), from \(O\) is less than \(r\) if \(0 < r\leqslant a,\) and show that if \(r\geqslant a\) the probability is \[ \left(\frac{r^{2}}{a^{2}}-1\right)^{\frac{1}{2}}+\frac{\pi r^{2}}{4a^{2}}-\frac{r^{2}}{a^{2}}\cos^{-1}\left(\frac{a}{r}\right). \] Find the median of \(R\) and probability density function of \(R\). The goat is then tethered to the corner \(O\) by a chain of length \(a\). Find the conditional probability that its distance from the fence \(OC\) is more than \(a/2\).
The continuous random variable \(X\) is uniformly distributed over the interval \([-c,c].\) Write down expressions for the probabilities that:
Solution: