Year: 2024
Paper: 3
Question Number: 12
Course: LFM Stats And Pure
Section: Continuous Probability Distributions and Random Variables
No solution available for this problem.
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item A point is chosen at random in the square $0 \leqslant x \leqslant 1$, $0 \leqslant y \leqslant 1$, so that the probability that a point lies in any region is equal to the area of that region. $R$ is the random variable giving the distance of the point from the origin.
Show that the cumulative distribution function of $R$ is given by
\[\mathrm{P}(R \leqslant r) = \sqrt{r^2 - 1} + \tfrac{1}{4}\pi r^2 - r^2 \cos^{-1}(r^{-1}),\]
when $1 \leqslant r \leqslant \sqrt{2}$. What is the cumulative distribution function when $0 \leqslant r \leqslant 1$?
\item Show that $\displaystyle\mathrm{E}(R) = \frac{2}{3}\int_1^{\sqrt{2}} \frac{r^2}{\sqrt{r^2-1}}\,\mathrm{d}r$.
\item Show further that $\mathrm{E}(R) = \frac{1}{3}\Bigl(\sqrt{2} + \ln\bigl(\sqrt{2}+1\bigr)\Bigr)$.
\end{questionparts}
A little over one fifth of the candidates attempted this, marginally less successfully than question 11 with a mean score of 10 marks. As with question 11, a significant number of candidates gained full or close to full credit. In the main, there was a dichotomy in student responses: for each of the parts, students were generally either unable to make any real progress with that part question or were able to produce a relatively full solution. Parts (i) and (ii) were generally well done, although quite a common error was to incorrectly differentiate the cumulative distribution function from (i) to find the probability distribution required for (ii). Another quite common error was attempting to use integration by change of variable rather than by parts to evaluate ∫r²cos⁻²(r⁻¹)dr in (ii). A number of students only attempted part (iii) of the question, in many of these cases, gaining full or close to full marks. For this part, by far the most common approach was to use the substitution r = cosh u to evaluate the integral. However, other solutions were also seen. Various different substitutions were used either successfully, or at least in some way productively, to evaluate the integral, including r = sin u, r = cosec u, r = cosh u, the double substitution u = √(r²−1) followed by u = sinh x, and the double substitution r = sin u followed by x = sin u. However, it was uncommon to see an unproductive substitution such as u = r²−1.