Year: 2014
Paper: 1
Question Number: 13
Course: LFM Stats And Pure
Section: Continuous Probability Distributions and Random Variables
More than 1800 candidates sat this paper, which represents another increase in uptake for this STEP paper. The impression given, however, is that many of these extra candidates are just not sufficiently well prepared for questions which are not structured in the same way as are the A-level questions that they are, perhaps, more accustomed to seeing. Although STEP questions try to give all able candidates "a bit of an intro." into each question, they are not intended to be easy, and (at some point) imagination and real flair (as well as determination) are required if one is to score well on them. In general, it is simply not possible to get very far into a question without making some attempt to think about what is actually going on in the situation presented therein; and those students who expect to be told exactly what to do at each stage of a process are in for a shock. Too many candidates only attempt the first parts of many questions, restricting themselves to 3-6 marks on each, rather than trying to get to grips with substantial portions of work – the readiness to give up and try to find something else that is "easy pickings" seldom allows such candidates to acquire more than 40 marks (as was the case with almost half of this year's candidature, in fact). Poor preparation was strongly in evidence – curve-sketching skills were weak, inequalities very poorly handled, algebraic capabilities (especially in non-standard settings) were often pretty poor, and the ability to get to grips with extended bits of working lacking in the extreme; also, an unwillingness to be imaginative and creative, allied with a lack of thoroughness and attention to detail, made this a disappointing (and, possibly, very uncomfortable) experience for many of those students who took the paper. On the other side of the coin, there was a very pleasing number of candidates who produced exceptional pieces of work on 5 or 6 questions, and thus scored very highly indeed on the paper overall. Around 100 of them scored 90+ marks of the 120 available, and they should be very proud of their performance – it is a significant and noteworthy achievement.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1483.3
Banger Comparisons: 3
A continuous random variable $X$ has a \textit{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 $\h(x)$.
\begin{questionparts}
\item Show that the mean of the distribution is $\frac13(a+b+c)$.
\item Find the median of the distribution in the different cases that arise.
\end{questionparts}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 $c = \frac{b+a}{2}$
Q13 elicited 320 attempts, and marks were around 5½/20. In Q13, apart from the usual sign errors, most candidates correctly found g(x) and h(x). A popular approach for the mean was to throw the word "centroid" at the problem and hope that this sufficed. For the median, around half of attempts failed to consider the easy (symmetric) case when m = c = ½(a + b). The case c > ½(a + b), corresponding to when m lies under the first line-segment, was handled very well; the other case very poorly, since most candidates tried to work from the LHS up rather than from the RHS down. A very few realised that working down from the top actually made this just a "write down" by switching a for b suitably.