Year: 2017
Paper: 2
Question Number: 12
Course: UFM Statistics
Section: Poisson Distribution
This year's paper was, perhaps, slightly more straightforward than usual, with more helpful guidance offered in some of the questions. Thus the mark required for a "1", a Distinction, was 80 (out of 120), around ten marks higher than that which would customarily be required to be awarded this grade. Nonetheless, a three‐figure mark is still a considerable achievement and, of the 1330 candidates sitting the paper, there were 89 who achieved this. At the other end of the scale, there were over 350 who scored 40 or below, including almost 150 who failed to exceed a total score of 25. As a general strategy for success in a STEP examination, candidates should be looking to find four "good" questions to work at (which may be chosen freely by the candidates from a total of 13 questions overall). It is unfortunately the case that so many low‐scoring candidates flit from one question to another, barely starting each one before moving on. There needs to be a willingness to persevere with a question until a measure of understanding as to the nature of the question's purpose and direction begins to emerge. Many low‐scoring candidates fail to deal with those parts of questions which cover routine mathematical processes ‐ processes that should be standard for an A‐level candidate. The significance of the "rule of four" is that four high‐scoring questions (15‐20 marks apiece) obtains you up to around the total of 70 that is usually required for a "1"; and with a couple of supporting starts to questions, such a total should not be beyond a good candidate who has prepared adequately. This year, significantly more than 10% of candidates failed to score at least half marks on any one question; and, given that Q1 (and often Q2 also) is (are) specifically set to give all candidates the opportunity to secure some marks, this indicates that these candidates are giving up too easily. Mathematics is about more than just getting to correct answers. It is about communicating clearly and precisely. Particularly with "show that" questions, candidates need to distinguish themselves from those who are just tracking back from given results. They should also be aware that convincing themselves is not sufficient, and if they are using a result from 3 pages earlier, they should make this clear in their working. A few specifics: In answers to mechanics questions, clarity of diagrams would have helped many students. If new variables or functions are introduced, it is important that students clearly define them. One area which is very important in STEP but which was very poorly done is dealing with inequalities. Although a wide range of approaches such as perturbation theory were attempted, at STEP level having a good understanding of the basics – such as changing the inequality if multiplying by a negative number – is more than enough. In fact, candidates who used more advanced methods rarely succeeded.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1563.6
Banger Comparisons: 7
Adam and Eve are catching fish. The number of fish, $X$, that Adam catches in any time interval is Poisson distributed with parameter $\lambda t$, where $\lambda$ is a constant and $t$ is the length of the time interval. The number of fish, $Y$, that Eve catches in any time interval is Poisson distributed with parameter $\mu t$, where $\mu$ is a constant and $t$ is the length of the time interval
The two Poisson variables are independent. You may assume that the expected time between Adam catching a fish and Adam catching his next fish is $\lambda^{-1}$, and similarly for Eve.
\begin{questionparts}
\item By considering $\P( X + Y = r)$, show that the total number of fish caught by Adam and Eve in time $T$ also has a Poisson distribution.
\item Given that Adam and Eve catch a total of $k$ fish in time $T$, where $k$ is fixed, show that the number caught by Adam has a binomial distribution.
\item Given that Adam and Eve start fishing at the same time, find the probability that the first fish is caught by Adam.
\item Find the expected time from the moment Adam and Eve start fishing until they have each caught at least one fish.
\end{questionparts}
[\textbf{Note} This question has been redrafted to make the meaning clearer.]\begin{questionparts}
\item $\,$ \begin{align*}
&& \mathbb{P}(X+Y=r) &= \sum_{k=0}^r \mathbb{P}(X = k, Y = r-k) \\
&&&= \sum_{k=0}^r \mathbb{P}(X = k)\mathbb{P}( Y = r-k) \\
&&&= \sum_{k=0}^r \frac{e^{-\lambda T} (\lambda T)^k}{k!}\frac{e^{-\mu T} (\mu T)^{r-k}}{(r-k)!}\\
&&&= \frac{e^{-(\mu+\lambda)T}}{r!}\sum_{k=0}^r \binom{r}{k}(\lambda T)^k (\mu T)^{r-k}\\
&&&= \frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^r}{r!}
\end{align*}
Therefore $X+Y \sim Po \left ( (\mu+\lambda)T \right)$
\item $\,$ \begin{align*}
&& \mathbb{P}(X = r | X+Y = k) &= \frac{\mathbb{P}(X=r, Y = k-r)}{\mathbb{P}(X+Y=k)} \\
&&&= \frac{\frac{e^{-\lambda T} (\lambda T)^r}{r!}\frac{e^{-\mu T} (\mu T)^{k-r}}{(k-r)!}}{\frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^k}{k!}} \\
&&&= \binom{k}{r} \left ( \frac{\lambda}{\lambda + \mu} \right)^r \left ( \frac{\mu}{\lambda + \mu} \right)^{k-r}
\end{align*}
Therefore $X|X+Y=k \sim B(k, \frac{\lambda}{\lambda + \mu})$
\item $P(X=1|X+Y = 1) = \frac{\lambda}{\lambda + \mu}$
\item Let $X_1, Y_1$ be the time to the first fish are caught by Adam and Eve, then
\begin{align*}
&& \mathbb{P}(X_1, Y_1 > t) &= \mathbb{P}(X_1> t) \mathbb{P}( Y_1 > t) \\
&&&= e^{-\lambda t}e^{-\mu t} \\
&&&= e^{-(\lambda+\mu)t} \\
\Rightarrow && f_{\max(X_1,Y_1)}(t) &= (\lambda+\mu)e^{-(\lambda+\mu)}
\end{align*}
Therefore the expected time is $\frac1{\mu+\lambda}$
\end{questionparts}
The statistics questions were attempted by only a small fraction of the cohort, with question 12 the least popular question on the paper, receiving fewer than 200 attempts, of which only about a third made any meaningful progress. Although the question had a small wording ambiguity this did not seem to have bothered any but a very few candidates. It was disappointing that even the fairly standard analysis in part (i) proved difficult for most candidates, with several claiming that the mean and the variance being equal was a sufficient condition for to follow a Poisson distribution.