Year: 2010
Paper: 1
Question Number: 13
Course: UFM Statistics
Section: Poisson Distribution
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1516.0
Banger Comparisons: 1
The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean $\lambda$ texts per hour. Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is
$p$, show that
\[
p\e^{2\lambda}-\e^{\lambda}+1=0.
\]
Given that $4p<1$, show that there are two positive values of $\lambda$ that satisfy this equation.
The number of texts that Mildred receives on each of her two mobile phones can be modelled by independent Poisson random variables with different means $\lambda_{1}$ and $\lambda_{2}$ texts per hour. Given that, for each phone, the probability that Mildred waits between 1 and 2 hours in the morning before she receives her first text is also $p$, find an expression for $\lambda_{1}+\lambda_{2}$
in terms of $p$. Find the probability, in terms of $p$, that she waits between 1 and 2 hours in the morning to receive her first text.Let $X_t$ be the number of texts he recieves before $t$ hours. So $X_t \sim P(t\lambda)$
\begin{align*}
&&\mathbb{P}(X_1 = 0 \, \cap \, X_2 > 0) &= e^{-\lambda} \cdot \left ( 1-e^{-\lambda}\right) = p \\
\Rightarrow && e^{2\lambda}p &= e^{\lambda} - 1 \\
\Rightarrow && 0 &= pe^{2\lambda}-e^{\lambda} + 1 \\
\Rightarrow && e^{\lambda} &= \frac{1 \pm \sqrt{1-4p}}{2p}
\end{align*}
Which clearly has two positive roots if $4p < 1$. We need to show both roots are $>1$. So considering the smaller one we are looking at:
\begin{align*}
&& \frac{1-\sqrt{1-4p}}{2p} & > 1 \\
\Leftrightarrow && 1-\sqrt{1-4p} &> 2p \\
\Leftrightarrow && 1-2p&> \sqrt{1-4p} \\
\Leftrightarrow && (1-2p)^2&> 1-4p \\
\Leftrightarrow && 1-4p+4p^2&> 1-4p \\
\end{align*}
which is clearly true.
We must have $e^{\lambda_1}\cdot e^{\lambda_2} = \frac{1}{p}$, so $\lambda_1 + \lambda_2 = -\ln p$ by considering the product of the roots in our quadratic. (Vieta).
Therefore the probability she waits between 1 and 2 hours in the morning is $e^{-(\lambda_1 + \lambda_2)} \cdot ( 1- e^{-(\lambda_1+\lambda_2)}) = p \cdot (1-p)$
This question was the most poorly answered on the paper, with over half of attempts scoring no marks. Nonetheless, most candidates were capable of writing down the pdf of a Poisson distribution, but only a minority understood that they needed to consider two different Poisson distributions to make any progress. Worse still, it was very common to see candidates writing things like: "Let X be the number of texts received. Then P(1 < X < 2) = · · · ." This shows a total lack of understanding of what the Poisson distribution is doing: there is no time period given in the definition of X, and how could the number of texts lie strictly between 1 and 2? Were the candidates to have let X be the waiting time until the first text, this would have made sense, but at this level, most candidates have not yet met this concept. Even those who progressed beyond this point and actually managed to reach the required quadratic in e^λ generally became stuck when trying to show that there are two positive values of λ: they showed (or tried to show) that e^λ > 0 rather than the necessary e^λ > 1. Those who tried the second part generally did not appreciate that e^(λ1) and e^(λ2) are the two roots of the equation pe^(2λ) − e^λ + 1 = 0, and often used the two possible roots for each of e^(λ1) and e^(λ2), leading to some nonsensical answers; very few reached the required expression for λ1 + λ2. Also, none of the candidates who reached this point seemed to know (or use) the result that the product of roots of ax² + bx + c = 0 is c/a; this is a very useful tool for students to have. Finally, in the last part of the question, very few of the candidates were capable of finding an event involving the two phones equivalent to "the first text arrives between 1 and 2 hours", leaving them unable to make any meaningful progress. Drawing a Venn diagram or listing possibilities would have been of help, but there was little evidence that candidates used techniques such as these.