Year: 2016
Paper: 3
Question Number: 12
Course: UFM Statistics
Section: Poisson Distribution
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Let $X$ be a random variable with mean $\mu$ and standard deviation
$\sigma$. \textit{Chebyshev's inequality}, which you may use without proof,
is
\[
\P\left(\vert X-\mu\vert > k\sigma\right) \le \frac 1 {k^2}
\,,
\]
where $k$ is any positive number.
\begin{questionparts}
\item
The probability of a biased coin landing heads up is $0.2$. It is thrown $100n$ times, where $n$ is an integer greater than 1. Let $\alpha $ be the probability that the coin lands heads up $N$ times, where $16n \le N \le 24n$.
Use Chebyshev's inequality to show that
\[
\alpha \ge 1-\frac 1n
\,.
\]
\item
Use Chebyshev's inequality to show that
\[
1+ n + \frac{n^2}{ 2!} + \cdots + \frac {n^{2n}}{(2n)!} \ge
\left(1-\frac1n\right) \e^n
\,.
\]
\end{questionparts}\begin{questionparts}
\item Let $N$ be the number of times the coin lands heads up, ie $N \sim Binomial(100n, 0.2)$, then $\mathbb{E}(N) = \mu = 20n, \mathrm{Var}(N) = \sigma^2 = 100n \cdot 0.2 \cdot 0.8 = 16n \Rightarrow \sigma = 4\sqrt{n}$.
\begin{align*}
&& \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\
\Rightarrow && 1 - \mathbb{P}(|X - \mu| \leq k\sigma) &\leq \frac1{k^2} \\
\Rightarrow && 1 - \mathbb{P}(|X - 20n| \leq \sqrt{n} \cdot 4\sqrt{n}) &\leq \frac1{{\sqrt{n}}^2} \\
\Rightarrow && 1 - \mathbb{P}(16n \leq N \leq 24n) &\leq \frac{1}{n} \\
\Rightarrow && 1 - \frac1n &\leq \alpha
\end{align*}
\item Suppose $X \sim Pois(n)$, then $\mathbb{E}(X) = n, \mathrm{Var}(X) = n$. Therefore
\begin{align*}
&& \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\
\Rightarrow && 1-\mathbb{P}(|X - n| \leq \sqrt{n} \cdot \sqrt{n}) &> \frac{1}{\sqrt{n}^2} \\
\Rightarrow && 1 - \sum_{i=0}^{2n} \mathbb{P}(X = i) & \leq \frac{1}{n} \\
\Rightarrow && \sum_{i=0}^{2n} e^{-n} \frac{n^i}{i!} \geq 1 - \frac{1}{n} \\
\Rightarrow && \sum_{i=0}^{2n} \frac{n^i}{i!} \geq \left ( 1 - \frac1n \right)e^n
\end{align*}
\end{questionparts}
Whilst as popular (or rather unpopular) as question 10, attempts at question 12 were more successful than all but questions 1 and 8 with on average half marks being scored. Part (i) was generally well done using the binomial distribution, and those that spotted they should use the Poisson distribution in part (ii) usually did well too. However, candidates were often sloppy in their explanations of the rearrangements of Chebyshev, and also quite often candidates had a fixation with the Normal distribution which did not help.