Year: 2016
Paper: 2
Question Number: 13
Course: UFM Statistics
Section: Approximating the Binomial to the Poisson distribution
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item
The random variable $X$ has a binomial distribution with parameters $n$ and $p$, where $n=16$ and
$p=\frac12$. Show, using an approximation in terms of the standard normal
density function $\displaystyle
\tfrac{1}{\sqrt{2\pi}} \, \e ^{-\frac12 x^2}
$, that \[
\P(X=8) \approx \frac 1{2\sqrt{2\pi}}
\,.
\]
\item By considering a binomial distribution with parameters $2n$ and $\frac12$, show that
\[
(2n)! \approx \frac {2^{2n} (n!)^2}{\sqrt{n\pi}} \,.
\]
\item By considering a Poisson distribution with parameter $n$, show that
\[
n! \approx \sqrt{2\pi n\, } \, \e^{-n} \, n^n \,.
\]
\end{questionparts}\begin{questionparts}
\item $X \sim B(16, \tfrac12)$, then $X \approx N(8, 2^2)$, in particular
\begin{align*}
&& \mathbb{P}(X = 8) &\approx \mathbb{P} \left ( 8 - \frac12 \leq 2Z + 8 \leq 8 + \frac12 \right) \\
&&&= \mathbb{P} \left (-\frac14 \leq Z \leq \frac14 \right) \\
&&&= \int_{-\frac14}^{\frac14} \frac{1}{\sqrt{2 \pi}}e^{-\frac12 x^2} \d x \\
&&&\approx \frac{1}{\sqrt{2\pi}} \int_{-\frac14}^{\frac14} 1\d x\\
&&&= \frac{1}{2 \sqrt{2\pi}}
\end{align*}
\item Suppose $X \sim B(2n, \frac12)$ then $X \approx N(n, \frac{n}{2})$, and
\begin{align*}
&& \mathbb{P}(X = n) &\approx \mathbb{P} \left ( n - \frac12 \leq \sqrt{\frac{n}{2}} Z + n \leq n + \frac12 \right) \\
&&&= \mathbb{P} \left ( - \frac1{\sqrt{2n}} \leq Z \leq \frac1{\sqrt{2n}}\right) \\
&&&= \int_{-\frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2 \pi}} e^{-\frac12 x^2} \d x \\
&&&\approx \frac{1}{\sqrt{n\pi}}\\
\Rightarrow && \binom{2n}{n}\frac1{2^n} \frac{1}{2^n} & \approx \frac{1}{\sqrt{n \pi}} \\
\Rightarrow && (2n)! &\approx \frac{2^{2n}(n!)^2}{\sqrt{n\pi}}
\end{align*}
\item $X \sim Po(n)$, then $X \approx N(n, (\sqrt{n})^2)$, therefore
\begin{align*}
&& \mathbb{P}(X = n) &\approx \mathbb{P} \left (-\frac12 \leq \sqrt{n} Z \leq \frac12 \right) \\
&&&= \int_{-\frac{1}{2 \sqrt{n}}}^{\frac{1}{2 \sqrt{n}}} \frac{1}{\sqrt{2\pi}}e^{-\frac12 x^2} \d x \\
&&&\approx \frac{1}{\sqrt{2 \pi n}} \\
\Rightarrow && e^{-n} \frac{n^n}{n!} & \approx \frac{1}{\sqrt{2 \pi n}} \\
\Rightarrow && n! &\approx \sqrt{2 \pi n} e^{-n}n^n
\end{align*}
\end{questionparts}
This question received only a small number of attempts, with a significant number of candidates not identifying that a rectangle could be used to approximate the area in each of the cases and so unable to make much progress on the question. The average mark per candidate for this question was the lowest on the paper, with a significant number of attempts not progressing beyond the first step.