Year: 2019
Paper: 2
Question Number: 12
Course: LFM Stats And Pure
Section: Continuous Probability Distributions and Random Variables
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The random variable $X$ has the probability density function on the interval $[0, 1]$:
$$f(x) = \begin{cases}
nx^{n-1} & 0 \leq x \leq 1, \\
0 & \text{elsewhere},
\end{cases}$$
where $n$ is an integer greater than 1.
\begin{questionparts}
\item Let $\mu = E(X)$. Find an expression for $\mu$ in terms of $n$, and show that the variance, $\sigma^2$, of $X$ is given by
$$\sigma^2 = \frac{n}{(n + 1)^2(n + 2)}.$$
\item In the case $n = 2$, show without using decimal approximations that the interquartile range is less than $2\sigma$.
\item Write down the first three terms and the $(k + 1)$th term (where $0 \leq k \leq n$) of the binomial expansion of $(1 + x)^n$ in ascending powers of $x$.
By setting $x = \frac{1}{n}$, show that $\mu$ is less than the median and greater than the lower quartile.
Note: You may assume that
$$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots < 4.$$
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& \mu &= \E[X] \\
&&&= \int_0^1 x f(x) \d x \\
&&&= \int_0^1 nx^n \d x \\
&&&= \frac{n}{n+1} \\
\\
&& \var[X] &= \sigma^2 \\
&&&= \E[X^2] - \mu^2 \\
&&&= \int_0^1 x^2 f(x) \d x - \mu^2 \\
&&&= \int_0^1 nx^{n+1} \d x - \mu^2 \\
&&&= \frac{n}{n+2} - \frac{n^2}{(n+1)^2} \\
&&&= \frac{n(n+1)^2 - n^2(n+2)}{(n+1)^2(n+2)} \\
&&&= \frac{n}{(n+1)^2(n+2)}
\end{align*}
\item $\,$
\begin{align*}
&& \frac14 &= \int_0^{Q_1} 2x \d x \\
&&&= Q_1^2 \\
\Rightarrow && Q_1 &= \frac12 \\
&& \frac34 &= \int_0^{Q_3} 2x \d x \\
&&&= Q_3^2 \\
\Rightarrow && Q_3 &= \frac{\sqrt{3}}2 \\ \\
\Rightarrow && IQR &= Q_3 - Q_1 = \frac{\sqrt{3}-1}{2} \\
&& 2 \sigma &= 2\sqrt{\frac{2}{3^2 \cdot 4}} \\
&&&= \frac{\sqrt{2}}{3} \\
\\
&& 2\sigma - IRQ &= \frac{\sqrt{2}}{3} - \frac{\sqrt{3}-1}{2} \\
&&&= \frac{2\sqrt{2}-3\sqrt{3}+3}{6} \\
&& (3+2\sqrt{2})^2 &= 17+12\sqrt{2} > 29 \\
&& (3\sqrt{3})^2 &= 27
\end{align*}
Therefore $2\sigma > IQR$
\item \[ (1+x)^n = 1 + nx + \frac{n(n-1)}2 x^2 + \cdots + \binom{n}{k} x^k+ \cdots \]
\begin{align*}
&& Q_1^{-n} &= 4 \\
&& Q_2^{-n} &= 2\\
&& \mu &=\frac{n}{n+1} \\
\Rightarrow && \mu^{-n} &= \left (1 + \frac1n \right)^n\\
&&&\geq 1 + n \frac1n + \cdots > 2 \\
\Rightarrow && \mu &< Q_2 \\
\\
&& \mu^{-n} &= \left (1 + \frac1n \right)^n\\
&&&= 1 + n \frac1n + \frac{n(n-1)}{2!} \frac{1}{n^2} + \cdots + \frac{n(n-1) \cdots (n-k+1)}{k!} \frac{1}{n^k} + \cdots \\
&&&= 1 + 1 + \left (1 - \frac1n \right ) \frac1{2!} + \cdots + \left (1 - \frac1n \right)\cdot\left (1 - \frac2n \right) \cdots \left (1 - \frac{k-1}n \right) \frac{1}{k!} + \cdots \\
&&&< 1 + 1 + \frac1{2!} + \cdots + \frac1{k!} \\
&&&< 4 \\
\Rightarrow && \mu &> Q_1
\end{align*}
\end{questionparts}
Almost all candidates who attempted this question were able to achieve full marks on the first part. In the second part, the values of the interquartile range and 2σ were generally found correctly, but then many candidates did not realise that squaring would eliminate the square roots from the values to be compared. In the final part of the question some candidates failed to recognise that the k+1 term of the expansion was the term in xᵏ and gave the term in xᵏ⁺¹ instead. A good number were successful in finding the lower quartile and the median, but only a minority realised that μ = 1 − (½)^(1/n). Those that did were more successful in proving that μ < median than μ > lower quartile.