Problem source

Suppose $X$ is a random variable with probability density 
\[
\mathrm{f}(x)=Ax^{2}\exp(-x^{2}/2)
\]
for $-\infty < x < \infty.$ Find $A$.
You belong to a group of scientists who believe that the outcome of a certain experiment is a random variable with the probability density just given, while other scientists believe that the probability density is the same except with different mean (i.e. the probability density is $\mathrm{f}(x-\mu)$ with $\mu\neq0$). In each of the following two cases decide whether the result given would shake your faith in your hypothesis, and justify your answer. 
\begin{questionparts}
\item A single trial produces the result 87.3. 
\item 1000 independent trials produce results having a mean value $0.23.$
\end{questionparts}
{[}Great weight will be placed on clear statements of your reasons and none on the mere repetition of standard tests, however sophisticated, if unsupported by argument. There are several possible approaches to this question. For some of them it is useful to know that if $Z$ is normal with mean 0 and variance 1 then $\mathrm{E}(Z^{4})=3.${]}

Solution source

Let $Z \sim N(0,1)$, with a pdf of $f(x) = \frac{1}{\sqrt{2\pi}} \exp(-x^2/2)$
\begin{align*}
&& 1 &= \int_{-\infty}^\infty Ax^2 \exp(-x^2/2) \d x \\
&&&=  A\sqrt{2\pi} \int_{-\infty}^\infty x^2 \frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \d x \\
&&&= A\sqrt{2\pi} \E[Z^2] = A\sqrt{2\pi} \\
\Rightarrow && A &= \frac{1}{\sqrt{2\pi}}
\end{align*}

\begin{questionparts}
\item The probability of seeing a result as extreme as $87.3$ is \begin{align*}
\mathbb{P}(X > 87.3) &=  \frac{1}{\sqrt{2\pi}}\int_{87.3}^{\infty} x^2 \exp(-x^2/2) \d x \\
&= \left [ -\frac{1}{\sqrt{2\pi}}x \exp(-x^2/2)\right]_{87.3}^{\infty}+\int_{87.3}^{\infty}\frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \d x \\
&\approx 0 +(1- \Phi(87.3)) \\
&\approx 0
\end{align*}
It is very unlikely this data point has come from our distribution rather than one with a higher mean, therefore our faith is very shaken.

\item If there are 1000 trials of this, we would expect the sample mean to be distributed according to the CLT. Each sample has mean $0$ and variance $\E[X^2] = \int_{-\infty}^\infty x^4 \frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \d x = \E[Z^4] = 3$, therefore the sample mean is $N(0, 3/1000)$. Therefore the probability of being $0.23$ away is

\begin{align*}
&& \mathbb{P}(S > 0.23) &= \mathbb{P}\left (Z > \frac{0.23}{\sqrt{3/1000}} \right) \\
&&&=  \mathbb{P}\left (Z > \frac{0.23}{\sqrt{30}/100} \right) \\
&&&\approx   \mathbb{P}\left (Z > \frac{0.23}{0.055} \right) \\
&&& \approx 0
\end{align*}

again our faith should be shaken

\end{questionparts}