= Solution
For the <standard normal distribution>, $m=0$ and $f(0)=1/\sqrt{2\pi}$, so the <sample median> has <asymptotic variance> $\pi/2$ under scaling by $\sqrt n$. The <central limit theorem> gives $\sqrt n\bar X_n\xrightarrow{d}N(0,1)$ for the <sample mean>. Thus \b[the <sample median> has $\pi/2$ times the <asymptotic variance> of the <sample mean>], and its <asymptotic relative efficiency> against the <sample mean> is
$$
\boxed{\operatorname{ARE}(\text{median},\text{mean})=\frac2\pi.}
$$
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