= Solution
{c}
For this rate parametrization of the <Laplace distribution>, $\mu$ is its <mean> and <median>. By <symmetry>, $\mathbb E(X-\mu)=0$, while $|X-\mu|$ has <exponential distribution> with rate $\lambda$, so
$$
\operatorname{Var}(X)=\mathbb E|X-\mu|^2
=\int_0^\infty r^2\lambda e^{-\lambda r}\,dr=\frac2{\lambda^2}.
$$
The <integral> follows by two applications of <integration by parts>. The <central limit theorem> therefore gives
$$
\frac{\sqrt n(\overline X-\mu)}{\sqrt2/\lambda}
\ \xrightarrow{d}\ N(0,1).
$$
With known $\lambda$ and the supplied cutoff 2, an approximate 95% <confidence interval> is
$$
\boxed{\left[\overline x-\frac{2\sqrt2}{\lambda\sqrt n},\
\overline x+\frac{2\sqrt2}{\lambda\sqrt n}\right].}
$$
This <known-rate Laplace location confidence interval> uses the <sample mean>; the later <maximum-likelihood estimator> of location need not be used to construct this valid asymptotic <confidence interval>.
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