= Solution
{c}
Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\geq2$. Two exact facts about a <normal sample> are
$$
\overline X\sim N(\mu,\mu^2/n),\qquad
\frac{(n-1)S^2}{\mu^2}\sim\chi^2_{n-1},
$$
and <independence of the normal sample mean and sample variance> holds. A <chi-squared distribution> with $n-1$ <statistical degrees of freedom> is a sum of $n-1$ <independent> $\chi^2_1$ variables, each with <mean> one and <variance> two. The <central limit theorem> therefore gives
$$
\sqrt{n-1}(S^2-\mu^2)\xrightarrow dN(0,2\mu^4).
$$
For $g(x)=\sqrt x$, the <derivative> at $\mu^2$ is $1/(2\mu)$ because $\mu>0$. The <delta method> gives the <asymptotic normality of a normal sample standard deviation>:
$$
\sqrt n(S-\mu)\xrightarrow dN(0,\mu^2/2).
$$
The exact independence of $\overline X$ and $S$ makes their limiting <normal distributions> independent. Consequently,
$$
\sqrt n(\widetilde\mu-\mu)
=\frac13\{\sqrt n(\overline X-\mu)+2\sqrt n(S-\mu)\}
\xrightarrow dN\!\left(0,\frac19\left\{\mu^2+4\frac{\mu^2}2\right\}\right).
$$
Thus
$$
\boxed{\sqrt n(\widetilde\mu-\mu)\xrightarrow dN(0,\mu^2/3),\qquad
\widetilde\mu\mathrel{\dot\sim}N\!\left(\mu,\frac{\mu^2}{3n}\right).}
$$
This has the same leading <asymptotic variance> as $\widehat\mu$. The weights $1/3$ and $2/3$ are those in the <optimal asymptotic combination of a normal sample mean and standard deviation>. This does not require $S$ to be an exactly <unbiased estimator> of $\mu$.
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