An apparent magnitude is the sum of absolute magnitude and <distance modulus>. Thus
$$
\bar m_{\mathrm{LMC}}
\sim N\!\left(M_0^*+\mu_{\mathrm{LMC}},
\frac{\sigma_*^2}{N_{\mathrm{LMC}}}\right).
$$
The independent unbiased distance-modulus estimate is $N(\mu_{\mathrm{LMC}},\sigma_{\mathrm{LMC}}^2)$, so closure of independent <normal distributions> under linear combinations gives
$$
\widehat S_1\sim N(M_0^*,V_1),
\qquad
V_1=\frac{\sigma_*^2}{N_{\mathrm{LMC}}}+\sigma_{\mathrm{LMC}}^2.
$$
Hence $\mathbb E\widehat S_1=M_0^*$ and $\operatorname{Var}(\widehat S_1)=V_1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 219 / 1 / a / Solution

{title2=$\mu$}
{wiki}

The distance modulus is the difference between apparent and absolute astronomical magnitude. For distance $d$ in parsecs,
$$
\mu=m-M=5\log_{10}(d/10\,\mathrm{pc}).
$$


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