Solution
= Solution
The <Hubble law> gives $d_i=cz_i/(100h)$ Mpc, so its distance modulus is
$$
25+5\log_{10}\frac{cz_i}{100h}
=f(z_i)-5\log_{10}h=f(z_i)-\theta.
$$
Consequently
$$
\widehat S_3^i=m_{\mathrm{SN}}^i-f(z_i)
=M_i^{\mathrm{SN}}-\theta
\sim N(\mathcal M,\sigma_{\mathrm{SN}}^2),
\qquad
\mathcal M=M_0^{\mathrm{SN}}-\theta.
$$