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
$$
\widehat S_3^i=m_{\mathrm{SN}}^i-f(z_i)
=M_i^{\mathrm{SN}}-\theta
\sim N(\mathcal M,\sigma_{\mathrm{SN}}^2),
\qquad
\mathcal M=M_0^{\mathrm{SN}}-\theta.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-219/1/c/solution