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

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- Solution c

Solution

The Hubble law gives

d_{i} = c z_{i} / (100 h)

Mpc, so its distance modulus is

25 + 5 lo g_{10} \frac{c z _{i}}{100 h} = f (z_{i}) - 5 lo g_{10} h = f (z_{i}) - θ .

Consequently

S_{3}^{i} = m_{SN}^{i} - f (z_{i}) = M_{i}^{SN} - θ \sim N (M, σ_{SN}^{2}), M = M_{0}^{SN} - θ .

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