 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-1/7h/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

x_{1}, \dots, x_{n}

ℓ (μ, σ^{2}) = - \frac{n}{2} lo g (2 π σ^{2}) - \frac{1}{2 σ ^{2}} \sum_{i = 1}^{n} (x_{i} - μ)^{2} .

Differentiating first with respect to

μ

gives

μ = \overline{x}

. Substituting this value and differentiating with respect to

σ^{2}

μ = \overline{X} = \frac{1}{n} i = 1 \sum n X_{i}, σ^{2} = \frac{1}{n} i = 1 \sum n (X_{i} - \overline{X})^{2} .

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