 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/20h/c/iii/solution

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

Let

χ_{ν, p}^{2}

denote the

p

-quantile of the chi-squared distribution. Since

S_{XX} / σ^{2} \sim χ_{n - 1}^{2}

, inversion of the central probability event gives

[\frac{S _{XX}}{χ _{n - 1, 1 - α /2}^{2}}, \frac{S _{XX}}{χ _{n - 1, α /2}^{2}}]

100 (1 - α) %

σ^{2}

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