 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-2/6h/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For observed counts

O_{j}

and null expected counts

E_{j} = n p_{j}

X^{2} = \sum_{j = 1}^{k} \frac{( O _{j} - E _{j} ) ^{2}}{E _{j}} .

When the null cell probabilities are specified and the expected counts grow, its limiting null distribution is

X^{2} d χ_{k - 1}^{2} .

Here

k = 6

, so the limit is

χ_{5}^{2}

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