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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-29/3/b/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 29 / 3 / b / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

The one-day slaughter-inspection yield is

p_{inspection} = \frac{15}{30000} = 0.0005,

(1)

or 500 detections per million inspected adult cattle. The routine field detections correspond, using a stable herd size of five million, to

r_{field} = \frac{14600}{5 \times 1 0 ^{6} \times 365} = 8.0 \times 1 0^{- 6}

(2)

per adult-cattle day, or 8 detections per million adult-cattle days. Therefore the numerical one-day comparison is

\frac{0.0005}{8.0 \times 1 0 ^{- 6}} = 62.5.

(3)

The annual field rate is

14600/ (5 \times 1 0^{6}) = 0.00292

per cattle-year; comparing the inspection probability directly with this annual proportion would mix observation periods. Applying the random inspection-sample proportion to all 150000 slaughtered adults would predict about 75 detected cases in that slaughter cohort. These arithmetic comparisons describe detection yields, not a 62.5-fold biological disease-risk difference: a cross-sectional inspection estimates a prevalence yield, whereas daily field reporting is an ascertainment rate over animal-time.

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