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Practice question

Question

A student marks 10 fish and releases them in a lake. On recapture, 15 fish are caught, 5 of which are marked. The estimated population size is:

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Explanation

The Lincoln–Petersen mark–recapture estimator assumes that the fraction of marked individuals in the second sample approximates the marked fraction in the whole population. If M individuals were initially marked, C individuals are caught later, and R of that second catch carry marks, then M/N ≈ R/C. Rearranging gives N ≈ MC/R. Here M = 10, C = 15, and R = 5, so N = (10 × 15)/5 = 30 fish. The estimate relies on marked fish mixing thoroughly with unmarked fish, marks being retained and recognized, equal capture probabilities, and negligible births, deaths, immigration, or emigration between samples. Violation of these assumptions biases the result. For example, mark loss lowers R and inflates the estimate, while trap-happy marked animals raise R and depress it. With small samples, adjusted estimators such as Chapman’s can reduce bias. The calculation nevertheless illustrates the core proportional principle: one-third of the recaptured sample is marked, so the 10 marked fish are inferred to represent one-third of the lake population.

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