## Related questions with answers

A common technique for estimating animal populations is to tag and release individual animals in two different outings. This procedure is called catch and release. If the wildlife remain in the sampling area and are randomly caught, a fraction of the animals tagged during the first outing are likely to be caught again during the second outing. Based on the number tagged and the fraction caught twice, the total number of animals in the area can be estimated. a. Consider a case in which $200$ fish are tagged and released during the first outing. During a second outing in the same area, $200$ fish are again caught and released, of which one-half are already tagged. Estimate $N$, the total number of fish in the entire sampling area. Explain your reasoning, b. Consider a case in which $200$ fish are tagged and released during the first outing. During a second outing in the same area, $200$ fish are again caught and released, of which one-fourth are already tagged. Estimate $N$ the total number of fish in the entire sampling area. Explain your reasoning. c. Generalize your results from parts (a) and (b) by letting $p$ be the fraction of tagged fish that are caught during the second outing. Find a formula for the function $N=f(p)$ that relates the total number of fish, $N$, to the fraction tagged during the second outing, $p$. d. Graph the function obtained in part (c). What is the domain? Explain. e. Suppose that $15 \%$ of the fish in the second sample are tagged. Use the formula from part (c) to estimate the total number of fish in the sampling area. Confirm your result on your graph. f. Locate a real study in which the catch and release method was used. Report on the specific details of the study and describe how closely it followed the theory outlined in this problem.

Solution

Verified$(a)$ As $200$ fish have been tagged in the first outing and in the next outing, only $50\%$ of caught fishes are already tagged, we can assume that the $200$ is half of the total population, so the total population is $400$ fishes.

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