Counting Fish

6060 /www/nrich/html/content/id/6060/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <div style="text-align: center;"><mdo:image alt="Fish Tessellation" height="329" src="FishTess.JPG" width="423"></mdo:image></div> <div style="text-align: left;"> </div> <div style="text-align: left;"> </div> <div style="text-align: left;"> </div> I need a figure for the fish population in a lake. I catch 40 fish and mark a scale on each fish so that they can be identified if caught again. The fish are then released and one week later I once more catch 40 fish and look to see which of these I caught before. How does this help me come up with a figure for the fish population in a lake? <a href="http://nrich.maths.org/public/viewer.php?obj_id=6614&part=">Click here for a poster of this problem</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Here's a couple of nice explanations : From Francesca at Wimbledon High School : by the end of that week, the $40$ marked fish would have spread out and integrated with the rest of the fish population. So that the second lot of $40$ fish that are caught would be a thoroughly mixed group of marked and unmarked fish. Count the number of marked and unmarked fish, get a ratio, then reason like this : For example if in the second lot of $40$ fish, the marked to unmarked ratio was $2 : 38$ and we know that there are $40$ marked fish altogether, we might assume that the ratio in the second lot of fish is close to the ratio for all the fish together. So $2 : 38$ matches $40 : ?$ $40/2$ is $20$ and then scaling that up by $38$ to get $760$ ($20\times38$) And just to make sure... $$38/2=19 $$ $$760/40=19 $$ then, we must add that ratio together: $$40+760=800$$ Hey presto!!! There are $800$ fish altogether! Thanks Francesca. And from Stephen at Blatchington Mill School : If you mark $40$ fish and release them. Then, when those fish have had time to mix back in with the whole population, catch another $40$. If you count the number ($c$) in that second catch which are already marked, you can estimate the size of the whole population ($T$) by saying that $40 = T(c/40) $ For example if when you looked at the second catch and $10$ were already marked then you could estimate that you had marked a quarter of the entire population of fish, because that's the proportion in the sample, so the original $40$ were about one quarter of the whole population and there are therefore approximately $120$ fish. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem :</h3> Beyond the reasoning challenge to sort out how the fraction re-caught suggests an estimate for the size of the whole population, this problem introduces thoughtful students to the concept of 'confidence' in hypothesis testing ' consideration of how likely an estimate is to be wrong by some specified amount. <h3>Possible approach :</h3> Present the group with this problem on paper and ask them to read it and discuss in pairs what the situation is and what is asked for. This may lead some pairs to successfully solve the problem, in which case the main activity now becomes the task of explaining not just the calculation, but also the justification, to the other students in the group. If however this is a problem that isn't quickly solved a simulation with counters or coloured cubes is an excellent aid to visualisation. <h3>Key questions :</h3> <ul> <li style="margin-left: 40px;">Describe the procedure used.What is this procedure trying to do?</li> </ul> <ul> <li style="margin-left: 40px;">What do we know?</li> </ul> <ul> <li style="margin-left: 40px;">What do we need to find?</li> </ul> <ul> <li style="margin-left: 40px;">Is it the actual population or an estimate?</li> </ul> <ul> <li style="margin-left: 40px;">How close do you think it is?</li> </ul> <h3>Possible extension :</h3> Conduct the same simulation as below for 'Possible Support' but draw attention to the variation that occurs as the simulation is repeated, and invite students to investigate how much their calculated estimates vary and in general use of the term how confident they might think it safe to be with their estimate. For example what 'plus or minus' amount might they attach to their answer. This situation is then gradually generalised to different size populations and different relative size of sample. <h3>Possible support :</h3> Simulation with counters or coloured cubes is the most useful aid to visualisation. For example put 20 counters into a bag and explain that the bag is the pond and the whole fish population in this instance is 20. Remove 5 counters and replace them with counters of a different colour, explaining that these five are the first sample, and the different colour allows the counters 'caught' for a second time to be identified. Now make the second sample of five. This establishes the context or procedure being discussed so that attention can now rest on solving the problem. A population of 100, with a sample size of 20, might give estimates closer to the actual population, and this may perhaps help students to see how to use the fraction re-caught. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Suppose 20 fish are re-caught - that's 20 of the 40 caught first - what would be a good guess for the whole lake fish population in this case? </mdoxml> 2 3 0 0 0 1 0 Counting Fish I need a figure for the fish population in a lake, how does it help to catch and mark 40 fish ? Fractions, Decimals, Percentages, Ratio and Proportion Fractions Advanced Probability and Statistics Sampling and hypothesis tests Admin Short problems