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Either paper 3 marks
Amara wants to estimate how many trout there are in the reservoir.
She catches 198 trout, tags them and puts them back.
A week later she catches 77 trout. Of these, 11 are tagged.
Use Amara's results to estimate the number of trout in the reservoir.
Show answer
\(1386\)Assume the fraction tagged in the second sample matches the fraction tagged in the whole population:
\(\frac{ 11 }{ 77 } = \frac{ 198 }{ \text{population} }\)
So the estimate is
\(\frac{ 198 \times 77 }{ 11 } = \frac{ 15246 }{ 11 } = 1386\)
About 1386 trout.
- M1equates the fraction tagged in the second sample with the fraction tagged in the whole population
- M1\(198 \times 77 \div 11\), or \(\frac{ 15246 }{ 11 }\)
- A1estimate \(1386\)
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Either paper 4 marks
Reuben is estimating the number of ladybirds in the hedgerow.
He catches 54 ladybirds, marks them and releases them.
Later that month he catches 78 ladybirds, of which 6 are marked.
(a)Use Reuben's results to estimate the number of ladybirds in the hedgerow.
(b)Write down one assumption you have made.
Show answer
(a) \(702\)
(b) e.g. no ladybirds entered or left the hedgerow between the two samples\(\frac{ 6 }{ 78 } = \frac{ 54 }{ \text{population} }\)
\(\frac{ 54 \times 78 }{ 6 } = \frac{ 4212 }{ 6 } = 702\)
About 702 ladybirds.
Any one of:
no ladybirds were born, died, entered or left the hedgerow between the two samples;
the marked ladybirds mixed back evenly with the rest;
every one of the ladybirds was equally likely to be caught the second time.
- (a) M1proportion set up, \(\frac{ 6 }{ 78 } = \frac{ 54 }{ \text{population} }\)
- (a) M1rearranging to \(\frac{ 54 \times 78 }{ 6 }\)
- (a) A1estimate \(702\)
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Either paper 3 marks
There are thought to be about 280 carp in the lake. 40 of them are tagged.
Joel catches a sample of 56 carp.
Estimate how many of his sample are tagged.
Show answer
\(8\)Fraction of the population that is tagged:
\(\frac{ 40 }{ 280 }\)
Apply that fraction to the sample:
\(56 \times \frac{ 40 }{ 280 } = 8\)
About 8 of them.
- M1fraction tagged in the population \(\frac{ 40 }{ 280 }\)
- M1fraction applied to the sample of \(56\)
- A1\(8\) tagged
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Either paper 4 marks
Nadia wants to estimate how many ladybirds there are in the hedgerow.
She catches 171 ladybirds, marks them and puts them back. Later she catches 36 ladybirds. Of these, 9 are marked.
Using these numbers, Nadia estimates that there are 684 ladybirds in the hedgerow.
Now think about two ways the study could have played out differently.
(a)Suppose she had found more than 9 marked ladybirds in the second catch — the size of each catch stays the same.
Would this make the estimate bigger or smaller than 684? Give a reason for your answer.
(b)Between the two catches, suppose the marked ladybirds become easier to catch a second time, drawn back to wherever they were caught before. Nadia still uses the same calculation as before.
Would this make Nadia's estimate of 684 ladybirds bigger or smaller than the true number of ladybirds actually living in the hedgerow? Give a reason for your answer.
Show answer
(a) Smaller — marked ladybirds would then be a bigger share of the second catch, so the same 171 marked ladybirds from the first catch must stand for a smaller population
(b) Smaller — the second catch ended up with more marked ladybirds than a fair sample would, so the 9 out of 36 makes the marked share of the second catch look bigger than it really is, so the calculation underestimates the number of ladybirds in the hedgerowThe method assumes \(\dfrac{ 9 }{ 36 } = \dfrac{ 171 }{ \text{population} }\).
If more of the second catch are marked, the left-hand fraction becomes bigger. \(171\) stays the same, so the right-hand fraction can only match it if the population is smaller.
The estimate would be smaller than 684.
The method assumes \(\dfrac{ 9 }{ 36 } = \dfrac{ 171 }{ \text{population} }\), which only works if the second catch is a fair sample of the hedgerow.
If the marked ladybirds become easier to catch a second time, drawn back to wherever they were caught before, the second catch ends up with more marked ladybirds than a fair sample would — so 9 out of 36 makes the left-hand fraction bigger than the true marked share of the hedgerow really is.
Solving the equation from a bigger fraction pushes the estimate the other way: Nadia's estimate of 684 would be smaller than the true number of ladybirds actually living in the hedgerow.
- (a) M1the second-catch fraction \(\dfrac{ 9 }{ 36 }\) becomes bigger
- (a) A1Smaller — 171 marked ladybirds from the first catch must then stand for a smaller population
- (b) M1the second catch ends up with more marked ladybirds than a fair sample, making \(\dfrac{ 9 }{ 36 }\) look bigger than the true marked share
- (b) A1Smaller — the calculation underestimates the number of ladybirds in the hedgerow
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Either paper 4 marks
Idris catches 60 sparrows in the reserve, rings them and puts them back.
(a)A month later he catches 50 sparrows from the reserve. 5 of them are ringed.
Estimate the number of sparrows in the reserve.
(b)What percentage of that estimate had been ringed?
Show answer
(a) \(600\)
(b) \(10\%\)\(\frac{ 60 \times 50 }{ 5 } = \frac{ 3000 }{ 5 } = 600\)
About 600 sparrows.
\(\frac{ 60 }{ 600 } \times 100 = 10\)
10% of the estimated population had been ringed.
- (a) M1equates the fraction ringed in the second sample with the fraction ringed in the whole population
- (a) M1\(60 \times 50 \div 5\), or \(\frac{ 3000 }{ 5 }\)
- (a) A1estimate \(600\)
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Either paper 3 marks
Yusuf is estimating the number of rabbits in a field.
He catches some rabbits, tagged all of them, and returns them to a field.
Later he catches \(50\) rabbits. \(5\) of them are tagged.
Using these results, he estimates that there are \(600\) rabbits in a field.
(a)Work out how many rabbits he tagged the first time.
(b)Write down one assumption that has been made.
Show answer
(a) \(60\ \text{rabbits}\)
(b) The tagged rabbits had mixed evenly back into the population before the second catchThe method assumes the fraction of the second catch that is tagged is the same as the fraction of all the rabbits that are tagged.
\(\frac{ 5 }{ 50 } = \frac{\text{number } tagged }{ 600 }\)
Multiply both sides by \(600\).
\(\frac{ 5 }{ 50 } \times 600 = 60\)
The estimate only works if the second catch is as likely to contain a tagged rabbits as any other handful would be. Standard answers are that the tagged rabbits mixed evenly back in, that none died or left a field between the two catches, that none were born into it, and that the tagged were no easier or harder to catch the second time.
- (a) M1proportion \(\frac{ 5 }{ 50 } = \frac{\text{number } tagged }{ 600 }\)
- (a) A1first sample \(60\)
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Either paper 3 marks
Isla is trying to estimate the number of beetles in the hedge bank.
She catches 45 beetles, marks them and puts them back. Later she catches 30 beetles. Of these, 5 are marked.
Here is her working:
\( 45 \times 30 = 1350 \)
She says there are about 1350 beetles in the hedge bank.
(a)Write down the mistake Isla has made.
(b)Find the correct estimate for the number of beetles in the hedge bank.
Show answer
(a) She multiplied the two catch sizes together but forgot to divide by the number that were marked
(b) \(270\)The correct method is \(\dfrac{ 45 \times 30 }{ 5 }\).
The working shown — \( 45 \times 30 \) — stops there, without dividing by the number that were marked, giving 1350 instead of 270.
\(\dfrac{ 45 \times 30 }{ 5 } = \dfrac{ 1350 }{ 5 } = 270\)
About 270 beetles.
- (b) M1\(\dfrac{ 45 \times 30 }{ 5 }\)
- (b) A1estimate \(270\)
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Calculator 5 marks
Hafsa estimates the number of stick insects in the greenhouse using capture-recapture.
She catches 180 stick insects, marks them and puts them back. Later she catches 72 stick insects. Of these, 9 are marked.
(a)Use Hafsa's results to estimate the number of stick insects in the greenhouse.
(b)A complete count later found 2648 stick insects in the greenhouse.
Work out the difference between Hafsa's estimate and the true number of stick insects, as a percentage of the true number. Give your answer correct to 1 decimal place.
Show answer
(a) \(1440\)
(b) \(45.6\%\)\(\dfrac{ 180 \times 72 }{ 9 } = \dfrac{ 12960 }{ 9 } = 1440\)
About 1440 stick insects.
Difference: \(|1440 - 2648| = 1208\)
As a percentage of the true number:
\(\dfrac{ 1208 }{ 2648 } \times 100 = 45.6\%\)
- (a) M1\(180 \times 72 \div 9\)
- (a) A1estimate \(1440\)
- (b) M1difference \(1208\)
- (b) M1\(1208 \div 2648 \times 100\)
- (b) A1\(45.6\%\)
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Either paper 6 marks
Amara wants to estimate the number of snails living across the whole of Hollow Wood, an area of 3 hectares (\(1\) hectare \(= 10\,000\) m²).
To do this, she fences off a smaller plot of 125 m² within Hollow Wood so the snails cannot move in or out, and studies them there.
Inside the plot, she catches 18 snails, marks them and releases them. Later she catches 12 snails in the same plot, of which 3 are marked.
(a)Work out Amara's estimate for the number of snails in the 125 m² plot.
(b)Hollow Wood covers 3 hectares in total. Using \(1\) hectare \(= 10\,000\) m², use Amara's plot estimate to estimate the number of snails across the whole of Hollow Wood, giving your answer to 2 significant figures.
(c)State one assumption Amara is making when scaling the plot estimate up to the whole of Hollow Wood.
Show answer
(a) \(72\)
(b) \(17000\)
(c) That snails are spread evenly across Hollow Wood, so the density found in the plot is typical of the wood as a whole\(\dfrac{ 18 \times 12 }{ 3 } = \dfrac{ 216 }{ 3 } = 72\)
About 72 snails in the plot.
Hollow Wood covers \(3 \times 10\,000 = 30000\) m².
The plot is \(\dfrac{1}{ 240 }\) of that area, since \(30000 \div 125 = 240\).
Scale up the plot estimate: \(72 \times 240 = 17280\).
To 2 significant figures, about 17000 snails across the whole of Hollow Wood.
Scaling up only works if the plot is typical of the rest of Hollow Wood. If snails were more crowded in some parts of Hollow Wood than others, one small plot could give a very misleading total.
- (a) M1\(18 \times 12 \div 3\)
- (a) A1plot estimate \(72\)
- (b) M1site area \(30000\) m² and scale factor \(240\)
- (b) M1\(72 \times 240 = 17280\)
- (b) A1\(17000\) to 2 significant figures
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Either paper 3 marks
Omar wants to estimate how many trout there are in the reservoir.
He catches 60 trout, tags them and puts them back.
A week later he catches 36 trout. Of these, 6 are tagged.
Use Omar's results to estimate the number of trout in the reservoir.
Show answer
\(360\)Assume the fraction tagged in the second sample matches the fraction tagged in the whole population:
\(\frac{ 6 }{ 36 } = \frac{ 60 }{ \text{population} }\)
So the estimate is
\(\frac{ 60 \times 36 }{ 6 } = \frac{ 2160 }{ 6 } = 360\)
About 360 trout.
- M1equates the fraction tagged in the second sample with the fraction tagged in the whole population
- M1\(60 \times 36 \div 6\), or \(\frac{ 2160 }{ 6 }\)
- A1estimate \(360\)