Sampling & capture-recapture – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

Ella wants to find out how often the adults in her town exercise.

She asks 20 people at her local gym.

(a) Give one reason why Ella might use a sample rather than asking every adult in the town. [1]

(b) Explain why Ella's sample is likely to be biased. [1]

(a) a sample is quicker / cheaper / easier than asking the whole population (accept: there are too many adults in the town to ask them all)C1
(b) people at a gym are likely to exercise more than other adults, so the sample does not represent the whole town (accept: she only asks people at the gym / people who do not go to the gym are left out; reject: the sample is too small)C1
(c) e.g. ask people in several different places or at different times, or choose people at random from the whole town, or use a larger sampleC1
Answer: (a) quicker/cheaper (b) gym users exercise more than most adults (c) choose a larger, random sample from across the town
Examiner tip: For ‘biased’, say WHO is over-represented and WHY that affects the answer to the question being asked.

2Grade 4–5Calculator3 marks

A factory makes 2400 light bulbs each day.

A quality checker tests a random sample of 50 of the bulbs. 3 of the bulbs in the sample are faulty.

(a) Work out an estimate for the number of faulty bulbs made each day. [2]

(b) Write down one assumption you have made. [1]

(a) $\frac{3}{50} \times 2400$ or $2400 \div 50 \times 3$ oeM1
(a) 144A1
(b) the sample is representative of all the bulbs made that day (e.g. the sample is random / the proportion of faulty bulbs is the same in the sample as in all the bulbs / the proportion is the same every day)C1
Answer: (a) 144 (b) e.g. the sample is representative of all the bulbs
Examiner tip: Scale up the sample proportion: $\frac{3}{50}$ of the sample, so $\frac{3}{50}$ of the population.

3Grade 4–5Calculator3 marks

Jodie wants to estimate the number of fish in a pond.

She catches 40 fish, puts a tag on each one and returns them to the pond.

The next day she catches 50 fish. 8 of these fish have a tag.

(a) Work out an estimate for the number of fish in the pond. [2]

(b) Write down one assumption you have made. [1]

(a) $\frac{8}{50} = \frac{40}{N}$ or $\frac{40 \times 50}{8}$ (oe)M1
(a) 250A1
(b) e.g. no fish were born, died, joined or left the pond between the two samples / tags are not lost / tagged fish mix fully and are equally likely to be caught as untagged fishC1
Answer: (a) 250 (b) e.g. the population did not change between the samples
Examiner tip: Set up the proportion: tagged in sample ÷ sample size = tagged in total ÷ population.

4Grade 4–5Calculator4 marks

The table shows the number of students in each year group at a school.

Year group	Year 7	Year 8	Year 9	Year 10	Year 11
Number of students	180	160	150	140	120

The headteacher takes a sample of 50 students. The number of students from each year group in the sample is in proportion to the number of students in that year group.

(a) How many Year 9 students should be in the sample? [2]

(b) How many Year 8 students should be in the sample? [2]

(a) $\frac{150}{750} \times 50$ oe, e.g. $150 \div 15$ (750 = total)M1
(a) 10 caoA1
(b) $\frac{160}{750} \times 50$ oe (= 10.66...)M1
(b) 11 cao (reject 10.6, 10.7 or 10)A1
Answer: (a) 10 (b) 11
Examiner tip: A sample must contain whole people – round 10.67 to the nearest whole number, 11.

5Grade 4–5Calculator4 marks

A scientist believes there are 600 rabbits in a wood.

She catches 75 rabbits, marks them and releases them.

A week later she catches 48 rabbits.

(a) If her belief is correct, how many of the 48 rabbits would you expect to be marked? [2]

(b) In fact, 4 of the 48 rabbits are marked. Work out a new estimate for the number of rabbits in the wood. [2]

(a) $\frac{75}{600} \times 48$ oe, e.g. $48 \div 8$M1
(a) 6A1
(b) $\frac{75 \times 48}{4}$ or $\frac{4}{48} = \frac{75}{N}$ oe, e.g. $75 \times 12$M1
(b) 900A1
Answer: (a) 6 (b) 900
Examiner tip: Fewer marked rabbits than expected in the second sample means the marked rabbits are more spread out, so the population is BIGGER than 600.

6Grade 6–7HigherCalculator5 marks

A sixth form has 600 students. The table shows some information about them.

	Boys	Girls
Year 12		140
Year 13	165

A sample of 40 students is taken. The number of students from each group (year group and gender) in the sample is in proportion to the number of students in that group in the sixth form.

There are 11 Year 12 boys in the sample.

(a) Work out an estimate for the number of Year 12 boys in the sixth form. [2]

(b) Use your answer to part (a) to work out the number of Year 13 girls in the sample. [3]

(a) $\frac{11}{40} \times 600$ or $11 \times \frac{600}{40}$ or $11 \times 15$ oeM1
(a) 165A1
(b) $600 - (\text{their }165 + 140 + 165)$ (= 130) for the number of Year 13 girls in the sixth form (Alt: $\frac{140}{600} \times 40$ (= 9.33…) and $\frac{165}{600} \times 40$ (= 11) for the other groups in the sample)M1
(b) $\frac{\text{their }130}{600} \times 40$ oe, e.g. $130 \div 15$ (= 8.66…) (Alt: $40 - 11 - 9.33… - 11$ (= 8.66…), or $40 - 11 - 9 - 11$)M1
(b) 9 (ft their (a): their number of Year 13 girls in the sample, rounded to a whole number; do not accept 8.66… or 8.7 unrounded for A1)A1
Answer: (a) 165 (b) 9
Examiner tip: Every student in the sample represents the same number of students in the sixth form ($600 \div 40 = 15$) – use that in both directions.

7Grade 6–7HigherCalculator3 marks

Sanjay uses the capture–recapture method to estimate the number of frogs in a pond.

He catches $n$ frogs, marks them and releases them. Later he catches 60 frogs, of which 12 are marked.

His estimate for the number of frogs in the pond is 450.

(a) Work out the value of $n$. [2]

(b) Some of the marks washed off the frogs before the second sample was taken. Explain what effect this would have on Sanjay's estimate. [1]

(a) $\frac{12}{60} = \frac{n}{450}$ or $\frac{12}{60} \times 450$ or $450 \div 5$ (oe)M1
(a) 90A1
(b) fewer marked frogs would be counted in the second sample, so the estimate would be too high (an overestimate) (both the reason and 'too high' needed; reject 'too low')C1
Answer: (a) 90 (b) the estimate would be too high
Examiner tip: In (b), follow the chain: fewer marks seen → smaller fraction marked → bigger estimate.

8Grade 1–3FoundationNon-calculator3 marks

A swimming club has 600 members. Each member has a membership number from 001 to 600.

Marcus wants to know what the members think of the new opening times. He is going to ask a sample of 30 members.

(a) Write down the population for Marcus's survey. [1]

(b) Describe how Marcus could use the membership numbers to choose a random sample of 30 members. [2]

(a) the 600 members of the swimming club (accept all the club's members; do not accept 30 or 'the sample')B1
(b) a method that gives every member an equal chance of being chosen, e.g. uses a random number generator / random number table / names (or numbers) drawn from a hat, linked to the membership numbers 001–600C1
(b) complete method: picks 30 different numbers (ignoring repeats and any numbers above 600 / not in use) and asks the members with those numbersC1
Answer: (a) all 600 members of the club (b) e.g. use a random number generator to pick 30 different numbers from 001 to 600 and ask those members
Examiner tip: For full marks say how the numbers are chosen at random AND that you ask the 30 members with those numbers – 'pick people randomly' alone scores nothing.

9Grade 1–3FoundationNon-calculator3 marks

A school has 800 students.

Nadia asks a random sample of 40 students to choose their favourite lunch. The bar chart shows her results.

(a) Work out an estimate for the number of students in the school whose favourite lunch is pasta. [2]

(b) Nadia's friend says, “Exactly 280 students in the school will choose pizza.”

Explain why her friend may be wrong. [1]

(a) $\frac{10}{40}$ oe, or $800 \div 40$ (= 20), or $\frac{10}{40} \times 800$ seenM1
(a) 200 caoA1
(b) the sample only gives an estimate: a different sample (or the whole school) could give a different number, e.g. the sample is small / only 40 of 800 asked (accept 'the results of a sample will not exactly match the whole population')C1
Answer: (a) 200 (b) a sample only gives an estimate; another sample of 40 could give a different result
Examiner tip: Read the pasta bar carefully (10, not 14) – the total of the sample is 40, so each sampled student represents 20 in the school.

10Grade 4–5Calculator4 marks

A college has 1120 students. Priya and Tom each want to estimate how many of the students walk to college.

Priya asks a random sample of 20 students. 7 of them walk to college.

Tom asks a random sample of 120 students. 48 of them walk to college.

No student is in both samples.

(a) Whose sample is likely to give the more reliable estimate? Give a reason for your answer. [1]

(b) Priya and Tom put their results together. Use all of their results to work out an estimate for the number of students in the college who walk to college. [3]

(a) Tom, because his sample is larger (allow 'more students asked'); do not award for 'Tom' with no reasonC1
(b) $7 + 48$ (= 55) and $20 + 120$ (= 140) seenM1
(b) $\frac{55}{140} \times 1120$ oe, e.g. $1120 \div 140$ (= 8) $\times 55$M1
(b) 440 cao (do not accept 420 from averaging $\frac{7}{20}$ and $\frac{48}{120}$)A1
Answer: (a) Tom – his sample is larger (b) 440
Examiner tip: Combine the totals (55 out of 140) – averaging the two proportions gives the wrong answer because the samples are different sizes.

11Grade 4–5Non-calculator3 marks

Lewis wants to estimate the number of ladybirds in a garden.

On Monday he catches 30 ladybirds, marks each one with a small dot of paint and lets them go.

On Tuesday he catches 45 ladybirds. 6 of these have a paint mark.

Here is Lewis's working.

$30 \times 6 \div 45 = 4$ so there are about 4 ladybirds in the garden.

(a) Explain why Lewis's answer cannot be correct. [1]

(b) Work out a correct estimate for the number of ladybirds in the garden. [2]

(a) e.g. he has already caught 30 (or 45) different ladybirds, so there must be at least 45 / the estimate must be more than the number caught (accept 'he has used the numbers the wrong way round – it should be $30 \times 45 \div 6$')C1
(b) $\frac{6}{45} = \frac{30}{N}$ oe, or $30 \times 45 \div 6$, or $45 \div 6 \times 30$ (= 7.5 × 30)M1
(b) 225 caoA1
Answer: (a) He caught 45 ladybirds on Tuesday alone, so there must be more than 4 (b) 225
Examiner tip: Sense-check a capture–recapture answer: it can never be smaller than the number of animals you have actually caught.

12Grade 4–5Calculator3 marks

The table shows the number of members in each age group of a walking club.

Age group	Under 18	18–40	41–60	Over 60
Number of members	145	268	191	96

The club secretary takes a sample of 60 members. The number of members from each age group in the sample is proportional to the number of members in that age group.

(a) Work out the number of members aged 41–60 in the sample. [2]

(b) The secretary works out the number in the sample for each age group and rounds each one to the nearest whole number. The four numbers add up to 59, not 60.

Explain why. [1]

(a) $\frac{191}{700} \times 60$ oe (= 16.37…); total 700 must be usedM1
(a) 16 cao (accept 16.37… → 16; do not accept 16.4 or 17 as a final answer)A1
(b) each value has been rounded, and the rounding errors do not cancel out / some were rounded down more than others were rounded up (e.g. 12.43, 16.37 and 8.23 all rounded down, only 22.97 rounded up)C1
Answer: (a) 16 (b) rounding each answer to the nearest whole number changes the total (more was lost rounding down than gained rounding up)
Examiner tip: Total the whole club first (700) – using 191 ÷ 4 groups or 191 ÷ 60 scores nothing; a number of people must be a whole number.

13Grade 6–7HigherCalculator5 marks

A biologist wants to estimate the number of newts in a lake.

On day 1 she catches 80 newts, marks them and releases them.

On day 2 she catches 50 newts. 9 of them are marked. She releases them all.

On day 3 she catches 70 newts. 11 of them are marked.

(a) Use the results from day 1 and day 2 only to work out an estimate for the number of newts in the lake. [2]

(b) Use the day 1 results and both recapture samples together to work out a better estimate for the number of newts in the lake. [2]

(a) $80 \times 50 \div 9$ oe, or $\frac{9}{50} = \frac{80}{N}$M1
(a) 444 or 445 (accept 444.4…; condone 440 only if 444.4… seen)A1
(b) $\frac{9+11}{50+70}$ (= $\frac{20}{120}$) used, e.g. $80 \times 120 \div 20$M1
(b) 480 cao (do not accept the mean of two separate estimates, e.g. (444.4 + 509.1) ÷ 2 ≈ 477)A1
(c) it is based on a larger sample (120 newts rather than 50) (accept 'more newts were checked for marks')C1
Answer: (a) 444 (or 445) (b) 480 (c) it uses a larger sample
Examiner tip: Combine the recapture samples (20 marked out of 120) – do not add 80 + 9 + 11, as the newts marked on day 1 are the only marked ones.

14Grade 6–7HigherCalculator4 marks

Kai uses the capture–recapture method to estimate the number of crabs in a rock pool area.

He catches 40 crabs, marks them and releases them.

A week later he catches some crabs. In this second sample the ratio of marked crabs to unmarked crabs is $2 : 13$.

(a) Work out an estimate for the number of crabs in the rock pool area. [3]

(b) Kai finds out that crabs which have been caught once are more likely to go back into the traps. Explain how this would affect his estimate. [1]

(a) proportion of second sample that is marked $= \frac{2}{15}$ (accept $\frac{2}{2+13}$)M1
(a) $40 \div \frac{2}{15}$ oe, e.g. $40 \div 2 \times 15$ or $\frac{2}{15} = \frac{40}{N}$M1
(a) 300 cao (do not accept 260 from using $\frac{2}{13}$)A1
(b) the estimate would be too low (an underestimate) with a reason, e.g. too many marked crabs are in the second sample, so the fraction marked is too bigC1
Answer: (a) 300 (b) too low – the second sample contains too high a proportion of marked crabs
Examiner tip: In a ratio $2 : 13$ the fraction marked is $\frac{2}{15}$, not $\frac{2}{13}$ – add the parts first.

15Grade 8–9HigherCalculator5 marks

A sports club has $x$ junior members and $3x + 20$ senior members.

A sample of 50 members is taken. The number of junior members and senior members in the sample is proportional to the number of junior and senior members in the club.

There are 12 junior members in the sample.

(a) Work out the value of $x$. [4]

(b) Work out the total number of members in the club. [1]

(a) total number of members $= 4x + 20$ seen or used, or $50 - 12 = 38$ seniors in the sampleM1
(a) $\frac{x}{4x+20} = \frac{12}{50}$ oe, e.g. $\frac{x}{4x+20} \times 50 = 12$, or $\frac{x}{3x+20} = \frac{12}{38}$ oeM1
(a) removes fractions correctly, e.g. $50x = 12(4x+20)$ or $50x = 48x + 240$, or $38x = 12(3x+20)$M1
(a) $x = 120$ caoA1
(b) 500 (ft $4 \times$ their $x + 20$ if M2 scored in (a))B1ft
Answer: (a) $x = 120$ (b) 500
Examiner tip: The fraction for juniors is $x$ out of the WHOLE club ($4x + 20$), not $x$ out of the seniors ($3x + 20$).

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