Using statistics to describe a population - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 18 exam-style questions on Using statistics to describe a population for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

18 questions
1–5 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

There are 1240 students at a school.
Ravi asks a random sample of 80 of these students how they travel to school.
28 of the students in his sample walk to school.

Work out an estimate for the number of students at the school who walk to school.

Model answer

Proportion who walk in the sample = \(\frac{28}{80} = 0.35\)
Estimate = \(0.35 \times 1240 = 434\) students

Explanation

A random sample should be representative of the whole school, so assume the same proportion of all 1240 students walk. Find the proportion in the sample and multiply by the population size.

Mark scheme

1 mark
M1 for \(\frac{28}{80} \times 1240\)
oe, e.g. 0.35 × 1240 or 1240 ÷ 80 × 28
1 mark
A1 for 434
cao; correct answer with no working scores full marks

Question 2

2 marks

A factory makes 5000 light bulbs each day.
To check how long the bulbs last, it tests a sample of 50 bulbs from each day until they stop working.

Write down the population for this test.
Give a reason why the factory tests a sample rather than every bulb.

Model answer

Population: all 5000 light bulbs made that day.
The test runs each bulb until it stops working, so every tested bulb is used up. Testing every bulb would leave no bulbs to sell (and would take a very long time).

Explanation

The population is the whole group you want to know about; the sample is the part you actually test. Sampling is essential when testing is destructive, and it also saves time and money.

Mark scheme

1 mark
B1 for the population: all 5000 bulbs made that day
accept all the bulbs the factory makes
1 mark
C1 for a reason, e.g. testing a bulb destroys it (it is used until it stops working), so testing every bulb would leave none to sell; or it would take too much time / cost too much
oe

Question 3

3 marks

A factory makes 25 000 bolts each week.
In a random sample of 80 bolts, 3 are faulty.
Each faulty bolt costs the factory £0.18 to replace.

Work out an estimate for the cost to the factory of replacing the faulty bolts made in one week.

Model answer

Estimated number of faulty bolts = \(\frac{3}{80} \times 25\,000 = 937.5\)
Cost ≈ \(937.5 \times 0.18 = 168.75\)
About £168.75

Explanation

Assume the sample is representative, so \(\frac{3}{80}\) of all the bolts are faulty. Multiply by 25 000 to estimate the number of faulty bolts, then by the cost of replacing each one. Rounding 937.5 to a whole number before multiplying changes the answer by only a few pence.

Mark scheme

1 mark
P1 for \(\frac{3}{80} \times 25\,000\) (= 937.5)
oe; correct answer with no working scores full marks
1 mark
P1 for their number of faulty bolts × 0.18
1 mark
A1 for £168.75
accept £168.66 to £168.84 (from 937 or 938 bolts)

Question 4

4 marks

A biologist wants to estimate the number of frogs in a pond.
She catches 80 frogs, tags them and releases them.
Later she catches 90 frogs. 6 of these frogs are tagged.

The pond has an area of 650 m².
The pond is overcrowded if there are more than 1.5 frogs per m².

Is the pond likely to be overcrowded?
You must show your working.

Model answer

Estimate of population: \(N = \frac{80 \times 90}{6} = 1200\) frogs
Frogs per m²: \(1200 \div 650 = 1.846...\)
1.85 > 1.5, so the pond is likely to be overcrowded.
(Or: the pond can hold \(1.5 \times 650 = 975\) frogs, and 1200 > 975.)

Explanation

Capture-recapture: the fraction of tagged frogs in the second sample (\(\frac{6}{90}\)) should equal the fraction of the whole population that is tagged (\(\frac{80}{N}\)). Then compare the population density with the limit. The conclusion is 'likely' because the population is only an estimate.

Mark scheme

1 mark
M1 for a correct method to estimate the population, e.g. \(\frac{80 \times 90}{6}\) or \(\frac{6}{90} = \frac{80}{N}\)
1 mark
A1 for 1200 frogs
cao
1 mark
M1 for \(1200 \div 650\) (= 1.846...) or \(1.5 \times 650\) (= 975)
ft their population
1 mark
C1 for Yes, with correct figures, e.g. 1.85 frogs per m² is more than 1.5 (or 1200 > 975)

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