What is S5: Describing populations?

Statistics can summarise a population or estimate its features from representative samples.

How many marks are S5 questions worth?

S5 practice drills on this page are worth 1–5 marks each; marks in a real exam vary by question.

Which past exam papers has S5 appeared in?

We have not yet indexed a verified past-paper appearance for S5 — check the Edexcel GCSE Maths 1MA1 past-papers hub directly.

Apply statistics to describe a population

Draft — not yet indexed

Describing populations

Worked answers and methods for S5 on Edexcel GCSE Maths 1MA1.

Explanation

Statistics can summarise a population or estimate its features from representative samples.
When groups have different sizes, combine their means using a weighted mean: add each group size multiplied by its mean, then divide by the total population.
When different subgroups have different sample proportions, estimate each subgroup separately before adding.
A simple average of group means is only valid when the groups are the same size.

Worked example

A town has $900$ northern and $600$ southern residents. In representative samples, $12$ of $40$ northern residents and $20$ of $50$ southern residents cycle to work. Estimate the town total and percentage.

1.North estimate $=900\times\dfrac{12}{40}=270$ .
2.South estimate $=600\times\dfrac{20}{50}=240$ .
3.Total $=270+240=510$ ; percentage $=\dfrac{510}{1500}\times100=34\%$ .

Answer: About $510$ residents, or $34\%$ of the town.

Common mistakes

Don't fall into the trap of averaging two group means without using the group sizes.
Don't fall into the trap of applying one subgroup's sample proportion to the whole population.

Exam tip

Turn each mean back into a total first; combine totals, then divide once.

Worked practice

Tier 1 · Easy

A representative sample of parcels has mean mass

2.4\,\text{kg}

. Estimate the total mass of

250

parcels in the population.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$600\,\text{kg}$	2	Apply the sample mean to all $250$ parcels: $2.4\times250=600\,\text{kg}$ .

Tier 2 · Standard

A population contains

120

junior members with mean attendance

6.5

sessions and

80

senior members with mean attendance

8

sessions. Work out the mean attendance for the whole population.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$7.1$ sessions	3	The total attendances represented are $120\times6.5=780$ and $80\times8=640$ . Divide their sum by all $200$ members: $(780+640)/200=7.1$ .

Tier 3 · Hard

A town has

1800

residents in the north and

1200

in the south. In representative samples,

15

60

northern residents and

20

50

southern residents cycle to work. Estimate the total number and percentage of the town's residents who cycle to work.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Estimated total $=930$ residents. Estimated percentage $=31\%$ .	5	For the north, estimate $1800\times15/60=450$ . For the south, estimate $1200\times20/50=480$ . The total is $450+480=930$ from a population of $3000$ , so the percentage is $930/3000\times100=31\%$ . Calculating by subgroup correctly accounts for their different sizes.

Tier 1 · Easy

The mean weekly electricity use for a representative sample of

64

households is

118\,\text{kWh}

. Estimate the mean weekly electricity use for all households in the village.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$118\,\text{kWh}$	1	Use the sample mean as the estimate of the population mean. Because the sample is representative, the estimated mean for all households is $118\,\text{kWh}$ .

Tier 2 · Standard

A company has

16

part-time employees whose mean working time is

15

hours per week. The full-time employees have a mean working time of

39

hours per week. The mean working time for all the employees is

23

hours per week. Work out the number of full-time employees.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$8$ full-time employees	3	Let $n$ be the number of full-time employees. The total weekly hours give $16\times15+39n=23(16+n)$ . Therefore $240+39n=368+23n$ , so $16n=128$ and $n=8$ .

Tier 3 · Hard

A company has

280

office employees and

420

warehouse employees. Overall,

32\%

of the employees work part-time. Of the office employees,

20\%

work part-time. Work out the percentage of the warehouse employees who work part-time.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$40\%$	4	There are $280+420=700$ employees, so $0.32\times700=224$ work part-time. Of the office employees, $0.20\times280=56$ work part-time. Therefore $224-56=168$ warehouse employees work part-time, giving $168/420\times100=40\%$ .

Tier 2 · Standard

A school has

2400

pupils. The mean number of days absent last term was

1.5

per pupil. Of the pupils,

25\%

had no absence. Work out the mean number of days absent for the pupils who had at least one day absent.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$2$ days	3	The total number of absence days is $2400\times1.5=3600$ . The number of pupils with at least one day absent is $75\%$ of $2400$ , which is $1800$ . Their mean is therefore $3600/1800=2$ days.

Tier 3 · Hard

A biologist records the masses of fish from a lake in two random samples. Sample A has

40

fish with a mean mass of

620

g. Sample B has

60

fish with a mean mass of

570

g. Work out the mean mass of all

100

sampled fish. Give one reason why this combined mean is a better estimate of the mean mass of fish in the lake than the mean of sample A alone. You must show all your working.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$590$ g The combined sample is larger, so its mean is less affected by sampling variation and is likely to be closer to the population mean.	4	The total mass in sample A is $40\times620=24{,}800$ g and in sample B is $60\times570=34{,}200$ g. The combined total is $59{,}000$ g across $100$ fish, so the combined mean is $59{,}000\div100=590$ g. A larger sample gives a more reliable estimate of the population mean.

Tier 3 · Hard

A survey estimates the number of oak trees affected by a disease in three woods. Wood A has

900

oak trees;

21

of a random sample of

30

are affected. Wood B has

600

oak trees;

16

of a random sample of

25

are affected. Wood C has

500

oak trees;

14

of a random sample of

n

are affected. Using these samples gives an estimated total of

1264

affected oak trees. Work out

n

. State which wood has the lowest estimated proportion of affected oak trees.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$n=28$ . Wood C has the lowest estimated proportion of affected oak trees.	4	The estimates for woods A and B are $900\times21/30=630$ and $600\times16/25=384$ . Wood C therefore contributes $1264-630-384=250$ to the estimated total. Its estimated affected proportion is $250/500=0.5$ , so $14/n=0.5$ and $n=28$ . The estimated proportions for A, B and C are $0.7$ , $0.64$ and $0.5$ , respectively, so Wood C has the lowest proportion.

Q10

Tier 3 · Hard

A representative random sample of

80

one-square-metre plots has a mean of

6.4

seedlings per plot. Use the sample to estimate the number of seedlings in a habitat of area

1250\,\text{m}^2

. A later census finds

7680

seedlings. Work out the percentage error in the estimate as a percentage of the census figure, and state whether the estimate is too high or too low. Write the percentage to two decimal places.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Estimated number $=8000$ seedlings. The percentage error is $4.17\%$ (or $4.2\%$ to 1 d.p.) and the estimate is too high.	4	The estimated density is $6.4$ seedlings per square metre, giving $6.4\times1250=8000$ seedlings. The estimate exceeds the census by $8000-7680=320$ . Relative to the census, the percentage error is $320/7680\times100=4.1666\ldots\%\approx4.17\%$ , so the estimate is too high.

Verified exam appearances

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Other points in S Statistics

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