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GCSE Maths · EdexcelChapter 24 · Statistics
Chapter 24 · Higher tier

GCSE Maths Statistics (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

  • Classify data as categorical, discrete or continuous, and collect it with tally charts, frequency tables, two-way tables and questionnaires
  • Spot bias, choose a fair sample, find a stratified sample and estimate a population with capture-recapture
  • Find the mean, median, mode and range from a list, a frequency table and a grouped frequency table
  • Draw and read bar charts, pictograms, vertical line charts, pie charts, line graphs, time series and stem-and-leaf diagrams
  • Draw scatter graphs, describe correlation, draw a line of best fit and use it to estimate (and know the dangers of extrapolating)
  • Compare distributions using an average and a measure of spread
  • Higher: find quartiles and the interquartile range, draw and use cumulative frequency graphs and box plots, spot outliers
  • Higher: draw and read histograms with unequal class widths using frequency density

You should already know

  • Fractions, decimals and percentages of amounts (Chapters 3, 4 and 5)
  • Reading scales and plotting coordinates (Chapter 12)
  • Angles at a point and using a protractor (Chapter 17)
  • Ratio and proportion (Chapter 15)

Where this leads

  • Probability and relative frequency (Chapter 23)
  • Interpreting graphs of real-life situations (Chapters 12 and 13)
  • Using data in exam problem solving across the course

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: S1, S2, S3, S4, S5, S6.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

Exercise P  Do you remember?

Non-calculator
1
FWork out 35 of 40.
2
FWork out 25% of 80.
3
FWork out 3.6 + 4.5 + 2.9 + 5.0.
4
FAngles at a point add up to 360°. Three angles at a point are 90°, 120° and 100°. Work out the fourth angle.
5
FWrite these numbers in order, smallest first: 7, −2, 0, 12, −5.
24.1

Types of data and collecting data

Edexcel 1MA1: S2, S1, S5
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Types of data and collecting dataComing soon

Statistics is about collecting facts and numbers (called data), organising them, showing them in charts and graphs, and using them to answer questions. Before we collect anything we must decide what kind of data we have, because the kind of data decides which charts and calculations make sense.

DataCategorical(words / labels)Numerical(numbers)Discrete(counted: 0, 1, 2 ...)Continuous(measured)e.g. colour, car make
The main types of data.

Categorical data is described in words: favourite colour, type of pet, make of car. Numerical data is numbers. Numerical data is either discrete or continuous. Discrete data is counted, so it takes separate values: the number of goals in a match (0, 1, 2, 3 …), shoe sizes, or the number of siblings. You cannot score 2.5 goals. Continuous data is measured, so it can take any value in a range: height, mass, time, temperature. Someone can be 1.6 m, 1.63 m or 1.6342 m tall, depending on how carefully we measure.

Primary data is data that you collect yourself, for example by asking questions or carrying out an experiment. Secondary data is data that someone else has already collected, such as figures from a website, a newspaper or a database. Secondary data is quick to get, but you cannot always be sure how it was collected.

Tally charts and frequency tables. To organise raw data we draw a table. We go through the data one item at a time and make one tally mark for each. Tally marks are drawn in groups of five (four upright lines with one line across them) so they are easy to count. The number of tally marks is the frequency. When there are lots of different numbers, we put the data in classes (groups) such as 10 ≤ x < 20. Classes must not overlap and must leave no gaps, so that every value has exactly one place.

Questionnaires. A questionnaire is a set of questions given to people. A good question is clear and short, is not leading (it must not push people toward one answer, like “Don’t you agree that football is the best sport?”), and has response boxes that cover every answer and do not overlap. For numbers of hours, use boxes like “0 to less than 2”, “2 to less than 4” rather than “0–2” and “2–4”, because 2 would fit in both. A time frame helps (“in a typical week”) and an “Other” or “None” box stops people being forced into a wrong answer.

Two-way tables. A two-way table shows two things at once, such as boys and girls and how they travel to school. Each row adds to its row total, each column adds to its column total, and the totals add to the grand total in the bottom right corner. If one number is missing, find a row or column with exactly one gap and use the total.

Key ideas

  • Categorical data is words. Numerical data is numbers: discrete (counted) or continuous (measured).
  • Primary data you collect yourself. Secondary data has already been collected by someone else.
  • A tally mark is one item. Groups of five are drawn as four upright lines with one across. Frequency = number of tally marks.
  • Class intervals must not overlap and must leave no gaps.
  • Good questions are clear, not leading, and have response boxes that cover everything without overlapping.
  • In a two-way table, rows add to row totals and columns add to column totals.
Worked example 24.1.1
Say whether each is categorical, discrete or continuous. (a) The number of text messages sent in a day. (b) The mass of a parcel. (c) The favourite film of each student.

Key words. Categorical data is described in words. Discrete data is numbers that you count. Continuous data is numbers that you measure.

Method. For each item ask: words or numbers? If numbers, counted or measured?

  1. (a) Number of text messages. It is a number, and you find it by counting. You cannot send 2.5 messages.
    Discrete
  2. (b) Mass of a parcel. It is a number, and you find it by measuring. It could be 1.2 kg, 1.25 kg, 1.253 kg …
    Continuous
  3. (c) Favourite film. The answers are names (words), not numbers.
    Categorical

Check. Ask whether a value halfway between two neighbours makes sense. 2.5 messages does not (discrete). 1.25 kg does (continuous). 2.5 films is not a favourite film name, so it is categorical. ✓

Watch out. A number does not automatically mean continuous. If you count it, it is discrete.

Worked example 24.1.2
Twenty students were asked for their favourite sport. Their answers were: Football, Netball, Football, Rugby, Netball, Football, Tennis, Football, Rugby, Netball, Football, Tennis, Netball, Football, Rugby, Football, Netball, Tennis, Football, Netball. Show this in a tally chart and frequency table.

Key words. A frequency table shows how many times each value occurs. The frequency (f) is the number of times. The total of the frequency column is the number of values in the data set.

Method. Go through the list once, putting a tally mark in the right row for every item. Then count the tally marks.

  1. Make a table with a row for each category. There is one row for each different answer.
    Football, Netball, Rugby, Tennis
  2. Go through the data one item at a time, making one tally mark each time. Tally marks are grouped in fives (four upright lines and one across), which makes them easy to count. Tick items off as you go so that none is missed or counted twice.
    SportTallyFrequency
    Football|||||||8
    Netball|||||6
    Rugby|||3
    Tennis|||3
    Total20
  3. Write the number of tally marks as the frequency. The frequency is how many times each answer occurs.
    Football: 8, Netball: 6, Rugby: 3, Tennis: 3
  4. Check the total. The frequencies must add up to the number of items in the list.
    8 + 6 + 3 + 3 = 20

Check. The list has 20 items and the frequencies total 20. ✓

Watch out. Check that the total frequency equals the number of items. Missing an item or counting one twice is the most common slip.

Worked example 24.1.3
The heights, in cm, of 15 students are: 152, 163, 158, 171, 149, 160, 165, 155, 168, 150, 162, 174, 157, 166, 159. Complete a grouped frequency table using the classes 140 ≤ h < 150, 150 ≤ h < 160, 160 ≤ h < 170 and 170 ≤ h < 180.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. Put each value in the class it belongs to, making a tally mark each time. Then count the tally marks.

  1. Look at the inequality signs for each class. The lower end is included in a class like 10 ≤ x < 20, but the upper end is not (it goes in the next class).
    140 ≤ h < 150 cm
    150 ≤ h < 160 cm
    160 ≤ h < 170 cm
    170 ≤ h < 180 cm
  2. Go through the data, one value at a time, putting a tally mark in its class. Every value belongs to exactly one class. Tick each value off as you go.
    HeightTallyFrequency
    140 ≤ h < 150 cm|1
    150 ≤ h < 160 cm|||||6
    160 ≤ h < 170 cm|||||6
    170 ≤ h < 180 cm||2
    Total15
  3. Write the number of tally marks as the frequency.
    1, 6, 6, 2
  4. Check the total. The frequencies must add up to the number of values.
    1 + 6 + 6 + 2 = 15

Check. There are 15 values and the frequencies total 15. ✓

Watch out. A value on the boundary goes in the class whose inequality includes it. Read the < and ≤ signs carefully.

Worked example 24.1.4
The two-way table shows how 59 students travel to school. Find the missing numbers.
TravelBusWalkCarTotal
Boys14   ?   730
Girls1112   ?   29
Total   ?   211359

Method. Find a row or column with only one missing number. Use the fact that rows and columns add to their totals.

  1. Boys, Walk. The boys’ row total is 30 and the other two cells are 14 and 7.
    30 − 14 − 7 = 9
  2. Girls, Car. The girls’ row total is 29 and the other cells are 11 and 12.
    29 − 11 − 12 = 6
  3. Total, Bus. The Bus column has 14 boys and 11 girls.
    14 + 11 = 25

Check. Check the bottom row: 25 + 21 + 13 = 59, which is the grand total. ✓

Watch out. Use the totals given in the table. Subtract the numbers you know from the total of that row or column.

Worked example 24.1.5
A questionnaire asks: “How much do you spend on snacks each week?   £0–£5   £5–£10   £10–£15”. Give two criticisms and write an improved question.

Method. Check the question is clear and that the response boxes cover every answer without overlapping.

  1. Look for overlap. A student who spends exactly £5 could tick the first or the second box.
    Overlapping boxes: £5 and £10 are in two boxes each.
  2. Look for gaps. Someone who spends nothing, or more than £15, has no box.
    No box for £0 (none) or for more than £15.
  3. Write an improved question. Add a time frame, remove overlaps, and cover every possibility.
    “In a typical week, how much do you spend on snacks?   None   Up to £5   More than £5 up to £10   More than £10”

Check. Test with £5, £0 and £20: each has exactly one box. ✓

Watch out. Response boxes must not overlap and must cover all possible answers.

Watch out

Do not call data continuous just because it is a number. Counted numbers (goals, siblings, shoe sizes, money in pounds and pence) are discrete. Measured quantities (time, height, mass) are continuous.

Exercise 24A  Types of data and collecting data

Non-calculator
1
FSay whether each type of data is categorical, discrete or continuous.
(a)The colour of cars in a car park
(b)The number of goals scored in a football match
(c)The time taken to run 100 m
(d)Shoe size
(e)The height of a tree
(f)A person’s favourite film
2
FSay whether each is primary data or secondary data.
(a)Sam asks every student in her class what they ate for breakfast.
(b)Ali looks up last year’s rainfall figures on the Met Office website.
(c)A shop counts its own sales from its till receipts.
(d)A student uses population figures from a government website.
3
FThe colours of 16 cars in a car park were: Red, Blue, Blue, Green, Red, Blue, Yellow, Red, Blue, Green, Blue, Red, Yellow, Blue, Red, Green. Complete a tally chart and frequency table.
4
FThe number of brothers and sisters of each of 24 students is: 2, 1, 0, 3, 1, 2, 2, 1, 0, 4, 2, 1, 1, 3, 2, 0, 1, 2, 3, 1, 2, 0, 1, 2. Complete a frequency table.
5
FThe ages, in years, of 20 people at a club are: 12, 25, 33, 18, 41, 27, 35, 14, 22, 30, 46, 19, 38, 24, 29, 11, 36, 43, 21, 28. Use the classes 10 ≤ a < 20, 20 ≤ a < 30, 30 ≤ a < 40 and 40 ≤ a < 50 to complete a grouped frequency table.
6
FThe two-way table shows how 59 students travel to school. Find the three missing numbers.
TravelBusWalkCarTotal
Boys14   ?   730
Girls1112   ?   29
Total   ?   211359
(a)Boys who walk
(b)Girls who come by car
(c)Total who come by bus
7
FUse the table in question 6 (once it is completed) to answer these questions.
TravelBusWalkCarTotal
Boys149730
Girls1112629
Total25211359
(a)How many girls walk to school?
(b)How many students do not come by car?
(c)What fraction of the boys come by bus? Give your answer in its simplest form.
8
PSThere are 80 students in a school club. 45 are girls. 38 students are in the choir, and 22 of the girls are in the choir. Use the table to work out the missing values.
In the choirNot in the choirTotal
Girls22   ?   45
Boys   ?      ?      ?   
Total38   ?   80
(a)Girls not in the choir
(b)Boys in the choir
(c)Boys not in the choir
9
RA student writes the question: “Everyone agrees that homework is far too long, don’t you?” Give a reason why this is a poor question and write a better version.
10
RA question asks: “How many hours of television do you watch each week?   0–5   5–10   10–15 ”. Explain two things wrong with the response boxes.
11
PSThe frequency table shows the number of pets owned by each student in a class.
Pets ownedFrequency
06
19
27
33
42
(a)How many students are in the class?
(b)How many students own at least 2 pets?
(c)How many pets do the students own altogether?
12
RPriya says: “The number of goals scored in a match is continuous because it is a number.” Is she correct? Explain.
13
RSam collects data on how long students take to travel to school. She says her data is secondary because she wrote it down. Explain whether she is correct, and say what kind of numerical data it is.
14
RThere are 50 students. 26 are boys. 30 prefer football. 8 of the girls prefer football. Complete the two-way table, then work out the fraction of the girls who do not prefer football. Give your answer in its simplest form.
24.2

Sampling and population estimates

Edexcel 1MA1: S1, S5
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Sampling and population estimatesComing soon

The population is the whole group that you want to find out about, such as every student in a school or every fish in a lake. Asking everyone is called a census. It is often too slow, too expensive or impossible (you cannot test every battery, because testing destroys it). So we take a sample: a smaller group chosen from the population. We then use the sample to infer (work out) things about the population.

PopulationSample
A sample is a small part of the population. Here 10 out of 60 are in the sample.

Random and biased samples. A sample is random if every member of the population has an equal chance of being chosen, for example by numbering everyone and using a random number generator or drawing names from a hat. A sample is biased if it unfairly favours some part of the population. Asking only the people leaving a gym about how often people exercise gives a biased result. A larger sample is usually more reliable, but only if it is chosen fairly: a large biased sample is still biased.

Estimating from a sample. If the sample is fair, the fraction of the sample with a property should be close to the fraction of the whole population. For example, if 14 out of a random sample of 50 students walk to school, then about 1450 of all the students walk. In a school of 1200, that is about 1450 × 1200 = 336 students. This is only an estimate. A different sample would give a slightly different answer.

Stratified sampling. If a population has clear groups (strata), such as year groups, a stratified sample takes the right number from each group so that the sample is a fair mini copy of the population. The number taken from each group is group size ÷ whole population × sample size. The method is fair because each group is represented in the same proportion as it appears in the population.

Capture–recapture. How can we count animals we cannot see? Catch a number of them, mark them harmlessly, release them, wait for them to mix back in, then catch a second sample. The fraction of the second sample that is marked should be about the same as the fraction of the whole population that is marked. So population estimate = (number marked first) × (second sample size) ÷ (number marked in the second sample).

Stage 1: catch and mark 40Stage 2: catch 30 moreall 40 are marked and released6 of the 30 are marked (blue)
Capture and recapture: 6 out of 30 are marked, which is one fifth. So the 40 marked animals are about one fifth of the population.

This works only if some assumptions hold: the animals mix thoroughly, no animals are born, die or move in or out between the two catches, the marks do not fall off or harm the animals, and marked animals are no easier or harder to catch. If these are not true, the estimate may be too high or too low.

Key ideas

  • Population = everyone or everything of interest. Sample = a smaller group chosen from it.
  • A random sample gives every member an equal chance. A biased sample favours some members.
  • Estimate from a sample: use the same fraction for the population.
  • Stratified sample: group size ÷ population × sample size.
  • Capture–recapture: population ≈ marked first × second sample ÷ marked in second sample.
  • All estimates rely on assumptions and are not exact.
Worked example 24.2.1
In a random sample of 50 students from a school of 1200 students, 14 walk to school. Estimate how many of the 1200 students walk to school.

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. A random sample should be like the whole population in miniature, so the same fraction of the population should have the property as in the sample.

  1. Write the fraction in the sample. Part over whole: the number with the property out of the sample size.
    1450 = 725
  2. Say that the same fraction applies to the whole population. This is the assumption that the sample is representative.
    725 of 1200
  3. Work out the fraction of the population. Multiply the fraction by the population size.
    0.28 × 1200 = 336

Check. The population is 24 times as big as the sample, and 14 × 24 = 336. ✓

Watch out. The answer is an estimate. A different sample would give a slightly different answer.

Worked example 24.2.2
A teacher takes a stratified sample of 30 students from a school. There are 120 students in Year 9, 100 in Year 10 and 80 in Year 11. How many students should be taken from each year?

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. Stratified sampling: take from each group (stratum) in proportion to its size.

  1. Find the size of the whole population. Add up the sizes of all the groups (strata).
    120 + 100 + 80 = 300
  2. Find the fraction of the population that is being sampled. In a stratified sample each group is sampled in the same proportion as its size in the population.
    30 out of 300: 30300 = 110
  3. Work out each group’s share of the sample. sample for the group = group size ÷ population size × sample size.
    Year 9: 120 ÷ 300 × 30 = 12
    Year 10: 100 ÷ 300 × 30 = 10
    Year 11: 80 ÷ 300 × 30 = 8
  4. Check the sample sizes add up to the sample. The groups together must make the whole sample.
    12 + 10 + 8 = 30

Check. Each group gets 110 of its size: 110 of 120 = 12, 110 of 100 = 10, 110 of 80 = 8. ✓

Watch out. Use the size of each group divided by the size of the whole population, not by the size of the sample.

Worked example 24.2.3
A park warden catches 40 ducks, marks them and releases them. Later she catches 30 ducks and finds that 6 are marked. Estimate the number of ducks in the park.

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. Capture–recapture: the fraction of the second sample that is marked is the same as the fraction of the whole population that is marked.

  1. Write down what we know. First we catch, mark and release some. Later we catch a second sample and count how many of those are marked.
    Marked and released: 40
    Second sample: 30
    Marked in the second sample: 6
  2. Write the proportion marked in the second sample. If the animals have mixed well, this is a fair estimate of the proportion marked in the whole population.
    6 out of 30 = 630
  3. Write the proportion marked in the whole population. Let N be the whole population. The 40 marked ones make up 40 out of N.
    40N
  4. Make the two proportions equal. We assume the second sample is representative of the population.
    630 = 40N
  5. Solve for N. Multiply both sides by N and by the second-sample size, then divide by the number marked in the second sample: N = marked × second sample ÷ marked in second sample.
    N = 40 × 30 ÷ 6 = 1200 ÷ 6 = 200

Check. Of the whole population of 200, 40 were marked: 40200 ≈ 0.200, and 630 = 0.200. The two proportions agree (to within rounding). ✓

Watch out. N = marked × second sample ÷ marked in the second sample. Do not divide by the second sample size or multiply by the number marked in it.

Worked example 24.2.4
A student wants a random sample of 5 students from a class of 28. Describe a fair method.

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. Give every student an equal chance of being picked, and make the choice by chance, not by choice.

  1. Number the students. Give each student a different number from 1 to 28, perhaps using the register order.
    1, 2, 3, …, 28
  2. Choose numbers at random. Use a random number generator or draw numbers from a hat. Ignore repeats.
    Choose 5 different numbers between 1 and 28.
  3. Pick the students with those numbers. These students form the sample.
    Sample of 5 students

Check. Every student had the same chance, 5 out of 28, of being picked. ✓

Watch out. Choosing friends or the first five in the register is not random.

Watch out

For stratified sampling, divide by the size of the whole population and multiply by the sample size. And in capture–recapture, the denominator is the number marked in the second sample, not the first.

Exercise 24B  Sampling and population estimates

Non-calculator
1
FFor each situation, say what the population is and what the sample is.
(a)A teacher asks 20 students in Year 10 whether they like the canteen food. There are 180 students in Year 10.
(b)A factory tests 50 light bulbs from a batch of 2000 bulbs.
2
RGive two reasons why a sample is used instead of asking the whole population.
3
RMia wants to find out how often people in her town use the bus. She asks the first 20 people she meets at the bus station. Explain why this sample may be biased.
4
RWhich of these is a random sample of 10 students from a school? A: the first 10 students to arrive on Monday. B: 10 students chosen by a computer from the full school list. C: the 10 students in the school football team.
5
FIn each case a random sample is taken. Estimate the number in the population who have the property.
(a)A random sample of 40 pupils from a school of 800: 10 are left-handed. How many left-handed pupils are there likely to be in the school?
(b)A random sample of 60 batteries from a batch of 3000: 3 are faulty. How many faulty batteries are there likely to be in the batch?
(c)A random sample of 25 customers from 400 customers: 9 prefer tea. How many prefer tea?
6
FA wildlife officer marks animals, releases them, and later catches a second sample. Estimate the population each time.
(a)Marked first: 30. Second sample: 40. Marked in the second sample: 6.
(b)Marked first: 50. Second sample: 60. Marked in the second sample: 10.
(c)Marked first: 24. Second sample: 50. Marked in the second sample: 8.
7
PSIn a lake, 120 fish are caught, tagged and returned. A week later, 150 fish are caught and 25 of them have tags. (a) Estimate the number of fish in the lake. (b) Give one assumption you made.
(a)Estimate the number of fish
(b)One assumption
8
FA school has the numbers of students shown. Work out how many students from each group should be in each stratified sample.
(a)Year 7: 90, Year 8: 60, Year 9: 150. Sample of 20.
(b)Boys: 240, Girls: 160. Sample of 50.
(c)Year 10: 125, Year 11: 100, Sixth form: 75. Sample of 60.
9
PSA club has 124 juniors, 96 seniors and 80 veterans. A stratified sample of 30 members is taken. How many members of each group are in the sample? Round to the nearest whole number.
10
RBen says: “In a stratified sample, you must take the same number from every group.” Explain why Ben is wrong.
11
RA scientist says: “My population estimate from capture–recapture may be too high because some of the marked animals died before the second sample was taken.” Explain whether she is right.
12
PSA population of 450 is estimated using capture–recapture. 60 animals were marked in the first catch and the second sample was 45. How many marked animals were in the second sample?
13
RIn a sample of 80 pupils from a school, 18 are in the school band. The school has 1050 pupils, and the headteacher says that about 236 pupils are in the band. Is the headteacher right to say “exactly 236”? Give your reasoning, and work out the estimate.
14
RZoe marks 40 deer and later catches 30, of which 5 are marked, giving an estimate of 240. Her friend takes a bigger second sample of 90 and finds 14 marked. Work out the second estimate, then say which estimate you would trust more and why.
24.3

Mean, median, mode and range

Edexcel 1MA1: S4
Grade 1-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Mean, median, mode and rangeComing soon

An average is a single number that tells us what is typical for a set of data. There are three averages, and each one answers the question “what is typical?” in a different way. We also need a measure of how spread out the data is. That measure is the range.

The mean is the total of all the values divided by how many values there are. It is the amount each value would be if the total were shared out equally. The picture shows data as dots on a number line. The mean is the balance point: the number line would balance on a pivot placed at the mean.

123456789mean = 4
The values 2, 3, 3, 4 and 8 have total 20 and there are 5 of them. The mean is 20 ÷ 5 = 4, the balance point.

The median is the middle value when the values are written in order. With an odd number of values it is the middle one. With an even number of values there are two middle ones and the median is halfway between them. The mode is the value that occurs most often. There can be more than one mode, or none if every value occurs once. The mode is the only average that works for categorical data (for example, the most popular colour).

The range is the biggest value minus the smallest value. A small range means the data is close together (consistent). A big range means the data is spread out.

Which average should we use? The mean uses every value, so it is affected by very big or very small values called outliers. If five people earn £18 000, £20 000, £21 000, £22 000 and £150 000, the mean is pulled up by the biggest salary and is not typical. The median is not affected by extreme values, so it is better when there are outliers. The mode is best for categorical data, or when we want the most popular choice.

Finding a missing value. If you know the mean of a set of numbers you know their total: total = mean × number of values. Then subtract the values you already know to find the missing one.

Key ideas

  • Mean = total of the values ÷ number of values.
  • Median = middle value when the values are in order. For an even number of values, halfway between the middle two.
  • Mode = most common value. There may be more than one.
  • Range = biggest value − smallest value.
  • Outliers pull the mean towards them, but they do not change the median much.
  • Total = mean × number of values (useful for missing values).
Worked example 24.3.1
Work out the mean of 12, 15, 9, 18, 14 and 11. Give your answer to 1 decimal place.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. To find the mean, add up all the values and divide by how many values there are.

  1. Add up all the values. The mean uses the total of all the data.
    12 + 15 + 9 + 18 + 14 + 11 = 79
  2. Count how many values there are. The mean shares the total equally between all the values, so we need to know how many there are.
    There are 6 values.
  3. Divide the total by the number of values. mean = total ÷ number of values.
    79 ÷ 6 = 13.1667… = 13.2 to 1 decimal place

Check. The mean must lie between the smallest (9) and the largest (18) value, and 13.2 does. Also 13.2 × 6 = 79.2, which is close to the total 79. ✓

Watch out. The mean is not the middle value. Add every value, and divide by how many values there are (not by the largest value).

Worked example 24.3.2
Work out the median of 7, 3, 9, 4, 12 and 6.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. The median is the middle value when the data is in order.

  1. Write the values in order, smallest to largest. The median is the middle value, so the data must be in order first.
    3, 4, 6, 7, 9, 12
  2. Count the values. We need to know where the middle is.
    There are 6 values.
  3. Find the two middle positions. With an even number of values there are two middle values. They are at positions n ÷ 2 and n ÷ 2 + 1.
    6 ÷ 2 = 3, so the 3rd and 4th values.
  4. Pick out the two middle values. Count along the ordered list.
    3, 4, 6, 7, 9, 12
  5. Find the number halfway between them. Add the two middle values and divide by 2 (their mean).
    (6 + 7) ÷ 2 = 13 ÷ 2 = 6.5

Check. 3 values are below 6.5 and 3 values are above it. ✓

Watch out. Put the numbers in order first. The middle of an unordered list is not the median.

Worked example 24.3.3
Find the mode and the range of 5, 8, 5, 9, 7, 5, 8.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Find the mode by counting and the range by subtracting. Both are done from the same list.

  1. Count how often each value occurs. The mode is the value that occurs most often.
    5 occurs 3 times
    7 occurs 1 time
    8 occurs 2 times
    9 occurs 1 time
  2. State the mode. 5 occurs most often.
    Mode = 5
  3. Find the biggest and smallest values. We need them for the range.
    Biggest = 9, smallest = 5
  4. Subtract. Range = biggest − smallest.
    9 − 5 = 4

Check. 5, 5, 5, 7, 8, 8, 9: the value 5 appears most. 5 + 4 = 9, the biggest value. ✓

Watch out. The mode is a value from the list (5), not the number of times it occurs (3).

Worked example 24.3.4
The mean of five numbers is 8. Four of the numbers are 6, 9, 11 and 7. Work out the fifth number.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. If we know the mean, we can find the total: total = mean × number of values. Then take away the values we already know.

  1. Find how many values there are altogether. The mean is shared between all the values, including the missing one.
    4 known values + 1 missing value = 5 values
  2. Find the total the values must make. total = mean × number of values.
    8 × 5 = 40
  3. Add up the known values.
    6 + 9 + 11 + 7 = 33
  4. Take the known total away from the full total. What is left is the missing value.
    40 − 33 = 7

Check. Add the missing value in: 33 + 7 = 40, and 40 ÷ 5 = 8. ✓

Watch out. Multiply the mean by the number of values including the missing one (not by the number you already know).

Worked example 24.3.5
Five people earn £18 000, £20 000, £21 000, £22 000 and £150 000 a year. The owner says: “The average wage is £46,200.” Is the mean a good average to use? Which average would be better?

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Work out the mean and the median, then decide which one is more typical.

  1. Work out the mean. Add the salaries and divide by 5.
    18 000 + 20 000 + 21 000 + 22 000 + 150 000 = 231,000
    231,000 ÷ 5 = £46,200
  2. Work out the median. The salaries are already in order. The middle (3rd) one is the median.
    Median = £21,000
  3. Compare. Four of the five people earn about £20 000, so £46,200 is far above what most people earn. The very large salary (an outlier) has pulled the mean up.
    The mean is not typical.
  4. Choose the better average. The median is not affected by the one very large value.
    The median (£21,000) is a better average.

Check. Four of the five wages are within £4 000 of the median, so the median describes most of the people. ✓

Watch out. When there is an extreme value (an outlier), the median is usually a better average than the mean.

Watch out

Always put the data in order before finding the median. And the mode is the value that occurs most often, not how many times it occurs.

Exercise 24C  Mean, median, mode and range

Non-calculator
1
FWork out the mean of each set of numbers.
(a)4, 7, 10
(b)12, 15, 9, 14
(c)3, 8, 5, 6, 8
(d)21, 18, 24, 17, 20
(e)2.5, 3.5, 4.5
(f)£2.40, £3.60, £1.50, £4.50
2
FWork out the median of each set of numbers.
(a)5, 9, 2, 7, 3
(b)11, 4, 8, 15, 6
(c)12, 7, 3, 9
(d)20, 14, 18, 22, 11, 16
(e)1.5, 0.8, 2.2, 1.1
(f)−3, 5, 0, −8, 2
3
FWrite down the mode of each set of data.
(a)3, 5, 2, 5, 7, 5, 2
(b)8, 6, 8, 9, 6, 8, 7
(c)red, blue, red, green, blue, red
(d)4, 7, 4, 9, 7, 2 (two modes)
4
FWork out the range of each set of numbers.
(a)4, 9, 3, 12, 7
(b)15, 22, 18, 30, 25
(c)1.2, 0.7, 2.5, 1.9
(d)−3, 5, 0, 8
5
FHere are the numbers of minutes that nine students took to eat lunch: 6, 9, 4, 9, 7, 5, 9, 8, 6. Work out (a) the mean, (b) the median, (c) the mode, (d) the range.
(a)Mean
(b)Median
(c)Mode
(d)Range
6
PSFour friends bought drinks costing £2.40, £1.80, £3.20 and £2.60. Work out the mean cost of a drink.
7
PSWork out the missing number each time.
(a)The mean of four numbers is 7. Three of them are 5, 8 and 9. Find the fourth number.
(b)The mean of five numbers is 12. Four of them are 10, 15, 11 and 14. Find the fifth number.
(c)The mean mass of six parcels is 4.5 kg. Five parcels weigh 3, 5, 4, 6 and 4.5 kg. Find the mass of the sixth parcel.
8
PSChloe sat five tests. Her scores were 14, 17, 12, 15 and 17 out of 20. She wants a mean of 15 after a sixth test. What score does she need in the sixth test?
9
RFive people earn £18 000, £20 000, £21 000, £22 000 and £150 000 a year. Which average, the mean or the median, gives a better idea of what the people typically earn? Give a reason.
10
RAisha has to find the median of 3, 9, 4, 7 and 5. She says the median is 4 because it is the middle number in the list. Explain her mistake and work out the correct median.
11
RThe list 4, 5, 6, 7, 8 has mean 6 and median 6. The number 8 is changed to 28, giving 4, 5, 6, 7, 28.
(a)Work out the new mean.
(b)Work out the new median.
(c)Which average changed more?
12
PSThe mean of 8 numbers is 12. A ninth number, 21, is added. Work out the mean of the nine numbers.
13
PSClass A has 10 students and a mean mark of 62. Class B has 15 students and a mean mark of 70. Work out the mean mark of all 25 students.
14
RWrite down five numbers with a mean of 6 and a range of 7.
15
RFour whole numbers have a mean of 5, a median of 4.5 and a mode of 3. Find the four numbers, written in order.
24.4

Averages from frequency tables

Edexcel 1MA1: S2, S4
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Averages from frequency tablesComing soon

When lots of values repeat, it is quicker to list them in a frequency table. The table shows each value and its frequency: how many times it occurs. For example, if 5 families have 2 children, the table shows the value 2 with frequency 5.

Goals scored in 20 matches0246801234Number of goalsFrequency
A vertical line chart for the table: the height of each line is the frequency.

The same data can be shown in a vertical line chart (above). The height of each line is the frequency. Use it for discrete numerical data. The total frequency, found by adding the frequencies, is the number of values in the data set.

The mean from a frequency table. We cannot just add the values in the first column, because they occur different numbers of times. Instead add a third column, frequency × value. For the row with value 2 and frequency 6, the 6 values of 2 add to 12. Adding the new column gives the total of all the values, and the mean is total of (frequency × value) ÷ total frequency.

The mode is the value with the highest frequency. The median is the middle value of the data in order. To find it, work out the position of the middle, (n + 1) ÷ 2, and add a running total (cumulative frequency) down the table to find which value is in that position. The range is the biggest value minus the smallest value.

Key ideas

  • Total frequency = number of values in the data set.
  • Mean = total of (frequency × value) ÷ total frequency.
  • Mode = the value with the highest frequency (not the frequency itself).
  • Median: position (n + 1) ÷ 2; use a running total of the frequencies to find the value in that position.
  • Range = biggest value − smallest value.
Worked example 24.4.1
The table shows the number of goals scored in each of 20 matches.
Goals01234
Frequency47621
Work out the mean number of goals.

Key words. A frequency table shows how many times each value occurs. The frequency (f) is the number of times. The total of the frequency column is the number of values in the data set.

Method. For a frequency table, multiply each value by its frequency, add these up, and divide by the total frequency.

  1. Add a column for frequency × value. Each value appears f times, so it contributes f × value to the total. This saves writing the value out f times.
    GoalsFrequency (f)f × goals
    040
    177
    2612
    326
    414
    Total2029
  2. Add up the frequency column. This gives the number of values in the data set.
    4 + 7 + 6 + 2 + 1 = 20
  3. Add up the (f × value) column. This gives the total of all the values in the data set.
    0 + 7 + 12 + 6 + 4 = 29
  4. Divide the total by the number of values. mean = total ÷ number of values.
    29 ÷ 20 = 1.45

Check. The mean must be between the smallest and largest values (0 and 4). 1.45 is. Also 1.45 × 20 = 29. ✓

Watch out. Divide by the total frequency (the number of values), not by the number of rows in the table.

Worked example 24.4.2
Use the table in the last example to find (a) the median, (b) the mode and (c) the range of the number of goals.

(a) Median.

Key words. A frequency table shows how many times each value occurs. The frequency (f) is the number of times. The total of the frequency column is the number of values in the data set.

Method. Use a running total of the frequencies to find which value is in the middle position.

  1. Add up the frequencies. This is the number of values, n.
    4 + 7 + 6 + 2 + 1 = 20
  2. Find the positions of the two middle values. With an even number of values we need the two middle ones, at n ÷ 2 and n ÷ 2 + 1.
    20 ÷ 2 = 10, so the 10th and 11th values.
  3. Add a running total of the frequencies. The running total tells us which values occupy which positions when the data is in order.
    GoalsFrequencyRunning total (cumulative)
    044
    1711
    2617
    3219
    4120
  4. Find the 10th value. The running total first reaches 10 or more on the row for 1: that row holds positions 5 to 11.
    10th value = 1
  5. Find the 11th value. The running total first reaches 11 or more on the row for 1: that row holds positions 5 to 11.
    11th value = 1

Check. There are 20 values, so 10 lie below the middle (or middle pair) and the same number above. ✓

Watch out. The median is the value in the middle position, not the middle frequency or the middle row of the table.

(b) Mode.

Key words. A frequency table shows how many times each value occurs. The frequency (f) is the number of times. The total of the frequency column is the number of values in the data set.

Method. The mode is the value with the highest frequency.

  1. Find the highest frequency. The value that occurs most often has the biggest frequency.
    Frequencies: 4, 7, 6, 2, 1. The highest is 7.
  2. Read the value that goes with it. The question asks for the mode, which is the value, not the frequency.
    Mode = 1

Check. 1 occurs 7 times, more than any other value. ✓

Watch out. Do not give the frequency 7 as the answer. The mode is the value that has that frequency.

(c) Range.

Key words. A frequency table shows how many times each value occurs. The frequency (f) is the number of times. The total of the frequency column is the number of values in the data set.

Method. Range = biggest value − smallest value. Only look at values that actually occur (frequency more than 0).

  1. Find the smallest value that occurs. Ignore any row with frequency 0.
    Smallest = 0
  2. Find the biggest value that occurs.
    Biggest = 4
  3. Subtract. Biggest minus smallest.
    4 − 0 = 4

Check. 0 + 4 = 4. ✓

Watch out. Do not subtract the frequencies. The range uses the values.

Worked example 24.4.3
The table shows the number of books read last month by 25 students.
Books012345
Frequency258631
Work out the mean number of books read.

Key words. A frequency table shows how many times each value occurs. The frequency (f) is the number of times. The total of the frequency column is the number of values in the data set.

Method. For a frequency table, multiply each value by its frequency, add these up, and divide by the total frequency.

  1. Add a column for frequency × value. Each value appears f times, so it contributes f × value to the total. This saves writing the value out f times.
    BooksFrequency (f)f × books
    020
    155
    2816
    3618
    4312
    515
    Total2556
  2. Add up the frequency column. This gives the number of values in the data set.
    2 + 5 + 8 + 6 + 3 + 1 = 25
  3. Add up the (f × value) column. This gives the total of all the values in the data set.
    0 + 5 + 16 + 18 + 12 + 5 = 56
  4. Divide the total by the number of values. mean = total ÷ number of values.
    56 ÷ 25 = 2.24

Check. The mean must be between the smallest and largest values (0 and 5). 2.24 is. Also 2.24 × 25 = 56. ✓

Watch out. Divide by the total frequency (the number of values), not by the number of rows in the table.

Worked example 24.4.4
The table shows the scores when a spinner is spun a number of times. The frequency for a score of 2 is missing. The mean score is 2.5. Work out the missing frequency.
Score1234
Frequency5x83

Method. Write the total frequency and the total of (frequency × score) using x, then use the mean to make an equation.

  1. Write the total frequency. Add the frequencies, including x.
    5 + x + 8 + 3 = 16 + x
  2. Write the total of the scores. Multiply each score by its frequency.
    1 × 5 + 2 × x + 3 × 8 + 4 × 3 = 5 + 2x + 24 + 12 = 41 + 2x
  3. Use the mean to write an equation. mean = total of scores ÷ total frequency.
    (41 + 2x) ÷ (16 + x) = 2.5
  4. Multiply both sides by (16 + x). This removes the fraction.
    41 + 2x = 2.5 × (16 + x) = 40 + 2.5x
  5. Collect the x terms and the numbers. Take 2x and 40 from both sides.
    41 − 40 = 2.5x − 2x
    1 = 0.5x
  6. Solve. Divide both sides by 0.5.
    x = 1 ÷ 0.5 = 2

Check. With x = 2: total frequency = 18, total of scores = 41 + 4 = 45, and 45 ÷ 18 = 2.5. ✓

Watch out. The mean is the total of (frequency × score) divided by the total frequency, and both contain x.

Watch out

For the mean, divide by the total frequency (the number of values), not by the number of rows in the table. And the mode is the value with the highest frequency, not the frequency.

Exercise 24D  Averages from frequency tables

Non-calculator
1
FThe vertical line chart shows the number of cars owned by each of 30 households.
(a) How many households own 2 cars? (b) What is the mode? (c) How many households have more than 2 cars? (d) Work out the mean number of cars.
024681012012345Number of carsFrequency
(a)Households with 2 cars
(b)Mode
(c)Households with more than 2 cars
(d)Mean number of cars
2
FWork out the mean from each frequency table.
(a)
Value1234
Frequency3683
(b)
Value0123
Frequency5843
(c)
Value10203040
Frequency2431
3
FWork out the median from each frequency table.
(a)
Value1234
Frequency3683
(b)
Value0123
Frequency5843
(c)
Value246810
Frequency14753
4
FWrite down the mode and the range from each frequency table.
(a)
Value1234
Frequency3683
Mode
(b)
Value5678
Frequency9425
Mode
(c)
Value1234
Frequency3683
Range
(d)
Value5678
Frequency9425
Range
5
PSThe table shows the shoe sizes of the 30 students in a class.
Shoe size345678
Frequency259743
Work out (a) the mode, (b) the median, (c) the range, (d) the mean, to 1 decimal place.
(a)Mode
(b)Median
(c)Range
(d)Mean
6
PSA hockey team played 30 matches. The table shows the goals scored.
Goals012345
Matches498531
Work out the mean number of goals per match, correct to 2 decimal places.
7
PSComplete the table, given that 40 people were surveyed.
Passengers per car1234
Cars914?6
Work out the missing frequency, then the number of passengers in total.
(a)Missing frequency
(b)Total passengers
8
RThe tables show the number of siblings of the students in two classes.
Class A
Siblings0123
Frequency6932
Class B
Siblings0123
Frequency4673
Which class has the higher mean number of siblings? Show your working.
9
RJon works out the mean of this table by adding the values and dividing by the number of rows: (1 + 2 + 3 + 4) ÷ 4 = 2.5.
Value1234
Frequency3863
Explain his mistake and work out the correct mean.
10
PSThe table shows the number of texts received by 20 students one evening.
Texts12345
Frequency46721
(a)Work out the mean number of texts.
(b)How many students received more texts than the mean?
11
PSThe table shows the scores of students in a quiz. The mean score is 2.4.
Score1234
Frequency4x63
Work out the value of x.
12
RThe frequencies of the scores 1 to 5 are 3, a, b, 4 and 2. There are 20 values in total and the mean is 2.85. Work out a and b.
13
RA frequency table has a mode of 3, but a mean that is bigger than 3. Give an example of a table where this happens and show that it works.
14
FThe table shows the number of rooms in each of 25 houses.
Rooms23456
Houses36952
Work out (a) the mean, (b) the median, (c) the mode and (d) the range of the number of rooms.
(a)Mean
(b)Median
(c)Mode
(d)Range
15
RTable A has values 2, 3, 4 with frequencies 1, 8, 1. Table B has values 1, 3, 5 with frequencies 3, 4, 3. Show that both tables have the same mean but different ranges.
24.5

Averages from grouped frequency tables

Edexcel 1MA1: S2, S4
Grade 3-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Averages from grouped frequency tablesComing soon

When data covers a wide range (heights, times, masses), it is put into classes (groups) in a grouped frequency table. For example, 12 students with heights between 150 cm and 160 cm are all in one class. The table tells us how many values are in each class, but we lose the exact values.

Because we no longer know the exact values, we cannot work out the exact mean, median or range. We can find: the modal class (the class with the highest frequency), the class that contains the median, and an estimate of the mean. To estimate the mean, we pretend that every value in a class is at the midpoint of the class. The midpoint is halfway between the two ends: add them and divide by 2. Then use the same method as for a frequency table: multiply the frequency by the midpoint, add up, and divide by the total frequency.

05101520Height (cm)Frequency140-1504150-16010160-17014170-1808180-1904
A bar chart of the grouped heights in the worked examples. The modal class is the tallest bar.

Median class. Add up the frequencies to find n, work out the position of the middle value (n + 1) ÷ 2, then add a running total of the frequencies. The median class is the class where the running total first reaches that position.

Key ideas

  • The modal class is the class with the highest frequency (when the classes have equal width).
  • Midpoint of a class = (lower end + upper end) ÷ 2.
  • Estimated mean = total of (frequency × midpoint) ÷ total frequency.
  • The median class contains the value at position (n + 1) ÷ 2: use a running total of frequencies.
  • Grouped data gives only estimates, because the exact values are not known.
Worked example 24.5.1
The table shows the heights of 40 students.
HeightFrequency
140 < h ≤ 150 cm4
150 < h ≤ 160 cm10
160 < h ≤ 170 cm14
170 < h ≤ 180 cm8
180 < h ≤ 190 cm4
Work out an estimate for the mean height.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.

  1. Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.
    (140 + 150) ÷ 2 = 145
    (150 + 160) ÷ 2 = 155
    (160 + 170) ÷ 2 = 165
    (170 + 180) ÷ 2 = 175
    (180 + 190) ÷ 2 = 185
  2. Multiply each frequency by its midpoint. This estimates the total of the values in that class.
    HeightFrequency (f)Midpointf × midpoint
    140 < h ≤ 150 cm4145580
    150 < h ≤ 160 cm101551550
    160 < h ≤ 170 cm141652310
    170 < h ≤ 180 cm81751400
    180 < h ≤ 190 cm4185740
    Total406580
  3. Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.
    Total frequency = 40
    Total of f × midpoint = 6580
  4. Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.
    6580 ÷ 40 = 164.5 cm

Check. The estimate should be inside the range of the classes, from 140 to 190, and 164.5 is. It is only an estimate because the real values are not known. ✓

Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.

Worked example 24.5.2
Use the table in the last example to write down (a) the modal class, (b) the class that contains the median.

(a) Modal class.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. The modal class is the class with the highest frequency.

  1. Look down the frequency column for the biggest number. The class that has the most values is the modal class.
    Frequencies: 4, 10, 14, 8, 4. The highest is 14.
  2. Read the class that goes with it. The answer is the class, not the frequency.
    Modal class = 160 < h ≤ 170 cm

Check. No other class has as many as 14 values. ✓

Watch out. Give the class (the group), not the frequency.

(b) Median class.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. The median class is the class that contains the middle value.

  1. Find the total frequency. This is the number of values.
    4 + 10 + 14 + 8 + 4 = 40
  2. Find the position of the middle value. Middle position = (n + 1) ÷ 2.
    (40 + 1) ÷ 2 = 20.5
  3. Add a running total column. The running total shows how many values are in each class and all the classes before it.
    ClassFrequencyRunning total
    140 < h ≤ 150 cm44
    150 < h ≤ 160 cm1014
    160 < h ≤ 170 cm1428
    170 < h ≤ 180 cm836
    180 < h ≤ 190 cm440
  4. Find the class where the running total first reaches that position. Positions 15 to 28 are in this class, and 20.5 is one of them.
    Median class = 160 < h ≤ 170 cm

Check. Before this class there are 14 values, and 20.5 is more than that, but not more than 28. ✓

Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.

Worked example 24.5.3
The table shows the number of text messages sent in one day by 30 students.
TextsFrequency
0–43
5–98
10–1412
15–195
20–242
Work out an estimate for the mean number of texts. Give your answer to 1 decimal place.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.

  1. Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.
    (0 + 4) ÷ 2 = 2
    (5 + 9) ÷ 2 = 7
    (10 + 14) ÷ 2 = 12
    (15 + 19) ÷ 2 = 17
    (20 + 24) ÷ 2 = 22
  2. Multiply each frequency by its midpoint. This estimates the total of the values in that class.
    TextsFrequency (f)Midpointf × midpoint
    0–4326
    5–98756
    10–141212144
    15–1951785
    20–2422244
    Total30335
  3. Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.
    Total frequency = 30
    Total of f × midpoint = 335
  4. Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.
    335 ÷ 30 = 11.1667… = 11.2 to 1 d.p.

Check. The estimate should be inside the range of the classes, from 0 to 24, and 11.2 is. It is only an estimate because the real values are not known. ✓

Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.

Worked example 24.5.4
Alex says: “The estimated mean of 164.5 cm for the heights is the exact mean.” Is Alex right?

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. Think about what information is missing when data is grouped.

  1. Recall what grouped data tells us. We know how many heights are in each class, but not where in the class each height is.
    For example, the 14 students in 160 < h ≤ 170 could be anywhere from just above 160 to 170.
  2. Recall what we did to find the mean. We used the midpoint of each class, which is a guess.
    The midpoint is an assumption.
  3. Conclude. The answer depends on the assumption, so it is an estimate, not the exact mean.
    Alex is wrong.

Check. If all 14 students in a class were at the bottom of the class, the true mean would be lower. ✓

Watch out. Grouped data gives an estimate. Say “estimate” in your answer.

Watch out

For grouped data multiply the frequency by the midpoint of the class, and divide by the total frequency. Do not use the class limits, and do not divide by the number of classes. The answer is only an estimate.

Exercise 24E  Averages from grouped frequency tables

Calculator allowed
1
FWrite down the modal class from each table.
(a)
ClassFrequency
0 < x ≤ 105
10 < x ≤ 209
20 < x ≤ 304
30 < x ≤ 402
(b)
ClassFrequency
10 < x ≤ 206
20 < x ≤ 3012
30 < x ≤ 408
40 < x ≤ 504
(c)
ClassFrequency
100 < x ≤ 1208
120 < x ≤ 14010
140 < x ≤ 16015
160 < x ≤ 1807
2
FWrite down the class that contains the median from each table.
(a)
ClassFrequency
0 < x ≤ 105
10 < x ≤ 209
20 < x ≤ 304
30 < x ≤ 402
(b)
ClassFrequency
10 < x ≤ 206
20 < x ≤ 3012
30 < x ≤ 408
40 < x ≤ 504
(c)
ClassFrequency
100 < x ≤ 1208
120 < x ≤ 14010
140 < x ≤ 16015
160 < x ≤ 1807
3
FWork out an estimate for the mean from each table. Give each answer to 1 decimal place where necessary.
(a)
ClassFrequency
0 < x ≤ 105
10 < x ≤ 209
20 < x ≤ 304
30 < x ≤ 402
(b)
ClassFrequency
10 < x ≤ 206
20 < x ≤ 3012
30 < x ≤ 408
40 < x ≤ 504
(c)
ClassFrequency
100 < x ≤ 1208
120 < x ≤ 14010
140 < x ≤ 16015
160 < x ≤ 1807
4
PSThe table shows the times, in minutes, that 40 customers waited at a bank.
Waiting timeFrequency
0 < t ≤ 10 min3
10 < t ≤ 20 min8
20 < t ≤ 30 min14
30 < t ≤ 40 min10
40 < t ≤ 50 min5
(a) Write down the modal class. (b) Which class contains the median? (c) Work out an estimate for the mean waiting time.
(a)Modal class
(b)Median class
(c)Estimated mean
5
FThe grouped table shows the number of goals scored by players in a season.
GoalsFrequency
0–46
5–99
10–147
15–193
Work out an estimate for the mean number of goals. Give your answer to 1 decimal place.
6
FThe table shows the lengths of phone calls. The classes have different widths.
LengthFrequency
0 < t ≤ 10 min6
10 < t ≤ 30 min10
30 < t ≤ 60 min4
Work out an estimate of the mean length of a call.
7
RTen people have ages 12, 18, 21, 25, 27, 31, 34, 36, 41, 45. (a) Work out the exact mean age. (b) Put the ages into the classes 10 ≤ a < 20, 20 ≤ a < 30, 30 ≤ a < 40 and 40 ≤ a < 50 and work out an estimate of the mean from the grouped table. (c) Compare your answers.
(a)Exact mean
(b)Grouped table and estimated mean
(c)Compare
8
RExplain why you cannot work out the exact median or range from a grouped frequency table.
9
PSThe table shows the masses of 50 parcels.
MassFrequency
0 < m ≤ 20 g8
20 < m ≤ 40 g18
40 < m ≤ 60 g14
60 < m ≤ 80 g10
(a) Which class contains the median? (b) How many parcels weigh more than 40 g? (c) What percentage of the parcels weigh 20 g or less?
(a)Median class
(b)Parcels heavier than 40 g
(c)Percentage 20 g or less
10
FA grouped table has classes 0 < x ≤ 20, 20 < x ≤ 40 and 40 < x ≤ 60 with frequencies 8, 14 and 18. Work out an estimate for the mean.
11
PSThe estimated mean of a grouped table is 20. The table has two classes, 0 < x ≤ 20 with frequency 10 and 20 < x ≤ 40 with frequency f. Work out f.
12
FThe table shows the times, in seconds, taken by 30 runners to complete a race.
TimeFrequency
40 < t ≤ 50 s3
50 < t ≤ 60 s9
60 < t ≤ 70 s12
70 < t ≤ 80 s6
(a) Write down the modal class. (b) Which class contains the median? (c) Work out an estimate for the mean time.
(a)Modal class
(b)Median class
(c)Estimated mean
13
PSUse the estimated mean time of the 30 runners in the last question to estimate the total time taken by all 30 runners together.
14
RLee says: “If the classes in a grouped table are very narrow, then the mean from the table is exactly the true mean.” Is Lee right? Explain.
24.6

Bar charts, pictograms and line charts

Edexcel 1MA1: S2
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Bar charts, pictograms and line chartsComing soon

Charts turn a table of numbers into a picture, so patterns are easier to see. The right chart depends on the type of data. Bar charts and pictograms are used for categorical data and for discrete data. A vertical line chart (see Section 24.4) is used for discrete numerical data.

051015Favourite fruitNumber of studentsApple12Banana9Orange8Grape6Pear5
A bar chart. The height of each bar is the frequency. The bars have equal widths and gaps between them.

Bar charts. The bars have equal widths with gaps between them, because the categories are separate. The height of a bar shows the frequency. Both axes must be labelled and the vertical scale needs equal steps starting at 0. To read a bar, find its top and read across to the scale.

024681012Favourite subjectNumber of studentsMathsScienceEnglishArtBoysGirls
A dual bar chart puts two bars side by side for each category so that two groups can be compared.

Dual bar charts show two groups side by side, and composite bar charts stack the groups on top of each other so that the total height shows the whole. Either way, a key shows what each colour means.

Books borrowed from a libraryMonTueWedThurepresents 4 books
A pictogram. Each symbol stands for 4 books, so half a symbol stands for 2 books.

Pictograms use a picture or symbol to stand for a number of items. There must be a key that says what one symbol stands for. A part of a symbol stands for a part of that number: a half symbol is half the value. To find the number of symbols for a frequency, divide the frequency by the value of one symbol.

Misleading charts. Charts can give a false impression. Watch for: a vertical axis that does not start at 0 (the bars look very different when the values are almost the same); a scale that is not evenly spaced; bars of different widths; missing labels or titles; and 3-D effects. Always look at the numbers on the axis before you decide what a chart shows.

Key ideas

  • Bar chart: equal-width bars with gaps; height = frequency; equal scale starting at 0; label both axes.
  • Dual and composite bar charts compare groups; use the key.
  • Pictogram: always use a key. Number of symbols = frequency ÷ value of one symbol.
  • Vertical line chart: discrete numerical data; the height of each line is the frequency.
  • Be suspicious of charts whose axes do not start at 0 or whose scales are uneven.
Worked example 24.6.1
The bar chart shows the favourite fruit of 40 students.
051015Favourite fruitNumber of studentsApple12Banana9Orange8Grape6Pear5
(a) Which fruit is the most popular? (b) How many more students chose apple than banana? (c) How many students chose pear or grape?

Method. Read the height of each bar against the vertical scale, then use the numbers.

  1. (a) Find the tallest bar. The tallest bar has the highest frequency, so it is the most popular.
    Apple is tallest, with frequency 12. The most popular fruit is apple.
  2. (b) Read the apple and banana bars. Read across from the top of each bar.
    Apple = 12, banana = 9
  3. Subtract. “How many more” means take away.
    12 − 9 = 3
  4. (c) Read the pear and grape bars and add. “Or” means add the two frequencies.
    Pear = 5, grape = 6
    5 + 6 = 11

Check. The bars add to 40, the number of students: 12 + 9 + 8 + 6 + 5. ✓

Watch out. Read the height on the vertical scale, not the position along the bottom.

Worked example 24.6.2
The pictogram shows the number of books borrowed from a library on four days. One symbol stands for 4 books.
MonTueWedThurepresents 4 books
(a) How many books were borrowed on Thursday? (b) How many books were borrowed altogether? (c) How many more were borrowed on Tuesday than on Wednesday?

Method. Use the key: each whole symbol is 4 books and each half symbol is 2 books.

  1. (a) Count the symbols on Thursday. Thursday has 3 whole symbols and one half symbol.
    3 × 4 + 2 = 12 + 2 = 14 books
  2. (b) Work out each day. Do the same for each row.
    Monday: 3 × 4 = 12
    Tuesday: 4 × 4 + 2 = 18
    Wednesday: 2 × 4 + 2 = 10
    Thursday: 14
  3. Add the days.
    12 + 18 + 10 + 14 = 54 books
  4. (c) Compare Tuesday and Wednesday. Subtract.
    18 − 10 = 8 books

Check. Total symbols: 3 + 4.5 + 2.5 + 3.5 = 13.5, and 13.5 × 4 = 54. ✓

Watch out. Do not forget the key. A half symbol is half of the value of a whole symbol, here 2.

Worked example 24.6.3
The dual bar chart shows the favourite subject of boys and of girls in a year group.
024681012Favourite subjectNumber of studentsMathsScienceEnglishArtBoysGirls
(a) How many boys chose science? (b) Which subject did more girls than boys choose by the largest amount? (c) How many students are in the year group?

Method. Read the blue bars for the boys and the orange bars for the girls. Use the key.

  1. (a) Read the blue bar for science. Blue stands for boys.
    10 boys
  2. (b) Find the differences girls − boys for each subject. Compare the two bars in each pair.
    Maths: 6 − 8 = -2
    Science: 9 − 10 = -1
    English: 9 − 5 = 4
    Art: 6 − 3 = 3
  3. Pick the largest positive difference. This is the subject where girls beat boys by the most.
    English (4 more girls)
  4. (c) Add every bar. Each student chose one subject, so every bar counts different students.
    Boys: 8 + 10 + 5 + 3 = 26
    Girls: 6 + 9 + 9 + 6 = 30
    26 + 30 = 56

Check. Girls − boys: Maths −2, Science −1, English +4, Art +3. The largest is English. ✓

Watch out. Check which colour is which in the key before you read the bars.

Worked example 24.6.4
The table shows how 34 students travel to school.
TransportFrequency
Bus14
Walk9
Car7
Cycle4
Draw a bar chart to show this information.

Method. Draw two axes, choose an even scale, then draw one bar for each category. All the bars must have the same width, with equal gaps between them.

  1. Choose a scale for the vertical axis. The biggest frequency is 14, so the axis must go up to at least 14. Use equal steps, such as 5.
    Vertical axis: 0 to 15 in steps of 5
  2. Label both axes and give the chart a title. The reader must know what each axis shows.
    Horizontal axis: Method of travel. Vertical axis: Number of students.
  3. Draw one bar for each category. The height of each bar is its frequency, read against the vertical scale.
    Bus: height 14
    Walk: height 9
    Car: height 7
    Cycle: height 4
  4. Check against the table. Every bar should match its frequency.
    05101520Method of travelNumber of studentsBus14Walk9Car7Cycle4

Check. Each bar’s height matches the table. ✓

Watch out. Bars must have equal widths and the vertical scale must start at 0 and have equal steps.

Worked example 24.6.5
A newspaper chart shows the sales of two phones: Phone A sold 82 000 and Phone B sold 85 000. The vertical axis starts at 80 000.
80818283848586Sales (thousands)Phone A82Phone B85
Explain why the chart is misleading.

Method. Compare what the chart looks like with what the numbers say.

  1. Look at the bars. The bar for Phone B looks much taller than the bar for Phone A. Measuring from the axis at 80 gives heights 2 and 5.
    Heights: Phone A = 82 − 80 = 2, Phone B = 85 − 80 = 5. The bar for B looks 2.5 times as tall.
  2. Look at the actual numbers. The sales are almost the same.
    85 ÷ 82 = 1.037: Phone B sold only about 4% more.
  3. Find the cause. The vertical axis does not start at 0, so the bars do not show the true sizes.
    The axis should start at 0.

Check. If the axis started at 0 the bars would be almost equal in height. ✓

Watch out. Always check where the scale starts.

Watch out

Always check the key and the scale before reading a chart. A pictogram symbol rarely stands for 1, and a chart with an axis that does not start at 0 can mislead.

Exercise 24F  Bar charts, pictograms and line charts

Non-calculator
1
FThe bar chart shows the pets owned by the students in a class.
(a) How many students have a cat? (b) Which is the most common pet? (c) How many more students have a dog than a rabbit? (d) How many students are there in the class? (e) What fraction of the class have a fish? Give your answer in its simplest form.
05101520Type of petNumber of studentsDog15Cat12Rabbit6Fish9Bird3
(a)Students with a cat
(b)Most common pet
(c)How many more dogs than rabbits
(d)Students in the class
(e)Fraction with a fish
2
PSThe pictogram shows the number of cakes sold in a shop. One symbol stands for 10 cakes.
(a) How many cakes were sold on Monday? (b) How many cakes were sold on Wednesday? (c) How many cakes were sold in the five days? (d) On which day were twice as many cakes sold as on Friday?
MonTueWedThuFrirepresents 10 cakes
(a)Cakes on Monday
(b)Cakes on Wednesday
(c)Total cakes
(d)Day with twice as many as Friday
3
FDraw a pictogram for the number of apples sold in a shop: Monday 30, Tuesday 45, Wednesday 20, Thursday 35. Use one symbol to stand for 10 apples.
4
FThe table shows the favourite sport of 40 students.
SportFrequency
Football11
Netball8
Rugby5
Tennis9
Hockey7
Draw a bar chart to show this information.
5
FThe dual bar chart shows the numbers of boys and girls who joined clubs in four year groups.
(a) How many girls joined in Year 8? (b) In which year group did more boys than girls join? (c) How many students joined altogether?
05101520Year groupNumber of studentsYear 7Year 8Year 9Year 10BoysGirls
(a)Girls in Year 8
(b)Year groups with more boys
(c)Total students
6
PSThe composite bar chart shows how students in three year groups travel to school.
(a) How many students are in Year 8? (b) How many Year 9 students come by car? (c) Which year group has the most students?
010203040Year groupNumber of studentsYear 7Year 8Year 9WalkBusCar
(a)Students in Year 8
(b)Year 9 by car
(c)Year group with most students
7
RA company draws this chart to show that its sales are “much higher” than its rival’s. Explain why the chart is misleading.
7880828486Sales (thousands)Our firm84Rival80
8
RSay which type of chart is best for each set of data and give a reason.
(a)The numbers of students choosing each of five school lunches
(b)The number of goals scored in each of 30 matches (0, 1, 2, 3, 4)
(c)How the temperature in a room changes over a day
9
PSIn a pictogram, one symbol stands for 8 students. Tom draws 3 whole symbols and 1 half symbol for the number of students who walk to school. How many students walk?
10
RClass A has 25 students and 10 of them chose pizza. Class B has 16 students and 8 of them chose pizza. Ana says: “More students in Class A chose pizza, so pizza is more popular in Class A.” Is she right? Explain.
11
RIn a pictogram, Mon shows 2 whole symbols and a half symbol. Tue shows 4 symbols. Wed shows 3 and a half symbols. The total for the three days is 100 items. Work out the number of items one symbol stands for, then the number of items on Tuesday.
12
PSThe bar chart shows the drinks sold at a kiosk one afternoon.
(a) Which drink sold least? (b) How many more cups of juice than coffee were sold? (c) What percentage of the drinks sold were tea?
01020304050DrinkNumber soldTea30Coffee20Juice40Water10
(a)Drink that sold least
(b)Juice minus coffee
(c)Percentage that were tea
13
RZoe draws a bar chart of the numbers of students in four clubs. She makes the bar for chess twice as wide as the others, and her vertical axis starts at 8, not 0. Give two reasons why her chart could mislead someone.
24.7

Pie charts

Edexcel 1MA1: S2
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Pie chartsComing soon

A pie chart is a circle cut into sectors. The whole circle stands for all the data, and each sector’s angle shows its share. The whole circle is 360°, so a sector of 90° is a quarter of the data, and a sector of 180° is a half. Pie charts are good for showing proportions, but they do not show the actual numbers unless we are told the total.

80°120°100°60°Walk: 8Bus: 12Car: 10Cycle: 6
The data 8, 12, 10, 6 has total 36, so each item is worth 360° ÷ 36 = 10°. The angles are 80°, 120°, 100° and 60°.

Drawing a pie chart. (1) Add the frequencies to find the total. (2) Divide 360° by the total to find the angle for one item. (3) Multiply each frequency by this angle. (4) Check that the angles add up to 360°. (5) Draw a circle, draw a radius, and use a protractor to measure each angle in turn. Label every sector.

Reading a pie chart. To find the frequency for a sector, write its angle as a fraction of 360° and find that fraction of the total: frequency = angle ÷ 360 × total. If you know the frequency of one sector but not the total, find how many degrees one item is worth: angle ÷ frequency, then divide 360 by it to find the total.

Comparing pie charts. A bigger sector in one pie chart does not necessarily mean more people, if the totals are different. Compare proportions (fractions or percentages), not sector sizes, unless the totals are given.

Key ideas

  • A full circle is 360°. Angle for one item = 360° ÷ total frequency.
  • Angle of a sector = frequency ÷ total × 360°.
  • Frequency of a sector = angle ÷ 360 × total.
  • The angles in a pie chart must add up to 360°.
  • Compare proportions, not raw sector sizes, when the totals differ.
Worked example 24.7.1
36 students were asked how they travel to school: walk 8, bus 12, car 10, cycle 6. Draw a pie chart to show this information.

Key words. A pie chart is a circle cut into sectors. The whole circle is 360° and stands for all the data. Each sector’s angle shows its share.

Method. Each category’s angle is its share of 360°: angle = frequency ÷ total frequency × 360°.

  1. Add up the frequencies. The total is the whole pie, which is 360°.
    8 + 12 + 10 + 6 = 36
  2. Find the angle for one item. Divide the whole circle by the total frequency.
    360° ÷ 36 = 10° for each item
  3. Multiply each frequency by the angle for one item. A category with f items gets f lots of this angle.
    Walk: 8 × 10° = 80°
    Bus: 12 × 10° = 120°
    Car: 10 × 10° = 100°
    Cycle: 6 × 10° = 60°
  4. Check that the angles make 360°. The angles of all the sectors must fill the circle.
    80 + 120 + 100 + 60 = 360°
  5. Draw the pie chart. Draw a circle with a compass and a radius. Then use a protractor to measure each angle in turn from the radius, going round the circle. Label each sector.
    Walk (80°)Bus (120°)Car (100°)Cycle (60°)

Check. Total = 360° = 360°. ✓

Watch out. Do not forget to check that the angles add up to 360°. If they do not, there is an arithmetic slip.

Worked example 24.7.2
120 students chose a favourite sport for a pie chart. The sector for football has angle 105°. How many students chose football?

Key words. A pie chart is a circle cut into sectors. The whole circle is 360° and stands for all the data. Each sector’s angle shows its share.

Method. A sector is a fraction of the whole pie. The fraction is angle ÷ 360°, and the same fraction of the total frequency belongs to the category.

  1. Write the angle as a fraction of 360°. The whole circle is 360°, so the sector is angle over 360.
    105360 = 724
  2. Find one part of the total. The denominator is 24, so divide the total (120) into 24 equal parts.
    120 ÷ 24 = 5
  3. Multiply to find the sector. The sector is 7 of those parts.
    7 × 5 = 35 students

Check. 1° stands for 120 ÷ 360 items, so 105° stands for 105 × 120 ÷ 360 = 35. ✓

Watch out. The pie shows a share of the total. Find the total first if it is not given.

Worked example 24.7.3
In a pie chart, 12 students chose tennis and the angle for tennis is 48°. How many students are represented by the whole pie chart?

Key words. A pie chart is a circle cut into sectors. The whole circle is 360° and stands for all the data. Each sector’s angle shows its share.

Method. If we know how many items match a sector, we can find how many items one degree stands for, and so find the total for the whole 360°.

  1. Find what 1 item is worth in degrees. 12 students have an angle of 48°, so divide the angle by the frequency.
    48° ÷ 12 = 4° for each item
  2. Find how many items fill the whole circle. The whole pie is 360°.
    360 ÷ 4 = 90
  3. State the answer. The question asks for the total of the whole pie chart, not just the sector.
    Total = 90

Check. 90 × 4° = 360°. ✓

Watch out. The total is for the whole pie, not for the sector you were told about.

Worked example 24.7.4
Pie chart A shows that 25% of a class of 40 walk to school (a sector of 90°). Pie chart B shows that 20% of a class of 100 walk to school (a sector of 72°). Lola says: “Class A has the bigger sector, so more students in Class A walk.” Is she right?

Key words. A pie chart is a circle cut into sectors. The whole circle is 360° and stands for all the data. Each sector’s angle shows its share.

Method. Work out the actual numbers using each pie chart’s total.

  1. Class A. Use fraction of the total = angle ÷ 360, then multiply by the class size.
    90360 = 14 and 14 of 40 = 10 students
  2. Class B. Do the same.
    72360 = 15 and 15 of 100 = 20 students
  3. Compare. Class B has more students who walk, even though its sector is smaller.
    20 > 10, so Lola is wrong about the numbers.

Check. 25% of 40 = 10 and 20% of 100 = 20. ✓

Watch out. Do not compare sector sizes without comparing the totals.

Watch out

The angles must add up to 360°, so always check. And when pie charts are for different totals, the same angle does not mean the same number of people.

Exercise 24G  Pie charts

Calculator allowed
1
FWork out the angle for each sector in a pie chart for each set of data.
(a)Pets: dog 8, cat 6, fish 4, other 2 (20 pets)
(b)Lessons: maths 9, science 6, art 3, PE 6 (24 students)
(c)Fruit: apple 30, pear 20, banana 25, orange 15 (90 pieces)
2
FThe table shows the colours of the cars in a car park.
ColourCars
Red14
Blue10
Silver22
Black14
Draw a pie chart to show this information.
3
FTwenty-five children voted for their favourite playground game. The results are: tag 6, football 9, skipping 4, hopscotch 6. Work out the angle for each sector, giving your answers to 1 decimal place if necessary.
4
FThe pie chart shows how 72 students spend their lunch break. The angle for “canteen” is 150°, for “playground” is 120° and for “library” is 90°. (a) What fraction of the students spend lunch in the canteen? (b) How many students spend lunch in the library? (c) How many students spend lunch in the playground?
150°120°90°CanteenPlaygroundLibrary
(a)Fraction in the canteen
(b)Students in the library
(c)Students in the playground
5
FEach pie chart has a sector with the given angle. Work out the number represented by that sector.
(a)Total 180 people; sector 60°
(b)Total 90 people; sector 140°
(c)Total 240 votes; sector 75°
(d)Total 54 cars; sector 100°
6
PSWork out the total number of items represented by each whole pie chart.
(a)A sector of 30° represents 5 people.
(b)A sector of 80° represents 16 students.
(c)A sector of 45° represents 10 votes.
7
PSA pie chart has sectors of 90°, 120°, 100° and one other sector. The total number of people is 72. (a) Work out the angle of the missing sector. (b) How many people does the missing sector represent?
(a)Missing angle
(b)People in the missing sector
8
FIn a pie chart, a sector has angle 54°. What percentage of the data does it represent?
9
RTwo pie charts show how students in Year 9 and in Year 11 travel to school. In Year 9 (120 students) the sector for “bus” is 60°. In Year 11 (180 students) the sector for “bus” is 50°. Which year group has more students travelling by bus?
10
RDev says: “A pie chart shows a sector of 90° for football, so 90 people chose football.” Explain his mistake.
11
RPie chart A has radius 3 cm and shows data for 100 people. Pie chart B shows data for 400 people and the area of each pie chart is in proportion to the number of people. Work out the radius of pie chart B.
12
PSThe frequency table shows the drinks chosen by 25 people. Work out the angle for each drink, and show that the angles add up to 360°.
DrinkFrequency
Tea7
Coffee11
Juice5
Water2
13
PSA pie chart shows the favourite ice cream flavours of 60 people. The angles are chocolate 150°, vanilla 90°, strawberry 72° and mint 48°.
(a) How many people chose vanilla? (b) What percentage chose strawberry?
150°90°72°48°ChocolateVanillaStrawberryMint
(a)People who chose vanilla
(b)Percentage who chose strawberry
14
RIn two pie charts the sector for “cycling” is 90° in both. The first pie chart shows 80 students and the second shows 120 students. How many more students cycle in the second pie chart than in the first?
24.8

Line graphs and time series

Edexcel 1MA1: S2
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Line graphs and time seriesComing soon

A time series is a set of measurements taken at regular times: the temperature every hour, the number of customers every month, a baby’s mass every week. We show a time series on a line graph: time goes on the horizontal axis, the measurement on the vertical axis. We plot a point for each reading and join the points with straight lines, from left to right.

Ice cream sales at a seaside kiosk020406080Quarter of the year (Q1 = Jan to Mar)Sales (hundreds)Q1 Y1Q2 Y1Q3 Y1Q4 Y1Q1 Y2Q2 Y2Q3 Y2Q4 Y2Q1 Y3Q2 Y3Q3 Y3Q4 Y3
A time series. Notice the pattern that repeats each year (low in Q1, high in Q3) and the way each year is higher than the one before.

What a line graph shows. Reading a line graph, look for three things. The trend is the general direction over a long time (here, sales rise from year to year). Seasonal variation is a pattern that repeats regularly (here, sales are highest in summer, Q3, every year). Unusual values do not fit the pattern and may need explaining.

Predicting. To predict the next value, continue the trend and the seasonal pattern. For example, Q3 sales were 58, 63, 68 in three years: they go up by 5 each year, so Year 4’s Q3 might be 73. This is an extrapolation (going beyond the data), so it is only an estimate and could be wrong if conditions change. Reading a value between two plotted points is interpolation: it is less risky, but it assumes the values change steadily between the readings.

Key ideas

  • A time series is data measured at regular times; show it on a line graph with time on the horizontal axis.
  • Trend = long-term direction. Seasonal variation = a pattern that repeats regularly.
  • Compare the same season in different years to see the trend.
  • Interpolation (between points) is more reliable than extrapolation (beyond the data).
  • A line graph that does not start at 0, or has an uneven scale, can mislead.
Worked example 24.8.1
The line graph shows the ice cream sales (in hundreds) for each quarter. (a) What were the sales in Q2 of Year 2? (b) In which quarter were sales highest? (c) How many more were sold in Q3 of Year 3 than in Q3 of Year 1?
020406080QuarterSales (hundreds)Q1 Y1Q2 Y1Q3 Y1Q4 Y1Q1 Y2Q2 Y2Q3 Y2Q4 Y2Q1 Y3Q2 Y3Q3 Y3Q4 Y3

Method. Find each quarter on the horizontal axis, go up to the plotted point, and read across to the vertical axis.

  1. (a) Find Q2 Y2 and read the point. The 6th point from the left is Q2 of Year 2.
    Sales = 45 hundred = 4500
  2. (b) Find the highest point. The highest point on the graph is the greatest sales.
    Highest point = 68 hundred, at Q3 of Year 3
  3. (c) Read the two Q3 points. Q3 Y3 and Q3 Y1.
    Q3 Y3 = 68, Q3 Y1 = 58
  4. Subtract. “How many more” means take away.
    68 − 58 = 10 hundred = 1000

Check. The Q3 points are 58, 63, 68, rising by 5 each year, so 68 is 10 more than 58. ✓

Watch out. Remember the units: the vertical axis is in hundreds.

Worked example 24.8.2
Describe the trend and the seasonal pattern in the ice cream sales.

Method. Compare the same quarter in different years for the trend, and compare the quarters within a year for the seasonal pattern.

  1. Look at the same quarter each year. This removes the seasonal effect so that only the trend is left.
    Q1: 24, 27, 30 (up 3 each year)
    Q3: 58, 63, 68 (up 5 each year)
  2. State the trend. The values for every quarter go up from year to year.
    Trend: sales are rising each year.
  3. Look within one year. Compare the four quarters of Year 1.
    Year 1: 24, 41, 58, 30: low, then rising to a peak in Q3, then falling.
  4. State the seasonal pattern. The same shape repeats in every year.
    Seasonal pattern: sales are highest in Q3 (summer) and lowest in Q1 (winter) every year.

Check. Each year looks like a bigger version of the year before. ✓

Watch out. The trend and the seasonal pattern are different: the trend compares years, the seasonal pattern compares seasons.

Worked example 24.8.3
The table shows the mass of a baby each month.
Age (months)0123456
Mass (kg)3.44.14.95.66.16.67
Draw a line graph.

Method. For a line graph, plot a point for each time and join the points with straight lines, from left to right.

  1. Draw and label the axes. Time goes on the horizontal axis. The measurement goes on the vertical axis, with an equal scale.
    Horizontal axis: Age (months). Vertical axis: Mass (kg).
  2. Plot a point for each pair of values. Go along to the time, then up to the value.
    0: 3.399999999999999911182158029987476766109466552734375
    1: 4.0999999999999996447286321199499070644378662109375
    2: 4.9000000000000003552713678800500929355621337890625
    3: 5.5999999999999996447286321199499070644378662109375
    4: 6.0999999999999996447286321199499070644378662109375
    5: 6.5999999999999996447286321199499070644378662109375
    6: 7
  3. Join the points with straight lines. The lines show the trend. Do not draw a line before the first point or after the last one.
    02468Age (months)Mass (kg)0123456

Check. Each point matches a value in the table. ✓

Watch out. Plot every point accurately before you join them.

Worked example 24.8.4
Use the pattern in the Q3 sales to estimate the Q3 sales in Year 4. Say why your answer might not be reliable.

Method. Find how the Q3 sales change each year, continue the pattern one more year, and comment on the risk.

  1. Write the Q3 sales for each year. Same quarter each year, so no seasonal effect.
    58, 63, 68
  2. Find the yearly change. Subtract each value from the next.
    63 − 58 = 5
    68 − 63 = 5
  3. Continue the pattern. Sales have gone up by 5 each year, so assume they go up by 5 again.
    68 + 5 = 73 hundred
  4. Comment on reliability. This is an extrapolation: it assumes the trend continues, but weather, prices or competition could change.
    It is only an estimate.

Check. The sequence 58, 63, 68, 73 has a constant difference of 5. ✓

Watch out. Say that the estimate assumes the trend continues.

Watch out

The trend compares the same season in different years. Do not compare Q1 with Q3 to find the trend: that mixes the seasonal pattern with the trend. And never say a prediction is certain.

Exercise 24H  Line graphs and time series

Non-calculator
1
FThe graph shows the ice cream sales (in hundreds) for each quarter over three years. (a) In which quarter were sales lowest? (b) Work out the total sales in Year 2. (c) What was the increase in sales in Q4 from Year 1 to Year 3?
020406080QuarterSales (hundreds)Q1 Y1Q2 Y1Q3 Y1Q4 Y1Q1 Y2Q2 Y2Q3 Y2Q4 Y2Q1 Y3Q2 Y3Q3 Y3Q4 Y3
(a)Lowest quarter
(b)Total sales in Year 2
(c)Increase in Q4
2
FThe table shows the number of visitors to a website each month.
MonthJanFebMarAprMayJun
Visitors120150210260240300
Draw a line graph to show this information.
3
RThe graph shows a family’s gas bill (in £) for each quarter over two years. Describe (a) the seasonal pattern and (b) the trend.
0255075100125150QuarterBill (&pound;)Q1 Y1Q2 Y1Q3 Y1Q4 Y1Q1 Y2Q2 Y2Q3 Y2Q4 Y2
(a)Seasonal pattern
(b)Trend
4
PSThe graph shows the monthly sales (in thousands of pounds) of two shops, A and B. (a) In which month did shop A first have higher sales than shop B? (b) How much higher were A’s sales than B’s in June? (c) Which shop has more steady sales? Give a reason.
01020304050MonthSales (&pound;000)JanFebMarAprMayJunShop AShop B
(a)First month A is higher
(b)June difference
(c)More steady
5
PSUse the ice cream data. The Q1 sales in the three years were 24, 27, 30 hundred. Estimate the Q1 sales in Year 4 if the pattern continues.
6
RA company shows this line graph of its profit and says “our profit is growing very fast”. Explain why the graph may be misleading.
100101102103104105YearProfit (&pound;000)20202021202220232024
7
RDev records the favourite colour of 5 students and draws a line graph joining the points for red, blue, green, yellow, black. Explain why this is not appropriate.
8
PSThe temperature was recorded at 12 noon (14°C) and at 2 pm (18°C). (a) Estimate the temperature at 1 pm. (b) Explain why your answer to (a) is only an estimate.
(a)Temperature at 1 pm
(b)Why only an estimate
9
PSWork out the mean quarterly sales in Year 2 from the ice cream graph, giving your answer to 2 decimal places.
10
RIce cream sales in Q3 were 58, 63, 68 hundred in Years 1, 2 and 3. If the trend continues, in which year will Q3 sales first be over 100 hundred? Comment on the reliability.
11
FThe table shows the temperature in a classroom at different times.
Time8am9am10am11am12pm1pm2pm3pm
Temperature (°C)1214171921222018
Draw a line graph to show this information.
12
RSam looks at a graph of umbrella sales. Sales were 40 in January and 28 in March. He says: “Sales are falling, so the shop will soon sell no umbrellas.” Give two reasons why this conclusion may be wrong.
24.9

Stem-and-leaf diagrams

Edexcel 1MA1: S2, S4
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Stem-and-leaf diagramsComing soon

A stem-and-leaf diagram shows the data in order and keeps every single value, so it is better than a grouped table when we want to find the median or range. Each number is split into a stem (the first digit or digits, usually the tens) and a leaf (the last digit, usually the units). The stems go down the left in order. The leaves go along each row in order. The diagram on its side looks like a bar chart, so we can see the shape of the data.

StemLeaf
28 9
33 4 5 8 8
41 1 1 4 7 9
52 6
Key: 3 | 4 means 34 marks
The test scores of 15 students: 28, 29, 33, 34, 35, 38, 38, 41, 41, 41, 44, 47, 49, 52, 56.

Always include a key that shows what a stem and leaf stand for, such as “3 | 4 means 34 marks”. A key is essential, because the same diagram could mean 3.4 or 340 if the key were different. To draw one: (1) write the stems in order; (2) go through the data and write each leaf on the right row; (3) redraw with the leaves in order; (4) add the key.

Back-to-back stem-and-leaf diagrams compare two sets of data using one set of stems in the middle. The left side leaves are written backwards so that the smallest leaf is nearest the stem on both sides. To find the median of ordered data, count to the middle value; the range is the biggest minus the smallest. The decimal version works the same way, with a key like “2 | 4 means 2.4 seconds”.

Key ideas

  • Stem = leading digit(s). Leaf = last digit. The leaves must be in order.
  • Always write a key: “3 | 4 means 34”.
  • The diagram keeps all the data, so you can find the median, mode and range exactly.
  • In a back-to-back diagram, the left leaves run outwards from the stem, with the smallest nearest the stem.
  • Compare two data sets by their medians (average) and ranges (spread).
Worked example 24.9.1
The test scores of 15 students are: 34, 41, 28, 35, 47, 52, 38, 41, 29, 44, 33, 41, 56, 38, 49. Draw an ordered stem-and-leaf diagram.

Method. Split each value into a stem (the tens) and a leaf (the units). Write the leaves next to the right stem, then put each row in order.

  1. Write the stems in order down the left. The stems are the tens digits. List every stem from the smallest to the largest, even if a stem has no leaves.
    2, 3, 4, 5
  2. Go through the data in the order given, writing each leaf next to its stem. The leaf is the units digit.
    2 | 8 9
    3 | 4 5 8 3 8
    4 | 1 7 1 4 1 9
    5 | 2 6
  3. Rewrite each row with its leaves in order, smallest first. An ordered diagram is much easier to read, and you need it for the median.
    StemLeaf
    28 9
    33 4 5 8 8
    41 1 1 4 7 9
    52 6
    Key: 3 | 4 means 34 marks
  4. Add a key. The key explains what the numbers mean, so that anyone can read the diagram.
    3 | 4 means 34 marks

Check. Count the leaves: there should be 15 leaves in total, one for each value. ✓

Watch out. Write the leaves in order and keep them evenly spaced, and always include a key.

Worked example 24.9.2
Use the diagram above (the scores of 15 students) to find (a) the median, (b) the mode, (c) the range, (d) the number of students who scored more than 45.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. The diagram is already in order, so we can read the values straight from it. First write them out as a list.

  1. Write the data as a list from the diagram. Read each row: stem then leaf.
    28, 29, 33, 34, 35, 38, 38, 41, 41, 41, 44, 47, 49, 52, 56
  2. (a) Median. There are 15 values, so the median is the (15 + 1) ÷ 2 = 8th value.
    8th value = 41
  3. (b) Mode. 41 appears three times, more than any other value.
    Mode = 41
  4. (c) Range. Biggest minus smallest.
    56 − 28 = 28
  5. (d) Count the values above 45. Look at the end of the list.
    47, 49, 52, 56: that is 4 students

Check. Count in from each end of the list: 7 values are below the median and 7 are above. ✓

Watch out. Count the position of the median in the list of values, not along the stems.

Worked example 24.9.3
The back-to-back diagram shows the scores of 9 boys and 10 girls in a test.
BoysStemGirls
28
8 5 231 3 6 8 9
7 5 4 142 3 6 9
5 25
Key: for boys 4 | 1 means 41 marks; for girls 1 | 3 means 31 marks
Compare the scores of the boys and the girls.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Compare an average (the median) and the spread (the range) for the two groups.

  1. Boys: median. There are 9 values, so the median is the 5th value.
    Median = 44
  2. Girls: median. There are 10 values, so the median is halfway between the 5th and 6th.
    (38 + 39) ÷ 2 = 38.5
  3. Boys: range.
    55 − 32 = 23
  4. Girls: range.
    49 − 28 = 21
  5. Write a comparison. Use the medians for the average and the ranges for the spread, and write in context.
    On average the boys scored higher (median 44 compared with 38.5). The spread is similar (range 23 for boys and 21 for girls).

Check. The boys’ leaves are in higher rows (stems 4 and 5) more often than the girls’. ✓

Watch out. Do not read the left side of a back-to-back diagram from the right. The smallest leaf is nearest the stem.

Worked example 24.9.4
The times, in seconds, taken by seven runners are 2.1, 2.4, 3.0, 2.7, 3.2, 2.4 and 3.5. Draw a stem-and-leaf diagram.

Method. Split each value into a stem (the tens) and a leaf (the units). Write the leaves next to the right stem, then put each row in order.

  1. Write the stems in order down the left. The stems are the tens digits. List every stem from the smallest to the largest, even if a stem has no leaves.
    2, 3
  2. Go through the data in the order given, writing each leaf next to its stem. The leaf is the units digit.
    2 | 1 4 7 4
    3 | 0 2 5
  3. Rewrite each row with its leaves in order, smallest first. An ordered diagram is much easier to read, and you need it for the median.
    StemLeaf
    21 4 4 7
    30 2 5
    Key: 2 | 4 means 2.4 seconds
  4. Add a key. The key explains what the numbers mean, so that anyone can read the diagram.
    2 | 4 means 2.4 seconds

Check. Count the leaves: there should be 7 leaves in total, one for each value. ✓

Watch out. Write the leaves in order and keep them evenly spaced, and always include a key.

Watch out

Do not forget the key, and put the leaves in order. Also, keep the leaves evenly spaced, like columns, so that the diagram is a fair picture of the data.

Exercise 24I  Stem-and-leaf diagrams

Non-calculator
1
FThe ages of 12 people at a party are: 23, 31, 19, 42, 35, 27, 31, 45, 22, 38, 29, 51. Draw an ordered stem-and-leaf diagram.
2
FThe heights, in cm, of 15 students are: 152, 147, 161, 158, 143, 166, 150, 155, 149, 163, 157, 171, 160, 154, 148. Draw an ordered stem-and-leaf diagram with the stems 14, 15, 16 and 17.
3
FThe stem-and-leaf diagram shows the number of points scored by 15 players in a season.
StemLeaf
12 5 7
21 3 3 6 8
30 1 4 5 8
41 5
Key: 2 | 3 means 23 points
(a) How many values are more than 30? (b) Find the mode. (c) Find the median. (d) Find the range.
(a)Values more than 30
(b)Mode
(c)Median
(d)Range
4
FThe masses, in kg, of nine parcels are shown.
StemLeaf
12 5 8
21 3 3 6
30 4
Key: 1 | 2 means 1.2 kg
Work out (a) the median, (b) the mode and (c) the range of the masses.
(a)Median
(b)Mode
(c)Range
5
PSThe back-to-back diagram shows the scores of nine boys and ten girls.
BoysStemGirls
28
8 5 231 3 6 8 9
7 5 4 142 3 6 9
5 25
Key: for boys 4 | 1 means 41; for girls 1 | 3 means 31
(a) Find the median score for each group. (b) Find the range for each group. (c) Compare the scores.
(a)Medians
(b)Ranges
(c)Comparison
6
RLook at this stem-and-leaf diagram for the ages of six people: 2 | 8 3 1, 3 | 5 2, 4 | 9. Give two things that are wrong with it.
7
FWork out the mean of the 12 ages in the first question of this exercise: 23, 31, 19, 42, 35, 27, 31, 45, 22, 38, 29, 51. Give your answer to 1 decimal place.
8
FWrite down the original list of data shown by this stem-and-leaf diagram.
StemLeaf
61 4 8
70 3 3 9
85
Key: 6 | 1 means 61 cm
9
FFind the lower quartile, the upper quartile and the interquartile range of the 15 scores in the diagram for question 3.
StemLeaf
12 5 7
21 3 3 6 8
30 1 4 5 8
41 5
Key: 2 | 3 means 23 points
10
RMaya says: “A stem-and-leaf diagram is better than a grouped frequency table if you want the exact median.” Is she right? Explain.
11
RA stem-and-leaf diagram for 11 values is shown.
StemLeaf
10 2 5 8
20 1 6 9
33 6
44
Key: 1 | 2 means 12
Show that the mean is greater than the median.
12
FThe scores of the students in two groups are: Group X: 22, 27, 31, 35, 38, 44, 46. Group Y: 25, 29, 32, 33, 40, 41, 48, 50. Draw a back-to-back stem-and-leaf diagram with Group X on the left.
13
RThe stem-and-leaf diagram shows the lengths of eight snakes, in metres.
StemLeaf
12 8
21 3 6 9
34 5
Key: 1 | 2 means 1.2 m
Show that the median length is 2.45 m.
24.10

Scatter graphs

Edexcel 1MA1: S6
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Scatter graphsComing soon

Bivariate data is data with two measurements for each person or object, such as the hours someone revised and their test mark. A scatter graph shows each pair as a point (a cross), with one measurement on the horizontal axis and the other on the vertical axis. Scatter graphs let us see whether the two things are related.

Positive0204060801000246810HoursMark
Negative02468101214024681012Age (yr)Value
No correlation0102030405060024681012xy
Positive correlation: as one goes up, the other goes up. Negative: as one goes up, the other goes down. None: no pattern.

Correlation describes how the two variables are related. With positive correlation the points slope upwards: when one variable increases, so does the other (more hours revising, higher marks). With negative correlation the points slope downwards: when one increases, the other decreases (an older car is worth less). With no correlation there is no pattern. The correlation is strong if the points are close to a straight line and weak if they are widely scattered.

Correlation is not causation. Two things can be correlated without one causing the other. Ice cream sales and sunburn cases are positively correlated, but ice cream does not cause sunburn: both go up when the weather is hot. Always say there is a relationship or link unless you know the cause.

Line of best fit. A line of best fit is a straight line drawn through the points that follows the trend. It does not have to pass through any particular point. Draw it so that there are about equal numbers of points on each side of the line, and it passes close to the mean point (the mean of the x-values, the mean of the y-values). Use a ruler, and extend it across the data.

0102030405060708090100012345678910Hours of revisionTest mark70
The line of best fit. To estimate the mark for 5.5 hours of revision, go up from 5.5 to the line, then across to the vertical axis.

Using the line. Reading a value from the line within the range of the data is interpolation: it is usually reliable if the correlation is strong. Reading a value outside the range of the data is extrapolation: it is much less reliable, because the relationship may not continue (for example, marks cannot go above 100). Say that your answer is an estimate.

Key ideas

  • Positive correlation: both increase. Negative: one increases as the other decreases. None: no pattern.
  • Strong correlation: points close to a line. Weak correlation: points widely scattered.
  • Correlation does not prove causation.
  • A line of best fit follows the trend, with about equal numbers of points on each side, near the mean point.
  • Interpolation (inside the data range) is more reliable than extrapolation (outside it).
  • An outlier is a point that does not fit the pattern.
Worked example 24.10.1
The scatter graph shows the hours of revision and the test marks of 12 students. Describe the correlation.
0102030405060708090100012345678910Hours of revisionTest mark

Method. Look at the direction of the points, and how close they are to a straight line.

  1. Look at the direction. The points go from the bottom left to the top right.
    As the hours increase, the marks increase: positive correlation.
  2. Look at the closeness. The points are close to a straight line.
    The correlation is strong.
  3. Write the answer in context. Use the words of the question.
    Strong positive correlation: students who revise for more hours tend to get higher marks.

Check. A line sloping up through the points would fit them closely. ✓

Watch out. Say positive or negative, and say strong or weak. Write it in context.

Worked example 24.10.2
A line of best fit is drawn on the graph in the last example. Use it to estimate the test mark of a student who revised for 5.5 hours.

Method. Read across or up from the line of best fit. The answer is an estimate, so a small tolerance is allowed.

  1. Find 5.5 hours on the horizontal axis. We know the hours-value, so start on the hours-axis.
    hours = 5.5 hours
  2. Go straight up to the line of best fit. The line of best fit shows the trend, so it gives us the best estimate.
    Draw a vertical line up from the axis until it meets the line.
    0102030405060708090100012345678910Hours of revisionTest mark70
  3. Go across to the vertical axis and read the mark-value. Read the scale carefully.
    mark ≈ 70

Check. The point you read should lie on the line of best fit, not on one of the crosses. ✓

Watch out. Read from the line, not from the nearest cross.

Worked example 24.10.3
Chen says: “A student who revises for 12 hours would get a mark of about 110.” Explain why this prediction is not sensible.

Method. Check whether 12 hours is inside the range of the data, and whether the answer makes sense.

  1. Find the range of the data. The hours in the graph go from 1 to 9.
    1 to 9 hours
  2. Compare 12 hours with the range. 12 hours is outside the range of the data, so reading from the line is extrapolation.
    12 > 9, so this is extrapolation.
  3. Check the answer. The line gives about 110, which is more than 100.
    A test mark out of 100 cannot be bigger than 100.
  4. Conclude. The relationship probably does not continue in a straight line, so the prediction is unreliable.
    Not sensible.

Check. Using the line: 35.7 + 6.2 × 12 = 110.1. ✓

Watch out. Extrapolation is risky. Always say why the estimate may not be reliable.

Worked example 24.10.4
The table shows the number of adverts shown and the number of sales for 8 weeks.
Adverts, x12345678
Sales, y254769811
Plot a scatter graph and draw a line of best fit.

Method. Plot every point, find the mean point to guide the line, then draw a straight line that follows the trend.

  1. Plot the points. Each pair is one point: go along to x, then up to y.
    (1, 2), (2, 5), (3, 4), (4, 7), (5, 6), (6, 9), (7, 8), (8, 11)
  2. Work out the mean point. The line of best fit should pass close to the mean point (mean of x, mean of y).
    Mean of x = 36 ÷ 8 = 4.5
    Mean of y = 52 ÷ 8 = 6.5
    Mean point = (4.5, 6.5)
  3. Draw a straight line with a ruler. It follows the trend of the points and passes close to the mean point. There should be about the same number of points above and below it: here 4 above and 4 below.
    0246810120123456789Adverts (x)Sales (y)

Check. There are 4 points above the line and 4 below it. ✓

Watch out. The line does not have to go through the first or last point, and does not need to go through any point at all.

Worked example 24.10.5
Ice cream sales and the number of people treated for sunburn are strongly positively correlated. A newspaper says: “Ice cream causes sunburn.” Is this a fair conclusion?

Method. Think about whether there could be another reason for both to increase.

  1. Describe the correlation. When ice cream sales rise, so do sunburn cases.
    Strong positive correlation.
  2. Look for another cause. Hot, sunny weather makes people buy ice cream and also causes sunburn.
    A third factor (hot weather) explains both.
  3. Conclude. Correlation shows a link, not that one thing causes the other.
    No. The correlation does not show that ice cream causes sunburn.

Check. Ice cream is eaten indoors in winter as well, with no sunburn. ✓

Watch out. Never write “this causes that” from a scatter graph alone.

Watch out

A line of best fit is not found by joining the first and last points, and it does not have to pass through the origin. Also, never use a line of best fit far outside the data: that is extrapolation and it may not be reliable.

Exercise 24J  Scatter graphs

Non-calculator
1
FDescribe the correlation shown by each scatter graph. Choose from: positive, negative, no correlation (and say strong or weak where it applies).
(a)0204060801000246810HoursMark
(b)02468101214024681012AgeValue
(c)0102030405060024681012xy
(a)Graph (a)
(b)Graph (b)
(c)Graph (c)
2
FThe scatter graph shows 12 pairs of values. Describe the correlation.
0102030405060024681012xy
3
FThe table shows the maths and science marks of eight students.
Maths mark3448556270758290
Science mark4052506668807992
Plot the scatter graph (use axes from 30 to 100) and describe the correlation.
4
FWork out the mean point of the data in question 3 (the mean of the maths marks and the mean of the science marks).
5
FA line of best fit has been drawn on the scatter graph of revision hours and test marks. Use it to estimate (a) the mark for a student who revised for 3 hours, (b) the mark for 7 hours, (c) the number of hours needed to get a mark of 80.
0102030405060708090100012345678910Hours of revisionTest mark
(a)Mark for 3 hours
(b)Mark for 7 hours
(c)Hours for a mark of 80
6
RThe data in question 5 range from 1 to 9 hours. For each estimate, say whether it is interpolation or extrapolation, and whether it is likely to be reliable.
(a)The mark for 4.5 hours
(b)The mark for 15 hours
(c)The mark for 0.5 hours
7
RA scatter graph shows the age and height of children aged 3 to 12 and the line of best fit goes up steeply. Charlie uses the line to say that a 40-year-old will be about 3 m tall. Explain what is wrong.
8
RThere is a positive correlation between the number of firefighters at a fire and the amount of damage. Does this show that firefighters cause damage? Explain.
9
RThe scatter graph shows revision hours and test marks. One point is circled. (a) Write down the coordinates of the circled point. (b) Suggest a reason why this student’s mark is so different from the pattern. (c) Should it be used to draw the line of best fit?
0102030405060708090100012345678910Hours of revisionTest mark
(a)Coordinates
(b)A reason
(c)Include it?
10
RDan draws a line of best fit by joining the first point on the left to the last point on the right. Explain why this may not be a good line of best fit.
11
PSA taxi company uses the line y = 2x + 3 to estimate the cost, £y, of a journey of x km. (a) Use the line to estimate the cost of a 6 km journey. (b) Write down what the number 3 and the number 2 mean in this situation.
(a)Cost of 6 km
(b)Meaning of 3 and 2
12
PSA line of best fit passes through the points (0, 35) and (10, 97). (a) Work out the gradient. (b) Write the equation of the line in the form y = mx + c. (c) Use it to estimate y when x = 7.5.
(a)Gradient
(b)Equation
(c)Estimate at x = 7.5
13
RThe equation of a line of best fit is y = 6x + 20, where x is the number of hours of practice per week and y is the score in a game. Explain what the gradient 6 tells you, and say why the line may not be sensible for x = 30.
14
RTwo scatter graphs show different data. In graph A the points lie very close to a straight line sloping down. In graph B the points are widely scattered but slope slightly downwards. Which graph shows stronger correlation? What type of correlation is each?
15
RThe mean point of a set of data is (6, 40). A student draws a line of best fit with equation y = 3x + 22. Show that the line passes through the mean point, and use it to estimate y when x = 8.
24.11

Comparing distributions

Edexcel 1MA1: S4
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Comparing distributionsComing soon

A distribution is all the values of a data set and how often they occur. To compare two distributions we use two kinds of measure: an average (what is typical) and a measure of spread (how consistent the data is). Comparing only an average, or only a spread, is not a full comparison.

Average. Use the mean if the data has no extreme values. Use the median if there are outliers, because the median is not affected by them. The group with the higher average is higher “on average”. Spread. The range (biggest − smallest) is the simplest measure of spread. A smaller range means the data is more consistent (closer together). Higher tier students also use the interquartile range (IQR), which ignores extreme values.

How to write a comparison. Always write your comparison in context, using the words of the question, and include the numbers. For example: “On average, the boys scored higher (mean 62 compared with 55), but the girls’ scores were more consistent (range 14 compared with 28).” A sentence such as “A has a bigger range” with no meaning is not enough.

0246810Goals scored in a matchMatches01234Team PTeam Q
Comparing two distributions: Team Q’s goals are centred higher (a higher average), and both are spread over the same range of 0 to 4.

Key ideas

  • Compare an average (mean or median) and a spread (range, or IQR).
  • Use the median if there are outliers.
  • A smaller range (or IQR) means more consistent data.
  • Write in context and include the numbers.
  • A larger average means “higher on average”, not that every value is bigger.
Worked example 24.11.1
The numbers of goals scored by two hockey teams in six matches are shown.
Team A358464
Team B291873
Compare the goals scored by the two teams.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Compare an average (the mean) and a measure of spread (the range) for each group, then write sentences in context.

  1. Work out the mean for Team A. An average tells us what is typical.
    3 + 5 + 8 + 4 + 6 + 4 = 30; 30 ÷ 6 = 5
  2. Work out the mean for Team B.
    2 + 9 + 1 + 8 + 7 + 3 = 30; 30 ÷ 6 = 5
  3. Work out the range for Team A. A spread tells us how consistent the data is.
    8 − 3 = 5
  4. Work out the range for Team B.
    9 − 1 = 8
  5. Write the comparison. Say something about the average and something about the spread, and use the context (the words of the question).
    On average, Team A and Team B are the same (mean 5).
    Team A is more consistent (smaller range: 5 compared with 8), so Team B is more spread out.

Check. Both sentences use numbers from the working. ✓

Watch out. A comparison needs both an average and a spread, in the context of the question. A smaller range (or IQR) means more consistent.

Worked example 24.11.2
Two machines fill bags of rice. The masses, in grams, of six bags from each machine are shown.
Machine A413844404239
Machine B305528603552
Which machine is more reliable? Use the mean and the range to explain.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Compare an average (the mean) and a measure of spread (the range) for each group, then write sentences in context.

  1. Work out the mean for Machine A. An average tells us what is typical.
    41 + 38 + 44 + 40 + 42 + 39 = 244; 244 ÷ 6 = 40.7
  2. Work out the mean for Machine B.
    30 + 55 + 28 + 60 + 35 + 52 = 260; 260 ÷ 6 = 43.3
  3. Work out the range for Machine A. A spread tells us how consistent the data is.
    44 − 38 = 6
  4. Work out the range for Machine B.
    60 − 28 = 32
  5. Write the comparison. Say something about the average and something about the spread, and use the context (the words of the question).
    On average, Machine B has the higher mass (mean 43.3 compared with 40.7).
    Machine A is more consistent (smaller range: 6 compared with 32), so Machine B is more spread out.

Check. Both sentences use numbers from the working. ✓

Watch out. A comparison needs both an average and a spread, in the context of the question. A smaller range (or IQR) means more consistent.

Worked example 24.11.3
The times, in seconds, taken by two runners in six races are shown. Runner P has one very slow race.
Runner P121413151460
Runner Q182022192123
Compare the times using the median and the range.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Because of the outlier (60 seconds), use the median rather than the mean. Compare an average (the median) and a measure of spread (the range) for each group, then write sentences in context.

  1. Work out the median for Runner P. An average tells us what is typical.
    In order: 12, 13, 14, 14, 15, 60; median = 14
  2. Work out the median for Runner Q.
    In order: 18, 19, 20, 21, 22, 23; median = 20.5
  3. Work out the range for Runner P. A spread tells us how consistent the data is.
    60 − 12 = 48
  4. Work out the range for Runner Q.
    23 − 18 = 5
  5. Write the comparison. Say something about the average and something about the spread, and use the context (the words of the question).
    On average, Runner Q has the higher time (median 20.5 compared with 14).
    Runner Q is more consistent (smaller range: 5 compared with 48), so Runner P is more spread out.

Check. Both sentences use numbers from the working. ✓

Watch out. A comparison needs both an average and a spread, in the context of the question. A smaller range (or IQR) means more consistent.

Worked example 24.11.4
Compare the ages of two clubs using the median and the interquartile range.
Club A12182022252730
Club B15161819212224

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Compare an average (the median) and a measure of spread (the interquartile range) for each group, then write sentences in context.

  1. Work out the median for Club A. An average tells us what is typical.
    In order: 12, 18, 20, 22, 25, 27, 30; median = 22
  2. Work out the median for Club B.
    In order: 15, 16, 18, 19, 21, 22, 24; median = 19
  3. Work out the interquartile range for Club A. A spread tells us how consistent the data is.
    Q1 = 18, Q3 = 27, IQR = 27 − 18 = 9
  4. Work out the interquartile range for Club B.
    Q1 = 16, Q3 = 22, IQR = 22 − 16 = 6
  5. Write the comparison. Say something about the average and something about the spread, and use the context (the words of the question).
    On average, Club A has the higher ages (median 22 compared with 19).
    Club B is more consistent (smaller interquartile range: 6 compared with 9), so Club A is more spread out.

Check. Both sentences use numbers from the working. ✓

Watch out. A comparison needs both an average and a spread, in the context of the question. A smaller range (or IQR) means more consistent.

Watch out

A comparison must include both an average and a spread, written in context, with numbers. Do not say “Team A is better” without saying what is better (higher average, more consistent).

Exercise 24K  Comparing distributions

Non-calculator
1
FCompare the two sets of data using the mean and the range. Write your answer in context.
(a)Scores of Amy: 7, 9, 6, 8, 10 and of Ben: 4, 10, 5, 9, 7 in five tests.
(b)Heights in cm of six girls: 150, 154, 148, 152, 156, 150 and six boys: 145, 160, 152, 148, 158, 157.
(c)Waiting times in minutes at two clinics, A: 12, 15, 9, 14, 10 and B: 20, 8, 25, 6, 11.
2
PSTwo players scored goals in seven matches. Player X: 1, 0, 3, 2, 1, 0, 2. Player Y: 0, 5, 0, 0, 1, 6, 0. Compare the goals scored using the mean and the range.
3
RThe ages of the people in two shops are shown.
Shop 1: 20, 22, 21, 23, 22, 65
Shop 2: 30, 32, 31, 33, 30, 34
Compare the ages using the median and the range. Explain why the median is better than the mean for Shop 1.
4
PSThe frequency tables show the number of goals scored in 20 matches by two teams.
Team P
Goals01234
Matches47621
Team Q
Goals01234
Matches13862
Which team scores more goals on average? Compare the means and give the range for each team.
(a)Mean for team P
(b)Mean for team Q
(c)Range for team P
(d)Range for team Q
(e)Comparison
5
RHannah says: “Class A has a mean mark of 64 and class B has a mean mark of 61, so everyone in class A did better than everyone in class B.” Explain why she is wrong.
6
FComplete the comparison. “The mean for team A is 12 and the mean for team B is 15. The range for A is 5 and the range for B is 14.” Write two sentences comparing the teams.
7
PSTwo archers shoot seven arrows each. Their scores out of 10 are: Archer 1: 6, 7, 7, 8, 8, 8, 9. Archer 2: 3, 6, 8, 8, 9, 10, 10. A coach must choose one archer for a team who needs consistent scores. Which archer should be chosen? Use averages and ranges to justify your answer.
8
RThe interquartile ranges of two groups are 4 and 9, and the medians are 52 and 50. Write a comparison of the two groups.
9
RSam says: “Class X has a range of 35 and class Y has a range of 20, so class X must have the higher mean.” Is he right? Explain.
10
RThe mean height of 5 players in team A is 180 cm. The mean height of 5 players in team B is 176 cm. One player from each team swaps teams. Team A’s new mean is 178 cm. How tall is the player who joined team A, if the player who left was 190 cm?
11
FThe numbers of points scored by two players in five games are: Player A: 4, 6, 5, 7, 8. Player B: 3, 9, 5, 4, 10. Compare the points using the median and the range.
12
PSThe battery lives, in hours, of five phones of each type are: Phone A: 14, 15, 13, 14, 16. Phone B: 10, 18, 12, 19, 16. A customer wants a phone with a reliable battery life. Which should the customer choose? Use the mean and the range.
13
FTeam X has a mean of 30 points and a range of 6. Team Y has a mean of 24 points and a range of 10. Write two sentences to compare the teams.
14
RWrite down two sets of five numbers, P and Q, so that P has the higher mean but the smaller range. Show that your sets work.
24.12

Quartiles and the interquartile range

Edexcel 1MA1: S4
Grade 5-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Quartiles and the interquartile rangeComing soon

The median splits ordered data into two halves. The quartiles go further and split it into four equal parts. The lower quartile (Q1) has a quarter of the data below it. The median is Q2, with half the data below it. The upper quartile (Q3) has three quarters of the data below it. These three numbers cut the ordered data into four groups with the same number of values in each.

From a list. (1) Put the data in order. (2) Find the median. (3) Split the data into a lower half and an upper half (if there is an odd number of values, leave out the median). (4) Q1 is the median of the lower half and Q3 is the median of the upper half. With an odd number of values this gives the same answer as the rule “Q1 is the (n + 1) ÷ 4th value and Q3 is the 3(n + 1) ÷ 4th value”.

The interquartile range (IQR) is Q3 − Q1. It is the spread of the middle half of the data. Unlike the range, it is not affected by extreme values, so it is a more reliable measure of spread when there are outliers. A small IQR means the middle half of the data is close together (consistent).

From a frequency table. Add a running total (cumulative frequency) column. Work out the positions (n + 1) ÷ 4, (n + 1) ÷ 2 and 3(n + 1) ÷ 4. The quartile is the value in that position, found from the running total, exactly as for the median.

51015202530Q1 = 6.5median = 15Q3 = 23
The nine values 3, 5, 8, 12, 15, 18, 21, 25, 30 in order. The median (15) splits them into two halves; Q1 = 6.5 and Q3 = 23 are the medians of the halves.

Key ideas

  • Q1 = lower quartile (a quarter of the data is below it). Q2 = median. Q3 = upper quartile (three quarters below it).
  • For a list: order the data, find the median, then the median of each half (leave out the median when n is odd).
  • IQR = Q3 − Q1: the spread of the middle 50% of the data.
  • The IQR is not affected by outliers, but the range is.
  • For a frequency table, use positions (n + 1) ÷ 4 and 3(n + 1) ÷ 4 with a running total.
Worked example 24.12.1
Find the lower quartile, the upper quartile and the interquartile range of 3, 5, 8, 12, 15, 18, 21, 25, 30.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. Quartiles split ordered data into four equal parts. Q1 is the median of the lower half, Q3 is the median of the upper half, and IQR = Q3 − Q1.

  1. Write the values in order. Quartiles are found from ordered data.
    3, 5, 8, 12, 15, 18, 21, 25, 30
  2. Find the median (Q2). The median splits the data into a lower half and an upper half.
    n = 9, median = 15 (the middle value)
  3. Split into two halves, leaving out the median. With an odd number of values the median belongs to neither half.
    Lower half: 3, 5, 8, 12
    Upper half: 18, 21, 25, 30
  4. Q1 is the median of the lower half. It is the value a quarter of the way through the data.
    Q1 = 6.5
  5. Q3 is the median of the upper half. It is the value three quarters of the way through the data.
    Q3 = 23
  6. Subtract to find the interquartile range. IQR = Q3 − Q1: the spread of the middle half of the data.
    23 − 6.5 = 16.5

Check. About a quarter of the values are below Q1 and a quarter are above Q3. Here 2 are below 6.5 and 2 are above 23. ✓

Watch out. Do not forget to put the data in order first, and leave out the median when you split an odd number of values.

Worked example 24.12.2
Find the median, the lower quartile and the interquartile range of 14, 9, 21, 17, 12, 25, 8, 19, 16, 11.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. Quartiles split ordered data into four equal parts. Q1 is the median of the lower half, Q3 is the median of the upper half, and IQR = Q3 − Q1.

  1. Write the values in order. Quartiles are found from ordered data.
    8, 9, 11, 12, 14, 16, 17, 19, 21, 25
  2. Find the median (Q2). The median splits the data into a lower half and an upper half.
    n = 10, median = 15 (halfway between the two middle values)
  3. Split into two equal halves. With an even number of values, the lower half is the first half of the list.
    Lower half: 8, 9, 11, 12, 14
    Upper half: 16, 17, 19, 21, 25
  4. Q1 is the median of the lower half. It is the value a quarter of the way through the data.
    Q1 = 11
  5. Q3 is the median of the upper half. It is the value three quarters of the way through the data.
    Q3 = 19
  6. Subtract to find the interquartile range. IQR = Q3 − Q1: the spread of the middle half of the data.
    19 − 11 = 8

Check. Half of the 10 values lie between Q1 and Q3. ✓

Watch out. Do not forget to put the data in order first, and leave out the median when you split an odd number of values.

Worked example 24.12.3
The table shows the number of calls received by a call centre in each of 31 hours.
Calls12345
Hours26986
Find the median, the lower quartile, the upper quartile and the interquartile range.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. For a frequency table, find the position of each quartile, then use a running total of the frequencies to find the value in that position.

  1. Add up the frequencies. This is the number of values, n.
    2 + 6 + 9 + 8 + 6 = 31
  2. Work out the positions of the quartiles. Q1 is at (n + 1) ÷ 4, the median at (n + 1) ÷ 2, and Q3 at 3(n + 1) ÷ 4.
    (31 + 1) ÷ 4 = 8
    (31 + 1) ÷ 2 = 16
    3 × (31 + 1) ÷ 4 = 24
  3. Add a running total column. The running total shows which positions each value occupies.
    CallsFrequencyRunning total
    122
    268
    3917
    4825
    5631
  4. Find the value at each position. Look for the first row where the running total reaches the position.
    Position 8: row for 2 (positions 3 to 8), so Q1 = 2
    Position 16: row for 3, so median = 3
    Position 24: row for 4 (positions 18 to 25), so Q3 = 4
  5. Subtract to find the interquartile range. IQR = Q3 − Q1.
    4 − 2 = 2

Check. The running total before the row for Q1 is less than 8, and in the row for Q1 it is at least 8. ✓

Watch out. Use (n + 1) ÷ 4 for a frequency table of single values. Give the value, not the position.

Worked example 24.12.4
The list 12, 14, 15, 17, 18, 20, 22 has a new value, 60, added to it. Show the effect on the range and on the interquartile range.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. Work out the range and IQR before and after adding the new value, then compare the changes.

  1. Before: range. Biggest minus smallest.
    22 − 12 = 10
  2. Before: IQR. The 7 values give Q1 = 14 and Q3 = 20.
    20 − 14 = 6
  3. After: range. The biggest value is now 60.
    60 − 12 = 48
  4. After: IQR. The 8 values give lower half 12, 14, 15, 17 (Q1 = 14.5) and upper half 18, 20, 22, 60 (Q3 = 21).
    21 − 14.5 = 6.5
  5. Compare. The range jumped from 10 to 48. The IQR changed from 6 to 6.5.
    The range is badly affected by the outlier. The IQR is hardly affected.

Check. The new value is far above the others, so it is at the very end of the data and only nudges the quartiles. ✓

Watch out. Use the IQR instead of the range when the data has extreme values.

Watch out

Put the data in order first. And with an odd number of values, leave out the median when you split the data into two halves.

Exercise 24L  Quartiles and the interquartile range

Non-calculator
1
FWork out the lower quartile, upper quartile and interquartile range of each set of numbers.
(a)2, 4, 7, 9, 12, 15, 20
(b)5, 8, 3, 12, 15, 7, 10, 20, 17
(c)18, 22, 15, 30, 25, 28, 12, 20, 24, 16, 26
(d)1.5, 2.1, 0.8, 3.2, 2.7, 1.9, 2.4
2
FWork out the median, lower quartile and interquartile range of each set of numbers (there are an even number of values).
(a)4, 7, 9, 12, 15, 18
(b)31, 25, 28, 40, 35, 22, 38, 30
3
FWork out the lower quartile, upper quartile and interquartile range from each frequency table.
(a)
Value12345
Frequency26986
(b)
Value10203040
Frequency4793
4
PSThe ages of 11 people on a bus are 23, 31, 19, 42, 35, 27, 38, 29, 22, 45, 33. Find the interquartile range of the ages.
5
RHannah found the interquartile range of 9, 3, 12, 6, 15, 4, 7 by working with the numbers in the order given. She said Q1 = 3 and Q3 = 15. Explain her mistake and work out the correct interquartile range.
6
FThe median of a set of data is 14, the lower quartile is 9 and the interquartile range is 12. Work out the upper quartile.
7
PSIn a data set of 80 values, how many values lie between the lower quartile and the upper quartile? Explain.
8
PSThe list 12, 14, 15, 17, 18, 20, 22 has the new value 60 added. Work out each of these for the new list of 8 values: (a) the range, (b) the interquartile range.
(a)Range
(b)Interquartile range
9
RZara says: “The interquartile range is the better measure of spread when there are outliers.” Explain why she is right.
10
RThe list 2, 5, 7, 10, 13, 16, 20 has median 10. A new value, 10, is added. Show that the median does not change but the interquartile range gets smaller.
11
PSCompare the ages of two clubs using the median and the interquartile range.
Club A: 12, 18, 20, 22, 25, 27, 30
Club B: 15, 16, 18, 19, 21, 22, 24
12
RA set of seven whole numbers has median 8, lower quartile 5 and upper quartile 12. The smallest number is 2 and the largest is 20. Write down the interquartile range and the range, and give one possible list of the seven numbers.
24.13

Cumulative frequency graphs

Edexcel 1MA1: S3, S4
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Cumulative frequency graphsComing soon

Cumulative frequency is a running total of the frequencies. It tells us how many values are up to and including the end of each class. A cumulative frequency graph plots each cumulative frequency against the upper boundary of its class and joins the points with a smooth S-shaped curve. It rises from 0 at the start and ends at the total number of values.

0204060800204060Time taken (minutes)Cumulative frequency
A cumulative frequency curve for the times taken by 80 workers: 4 took up to 10 minutes, 14 took up to 20 minutes, and so on.

Reading the graph. To find the median, go to half of the total (n ÷ 2) on the vertical axis, across to the curve and down to the horizontal axis. For the lower quartile use n ÷ 4 and for the upper quartile use 3n ÷ 4. The interquartile range is the upper quartile minus the lower quartile. A percentile works in the same way: the 90th percentile is at 90% of n. These readings are estimates. Notice that on a cumulative frequency graph we use n ÷ 2, n ÷ 4 and 3n ÷ 4, not (n + 1) ÷ 2 and so on.

To find how many values are less than a given value, go up from that value to the curve and across to the cumulative frequency axis. To find how many are more than a value, subtract that reading from the total.

Key ideas

  • Cumulative frequency = running total of the frequencies.
  • Plot (upper class boundary, cumulative frequency), starting with (lowest boundary, 0), and join with a smooth curve.
  • Median at n ÷ 2, lower quartile at n ÷ 4, upper quartile at 3n ÷ 4. IQR = Q3 − Q1.
  • Percentiles: the p-th percentile is at p% of n.
  • To find the number above a value, read the cumulative frequency and subtract it from the total.
  • Readings from a graph are estimates.
Worked example 24.13.1
The table shows the times taken by 80 workers to complete a task.
TimeFrequency
0 < t ≤ 10 min4
10 < t ≤ 20 min10
20 < t ≤ 30 min22
30 < t ≤ 40 min25
40 < t ≤ 50 min13
50 < t ≤ 60 min6
Complete the cumulative frequency column.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. Cumulative frequency is a running total: add each frequency to the total of all the earlier ones.

  1. Start the running total with the first frequency. Cumulative frequency means “how many values so far”.
    First class: 4
  2. Add the next frequency to the total so far (10 < t ≤ 20 min). Each cumulative frequency is the previous one plus this class’s frequency.
    4 + 10 = 14
  3. Add the next frequency to the total so far (20 < t ≤ 30 min). Each cumulative frequency is the previous one plus this class’s frequency.
    14 + 22 = 36
  4. Add the next frequency to the total so far (30 < t ≤ 40 min). Each cumulative frequency is the previous one plus this class’s frequency.
    36 + 25 = 61
  5. Add the next frequency to the total so far (40 < t ≤ 50 min). Each cumulative frequency is the previous one plus this class’s frequency.
    61 + 13 = 74
  6. Add the next frequency to the total so far (50 < t ≤ 60 min). Each cumulative frequency is the previous one plus this class’s frequency.
    74 + 6 = 80
  7. Check the last number. The last cumulative frequency must equal the total number of values.
    Total of frequencies = 80 = last cumulative frequency 80

Check. The last running total (80) equals the sum of all the frequencies. ✓

Watch out. The cumulative frequency never goes down, and the last one is the total number of data values.

Worked example 24.13.2
Use the table in the last example to draw a cumulative frequency graph.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. Make a cumulative frequency table, then plot each cumulative frequency against the upper end of its class, and join the points with a smooth curve.

  1. Work out the cumulative frequencies. Add each frequency to the total so far.
    TimeFrequencyCumulative frequency
    0 < t ≤ 10 min44
    10 < t ≤ 20 min1014
    20 < t ≤ 30 min2236
    30 < t ≤ 40 min2561
    40 < t ≤ 50 min1374
    50 < t ≤ 60 min680
  2. Write the points to plot. Plot each cumulative frequency at the upper class boundary, because that is the point by which that many values have been counted. Start at (0, 0): nothing is below the first lower boundary.
    (0, 0), (10, 4), (20, 14), (30, 36), (40, 61), (50, 74), (60, 80)
  3. Choose scales and plot the points. The horizontal axis goes from the lowest to the highest class boundary. The vertical axis goes from 0 to the total frequency.
    Horizontal axis: Time taken (minutes). Vertical axis: cumulative frequency up to 80.
  4. Join the points with a smooth curve. The curve is S-shaped and never goes down.
    0204060800204060Time taken (minutes)Cumulative frequency

Check. The curve starts at 0 and ends at the total frequency, 80. ✓

Watch out. Plot at the upper boundary of each class, not the midpoint and not the lower boundary.

Worked example 24.13.3
Use the cumulative frequency graph above to estimate the median, the lower quartile, the upper quartile and the interquartile range.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. On a cumulative frequency graph, the median is at height n ÷ 2, the lower quartile at n ÷ 4 and the upper quartile at 3n ÷ 4. Read across to the curve and down to the axis.

  1. Find the total number of values. The last cumulative frequency is the total, n.
    n = 80
  2. Work out the heights to read at. Lower quartile at n ÷ 4, median at n ÷ 2, upper quartile at 3n ÷ 4.
    n ÷ 4 = 20
    n ÷ 2 = 40
    3n ÷ 4 = 60
  3. Draw a horizontal line across from each height to the curve, then straight down to the axis. The value on the horizontal axis is the quartile.
    0204060800204060Time taken (minutes)Cumulative frequency23.531.539.5
  4. Read the three values. Read the scale carefully. The answers are estimates.
    Q1 ≈ 23.5 min
    Median ≈ 31.5 min
    Q3 ≈ 39.5 min
  5. Subtract to find the interquartile range. IQR = Q3 − Q1.
    39.5 − 23.5 = 16 min

Check. Going up from each reading to the curve and across should give the heights again. ✓

Watch out. Read heights from the cumulative frequency axis and answers from the data axis. Do not use n + 1 here.

Worked example 24.13.4
Estimate the number of workers who took more than 45 minutes.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. Read the cumulative frequency at the value. This is the number of values up to that value; subtract from the total to get the number above it.

  1. Find 45 minutes on the horizontal axis and go up to the curve. The curve shows how many values are at or below each value.
    0204060800204060Time taken (minutes)Cumulative frequency69
  2. Read across to the cumulative frequency axis. This gives the number of values up to 45.
    Cumulative frequency = 69
  3. Subtract from the total to find how many are more than that. The last cumulative frequency is the total.
    80 − 69 = 11

Check. 69 below plus 11 above makes 80, the total. ✓

Watch out. The reading gives the number below the value. For “more than” you must subtract from the total.

Worked example 24.13.5
Estimate the 90th percentile of the times.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. Cumulative frequency graphs are read by going across from the cumulative frequency to the curve, then down to the axis.

  1. Find the position on the cumulative frequency axis. The total cumulative frequency is n = 80. The 90th percentile is 90% of the way through the data: 90 ÷ 100 × 80 = 72.
    Position = 72
  2. Draw a horizontal line across from that height to the curve. We want the data value that has this many values at or below it.
    0204060800204060Time taken (minutes)Cumulative frequency48
  3. Draw a vertical line down to the horizontal axis and read the value. Read the scale carefully. The result is an estimate.
    90th percentile ≈ 48 min

Check. Going up from your answer to the curve and across should give the same height again. ✓

Watch out. Read across from the cumulative frequency axis. The frequency axis is the vertical one, the data axis is the horizontal one.

Watch out

Plot at the upper class boundary (not the midpoint), and start the curve at (lowest boundary, 0). When reading, use n ÷ 2, n ÷ 4 and 3n ÷ 4: do not add 1.

Exercise 24M  Cumulative frequency graphs

Non-calculator
1
FThe table shows the heights of 60 plants.
HeightFrequency
10 < h ≤ 20 cm5
20 < h ≤ 30 cm12
30 < h ≤ 40 cm24
40 < h ≤ 50 cm14
50 < h ≤ 60 cm5
Work out the cumulative frequency for each class.
2
FUse the table in question 1 to draw a cumulative frequency graph.
3
FThe graph shows the times, in minutes, taken by 80 workers to complete a task. Use the graph to estimate (a) the median, (b) the lower quartile, (c) the upper quartile, (d) the interquartile range.
0204060800204060Time taken (minutes)Cumulative frequency
(a)Median
(b)Lower quartile
(c)Upper quartile
(d)Interquartile range
4
PSUse the same graph to estimate (a) the number of workers who took more than 45 minutes, (b) the number of workers who took less than 25 minutes, (c) the percentage of workers who took less than 20 minutes.
0204060800204060Time taken (minutes)Cumulative frequency
(a)More than 45 minutes
(b)Less than 25 minutes
(c)Percentage less than 20 minutes
5
FUse the graph in question 3 to estimate the 90th percentile of the times.
6
FThe table shows the marks of 100 students in a test.
MarkFrequency
0 < m ≤ 204
20 < m ≤ 3010
30 < m ≤ 4018
40 < m ≤ 5024
50 < m ≤ 6022
60 < m ≤ 8016
80 < m ≤ 1006
Draw a cumulative frequency graph.
7
PSThe cumulative frequency graphs show the marks of 40 students in class A (blue) and class B (orange). (a) Estimate the median of each class. (b) Estimate the interquartile range of each class. (c) Compare the two classes.
05101520253035400102030405060708090100MarkCumulative frequency
(a)Medians
(b)Interquartile ranges
(c)Comparison
8
PSThe cumulative frequency graph in question 3 shows the times of 80 workers. The company gives a bonus to the fastest 15% of workers. Estimate the longest time that a worker can take and still get the bonus.
9
PSUse the graph in question 3 to estimate the number of workers who took between 25 and 45 minutes.
10
RExplain why cumulative frequency is plotted at the upper boundary of each class, not at the midpoint.
11
PSA cumulative frequency curve passes through (0, 0), (10, 12), (20, 40), (30, 70) and (40, 80). (a) How many values are in the class 20 < x ≤ 30? (b) What percentage of all the values are 20 or less?
(a)Values in 20 < x ≤ 30
(b)Percentage 20 or less
12
PSThe table is partly completed.
Time, t (minutes)FrequencyCumulative frequency
0 < t ≤ 1066
10 < t ≤ 2014?
20 < t ≤ 30?45
30 < t ≤ 409?
40 < t ≤ 50?60
Complete the table.
13
RA cumulative frequency graph for 120 values has median 38, lower quartile 30 and upper quartile 50. Estimate how many values are between 30 and 50. Explain your answer.
24.14

Box plots and outliers

Edexcel 1MA1: S4
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Box plots and outliersComing soon

A box plot (or box-and-whisker diagram) shows the main features of a distribution using five numbers, called the five-number summary: the smallest value, the lower quartile (Q1), the median (Q2), the upper quartile (Q3) and the largest value. The box goes from Q1 to Q3 with a line at the median. The whiskers are lines from the box to the smallest and the largest values. The box holds the middle half of the data. Each whisker and each half of the box holds one quarter of the data.

20304050607080Test markClass AClass B
Two box plots on the same scale. The median is the orange line. Class B has the smaller box, so its marks are more consistent.

Reading a box plot. The median is the line in the box. The range is the largest minus the smallest. The IQR is the length of the box (Q3 − Q1). About 25% of the data lies in each section: from the smallest value to Q1, from Q1 to the median, from the median to Q3, and from Q3 to the largest value.

Comparing box plots. Compare the medians (a typical value), and compare the IQRs (the lengths of the boxes) or the ranges (the overall spread). Always write your comparison in context.

Outliers. An outlier is a value that is far from the rest of the data. A common rule says a value is an outlier if it is more than 1.5 × IQR above Q3 or below Q1. In a box plot an outlier is shown as a cross, and the whisker then goes to the most extreme value that is not an outlier. Outliers might be genuine unusual values or mistakes, so we should check them rather than just deleting them.

Key ideas

  • Five-number summary: smallest, Q1, median, Q3, largest.
  • The box is the middle 50% of the data. Each of the four sections holds 25%.
  • Range = largest − smallest. IQR = length of the box = Q3 − Q1.
  • Compare box plots using medians and IQRs, in context.
  • Outlier rule: below Q1 − 1.5 × IQR or above Q3 + 1.5 × IQR. Show outliers as crosses.
Worked example 24.14.1
Draw a box plot for the marks 23, 34, 38, 41, 45, 48, 52, 55, 60, 66, 78.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. A box plot is drawn from five numbers: the smallest value, the lower quartile, the median, the upper quartile and the largest value.

  1. Write the data in order. We need the ordered data to find the quartiles.
    23, 34, 38, 41, 45, 48, 52, 55, 60, 66, 78
  2. Find the median, lower quartile and upper quartile. Q1 is the median of the lower half and Q3 is the median of the upper half.
    Median = 48, Q1 = 38, Q3 = 60
  3. Write down the smallest and largest values. These are the ends of the whiskers.
    Smallest = 23, largest = 78
  4. Draw a scale and then the box plot. Draw a box from Q1 to Q3 with a line at the median, then whiskers to the smallest and largest values.
    20406080100Class A

Check. The box holds the middle half of the data, and the line is inside the box. ✓

Watch out. The five numbers must be in order: smallest < Q1 < median < Q3 < largest, and the scale must be even.

Worked example 24.14.2
Use the box plots for class A and class B above to compare the marks of the two classes.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. Read the median and find the IQR of each box plot, then compare them.

  1. Read the medians. The median is the line inside each box.
    Class A: 48
    Class B: 49
  2. Find the IQRs. IQR = Q3 − Q1, the length of the box.
    Class A: 60 − 38 = 22
    Class B: 57 − 42 = 15
  3. Find the ranges. Largest minus smallest.
    Class A: 78 − 23 = 55
    Class B: 64 − 30 = 34
  4. Write the comparison. Use the medians for the average and the IQRs for the spread, in context.
    The classes have similar median marks (A: 48, B: 49) but class B is more consistent: its IQR is 15 compared with 22.

Check. Class B has the shorter box and the shorter whiskers. ✓

Watch out. Compare an average and a spread, and use the words of the question.

Worked example 24.14.3
The data set 12, 14, 15, 17, 18, 20, 22, 60 has Q1 = 14.5 and Q3 = 21. Use the rule “more than 1.5 × IQR from the quartiles” to find any outliers.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. An outlier is a value that is far from the rest. One rule says a value is an outlier if it is more than 1.5 × IQR below Q1 or above Q3.

  1. Work out the IQR. IQR = Q3 − Q1.
    21 − 14.5 = 6.5
  2. Work out 1.5 × IQR. This is the allowed distance from the quartiles.
    1.5 × 6.5 = 9.75
  3. Work out the lower limit. Take it away from Q1.
    14.5 − 9.75 = 4.75
  4. Work out the upper limit. Add it to Q3.
    21 + 9.75 = 30.75
  5. Compare each value with the limits. A value below the lower limit or above the upper limit is an outlier.
    Outliers: 60

Check. Every other value lies between 4.75 and 30.75. ✓

Watch out. The limits are measured from Q1 and Q3, not from the median.

Worked example 24.14.4
Draw the box plot for the data set in the last example, showing the outlier.
(Q1 = 14.5, median = 17.5, Q3 = 21, smallest = 12.)

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. The outlier is plotted as a cross. The upper whisker ends at the largest value that is not an outlier.

  1. Find the largest value that is not an outlier. The value 60 is an outlier. The next largest value is 22.
    Whisker goes up to 22.
  2. Write the five numbers for the box plot. Use the smallest value, Q1, median, Q3 and the largest non-outlier.
    12, 14.5, 17.5, 21, 22
  3. Draw the box plot and mark 60 with a cross. A cross shows that the value is an outlier.
    10203040506070ValueData

Check. The cross at 60 is far from the end of the whisker at 22. ✓

Watch out. Do not extend the whisker to the outlier: mark it separately with a cross.

Watch out

A box plot does not show the mean, and a longer box does not mean there is more data. A box plot also does not show how many values there are. Compare the box plots’ medians and IQRs, and mention both.

Exercise 24N  Box plots and outliers

Non-calculator
1
FDraw a box plot for these values: 8, 11, 14, 15, 17, 19, 22, 25, 30.
2
FDraw a box plot for these values: 41, 50, 38, 45, 52, 47, 60, 43.
3
PSThe box plots show the times, in minutes, taken by two groups to solve a puzzle. (a) Write down the median of group A. (b) Work out the interquartile range of each group. (c) Work out the range of group B. (d) Compare the two groups.
102030405060Time (minutes)Group AGroup B
(a)Median of group A
(b)IQRs
(c)Range of group B
(d)Comparison
4
RThe box plot shows the heights of some plants. (a) What fraction of the plants are taller than the upper quartile? (b) In a sample of 80 plants, how many have a height between the median and the upper quartile? (c) Explain why you cannot tell from the box plot how many plants there were.
10203040506070Height (cm)Plants
(a)Taller than the upper quartile
(b)Between the median and the upper quartile
(c)Why the number is unknown
5
FFind out whether there are any outliers using the rule “more than 1.5 × IQR from the quartiles”.
Data: 31, 33, 35, 37, 38, 40, 41, 42, 44, 46, 78. For this data Q1 = 35 and Q3 = 44.
6
FUse the same rule to check for outliers in this data: 44, 46, 47, 49, 50, 52, 53, 55, 56, 58, 59. For this data Q1 = 47 and Q3 = 56.
7
RTwo data sets have the same median. Data set P has an IQR of 6 and data set Q has an IQR of 15. Write down one thing the box plots would show about the two data sets.
8
PSIn a data set with 60 values, 25% of the values are bigger than 52. How many values are bigger than 52? What does 52 represent on a box plot?
9
PSA box plot has smallest value 10, range 40, median 28, lower quartile 20 and interquartile range 16. Work out the largest value and the upper quartile, then draw the box plot.
10
RAisha says: “The box plot for group A has a longer box than group B, so group A has more people in it.” Is she right? Explain.
11
PSThe box plot for the times of 40 runners has a median of 24 minutes, a lower quartile of 20 and an upper quartile of 31. Work out, giving reasons, how many runners ran faster than 24 minutes, and how many ran between 20 and 31 minutes.
12
PSThe cumulative frequency graph shows the times of 80 workers. The quickest took 2 minutes and the slowest took 58 minutes. Use the graph to draw a box plot of the times.
0204060800204060Time taken (minutes)Cumulative frequency
13
RIn a box plot the median is very close to the lower quartile, but far from the upper quartile. What does this tell you about the data?
24.15

Histograms

Edexcel 1MA1: S3
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: HistogramsComing soon

A histogram shows continuous data. It looks like a bar chart, but the bars can have different widths, and there are no gaps between bars because the classes touch. When the class widths are different, the height of a bar cannot be the frequency, because a wide class would look much bigger just because it is wide. Instead the height is the frequency density, so that the area of each bar stands for the frequency.

frequency density = frequency ÷ class width, and so frequency = frequency density × class width (the area of the bar). For example, 30 people in a class of width 20 gives frequency density 30 ÷ 20 = 1.5, and a bar 20 wide and 1.5 tall has area 30.

00.511.5201020405070Time reading (minutes)Frequency density
A histogram. The bars have different widths. The area of each bar is the frequency of its class.

Drawing a histogram. (1) Work out each class width. (2) Divide each frequency by its class width to get the frequency density. (3) Draw a horizontal axis for the data and a vertical axis for the frequency density. (4) Draw each bar from the lower to the upper class boundary, with height equal to its frequency density.

Reading a histogram. The frequency of a bar is its height times its width. For part of a bar, take the same fraction of the bar’s frequency. To estimate the mean, use the midpoints, as for any grouped data. To estimate the median, find the class holding the middle value and assume the values are spread evenly across it.

Key ideas

  • Frequency density = frequency ÷ class width.
  • Frequency = frequency density × class width = the area of the bar.
  • The vertical axis is frequency density, not frequency. There are no gaps between bars.
  • For part of a bar: part of the width × the frequency density.
  • Estimate the mean using midpoints; estimate the median by assuming an even spread within the median class.
Worked example 24.15.1
The table shows the times that 80 people spent reading.
TimeFrequency
0 < t ≤ 10 min8
10 < t ≤ 20 min14
20 < t ≤ 40 min30
40 < t ≤ 50 min16
50 < t ≤ 70 min12
Draw a histogram.

Key words. A histogram has bars of different widths. The height is the frequency density = frequency ÷ class width, so the area of a bar is the frequency.

Method. For a histogram, plot frequency density = frequency ÷ class width on the vertical axis.

  1. Find the width of each class. The bars have different widths, so the height cannot be the frequency. Width = upper end − lower end.
    10 − 0 = 10
    20 − 10 = 10
    40 − 20 = 20
    50 − 40 = 10
    70 − 50 = 20
  2. Divide each frequency by its class width. frequency density = frequency ÷ class width. This is the height of the bar, and it makes the area of the bar equal to the frequency.
    8 ÷ 10 = 0.8
    14 ÷ 10 = 1.4
    30 ÷ 20 = 1.5
    16 ÷ 10 = 1.6
    12 ÷ 20 = 0.6
  3. Write the results in a table.
    TimeFrequencyClass widthFrequency density
    0 < t ≤ 10 min8100.8
    10 < t ≤ 20 min14101.4
    20 < t ≤ 40 min30201.5
    40 < t ≤ 50 min16101.6
    50 < t ≤ 70 min12200.6
  4. Draw the histogram. Put the data on the horizontal axis and frequency density on the vertical axis. Draw one bar for each class with no gaps, because the classes touch. The width of each bar is its class width and the height is its frequency density.
    00.511.5201020405070Time spent reading (minutes)Frequency density

Check. Multiply back: 0.8 × 10 = 8, 1.4 × 10 = 14, 1.5 × 20 = 30, 1.6 × 10 = 16, 0.6 × 20 = 12. These are the frequencies. ✓

Watch out. The vertical axis is frequency density, not frequency.

Worked example 24.15.2
Use the histogram above to work out the number of people in each class, and the total number of people.

Key words. A histogram has bars of different widths. The height is the frequency density = frequency ÷ class width, so the area of a bar is the frequency.

Method. For each bar, frequency = frequency density × class width.

  1. Read the heights and widths. The heights are the frequency densities. The widths come from the class boundaries.
    0 to 10: height 0.8, width 10
    10 to 20: height 1.4, width 10
    20 to 40: height 1.5, width 20
    40 to 50: height 1.6, width 10
    50 to 70: height 0.6, width 20
  2. Multiply height by width. The area of each bar is its frequency.
    0.8 × 10 = 8
    1.4 × 10 = 14
    1.5 × 20 = 30
    1.6 × 10 = 16
    0.6 × 20 = 12
  3. Add the frequencies. This is the total number of people.
    8 + 14 + 30 + 16 + 12 = 80

Check. The total of 80 matches the table above. ✓

Watch out. Do not read the heights as frequencies.

Worked example 24.15.3
Use the histogram above to estimate the number of people who spent between 30 and 40 minutes reading.

Key words. A histogram has bars of different widths. The height is the frequency density = frequency ÷ class width, so the area of a bar is the frequency.

Method. The class 20 to 40 has frequency 30. We need the part of the bar from 30 to 40, which is half of the width of the bar.

  1. Find the frequency density of the bar. The bar 20 to 40 has frequency density 30 ÷ 20.
    30 ÷ 20 = 1.5
  2. Find the width of the part wanted. From 30 to 40.
    40 − 30 = 10
  3. Multiply to find its area. Area = height × width = frequency.
    1.5 × 10 = 15 people

Check. The part from 30 to 40 is half of the class 20 to 40, so half of 30 = 15. ✓

Watch out. This assumes that the values are spread evenly through the class.

Worked example 24.15.4
Estimate the mean time spent reading, using the histogram above.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.

  1. Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.
    (0 + 10) ÷ 2 = 5
    (10 + 20) ÷ 2 = 15
    (20 + 40) ÷ 2 = 30
    (40 + 50) ÷ 2 = 45
    (50 + 70) ÷ 2 = 60
  2. Multiply each frequency by its midpoint. This estimates the total of the values in that class.
    TimeFrequency (f)Midpointf × midpoint
    0 < t ≤ 10 min8540
    10 < t ≤ 20 min1415210
    20 < t ≤ 40 min3030900
    40 < t ≤ 50 min1645720
    50 < t ≤ 70 min1260720
    Total802590
  3. Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.
    Total frequency = 80
    Total of f × midpoint = 2590
  4. Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.
    2590 ÷ 80 = 32.3750… = 32.4 min to 1 d.p.

Check. The estimate should be inside the range of the classes, from 0 to 70, and 32.4 is. It is only an estimate because the real values are not known. ✓

Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.

Worked example 24.15.5
Estimate the median time spent reading.

Key words. A histogram has bars of different widths. The height is the frequency density = frequency ÷ class width, so the area of a bar is the frequency.

Method. Estimate the median by finding which class holds the middle value, then assuming the values are spread evenly across that class.

  1. Work out the frequencies and the total. Frequency = frequency density × class width for each bar.
    Frequencies: 8, 14, 30, 16, 12; total n = 80
  2. Find the position of the median. It is half way through the data.
    80 ÷ 2 = 40
  3. Use a running total to find the median class. The running total first reaches 40 in the class 20 < x ≤ 40 min.
    ClassFrequencyRunning total
    0 < x ≤ 10 min88
    10 < x ≤ 20 min1422
    20 < x ≤ 40 min3052
    40 < x ≤ 50 min1668
    50 < x ≤ 70 min1280
  4. Find how far into the class we need to go. Before this class there are 22 values, so we need 18 more from this class of 30.
    18 ÷ 30 = 0.6 of the class
  5. Find the estimate of the median. Start at the lower end and move that fraction of the class width.
    20 + 0.6 × 20 = 32 min

Check. The estimate lies inside the median class. ✓

Watch out. This is an estimate: it assumes the values are spread evenly within the class.

Watch out

The vertical axis of a histogram is frequency density, not frequency. If the bars have different widths, the tallest bar is not necessarily the class with the most values.

Exercise 24O  Histograms

Calculator allowed
1
FThe table shows the ages of the members of a club.
AgeFrequency
0 < a ≤ 20 years20
20 < a ≤ 30 years35
30 < a ≤ 40 years50
40 < a ≤ 60 years40
60 < a ≤ 100 years16
Work out the frequency density for each class.
2
FUse the table in question 1 to draw a histogram.
3
FThe histogram shows the waiting times at a clinic. (a) Work out the number of patients who waited between 10 and 20 minutes. (b) Work out the number of patients who waited between 0 and 5 minutes. (c) Work out the total number of patients.
01234560510203050Waiting time (minutes)Frequency density
(a)Waited 10 to 20 minutes
(b)Waited 0 to 5 minutes
(c)Total patients
4
PSUse the histogram in question 3 to estimate the number of patients who waited (a) between 24 and 30 minutes, (b) more than 40 minutes, (c) between 15 and 20 minutes.
01234560510203050Waiting time (minutes)Frequency density
(a)Between 24 and 30 minutes
(b)More than 40 minutes
(c)Between 15 and 20 minutes
5
FThe table shows the lengths of 120 phone calls.
LengthFrequency
0 < t ≤ 10 min12
10 < t ≤ 20 min28
20 < t ≤ 40 min50
40 < t ≤ 60 min30
Estimate the mean length of a call. Give your answer to 1 decimal place.
6
FEstimate the median length of a call using the table in question 5.
7
PSIn a histogram, the bar for 20 < x ≤ 30 is 3 cm wide and 4 cm tall, and represents 24 people. The bar for 30 < x ≤ 50 is 6 cm wide and 2 cm tall. How many people does the second bar represent?
8
PSThe bar for 0 < t ≤ 10 in a histogram has frequency density 4 and frequency 40. The next bar is for 10 < t ≤ 30 and has frequency 100. Work out the frequency density of the second bar and say how its height compares with the first bar.
9
RJon says: “The tallest bar in a histogram is always the class with the most values.” Explain why he is wrong, using an example.
10
PSThe histogram bars for four classes have the widths and frequencies shown. Complete the table of frequency densities and work out the probability that a person chosen at random from the data is in the class 25 to 30. Give your answer as a fraction in its simplest form.
ClassFrequency
0 < x ≤ 1015
10 < x ≤ 2545
25 < x ≤ 3010
30 < x ≤ 5050
(a)Frequency densities
(b)Probability
11
RA histogram shows the masses of 150 parcels. The class 2 < m ≤ 4 has frequency density 15, the class 4 < m ≤ 10 has frequency density 8 and the class 10 < m ≤ 20 has frequency density 4.6. Work out the number of parcels in the class 0 < m ≤ 2, and the mean mass to 1 decimal place if that class has midpoint 1.
(a)Parcels in the class 0 to 2
(b)Estimated mean
12
RThe histogram shows the lengths of 100 films, in minutes. Estimate the percentage of films that are longer than 30 minutes.
01234010204060Length (minutes)Frequency density

Chapter summary

SkillMethod
Types of dataCategorical (words), discrete (counted), continuous (measured)
SamplingRandom, unbiased sample; stratified: group ÷ population × sample size
Capture-recaptureN = marked first × second sample ÷ marked in second sample
Mean, median, mode, rangeTotal ÷ number; middle value in order; most common; biggest − smallest
Frequency tableMean: total of (f × value) ÷ total f; median from running total
Grouped dataMidpoints for the estimated mean; modal class; median class
ChartsBar chart, pictogram (use the key), pie chart (angle = f ÷ total × 360°), line graph, stem-and-leaf (with key)
Scatter graphsCorrelation, line of best fit through the mean point; interpolation is more reliable than extrapolation
ComparingCompare an average and a spread, in context, with numbers
Quartiles (H)Q1 and Q3 are the medians of the two halves; IQR = Q3 − Q1
Cumulative frequency (H)Plot at the upper boundary; read n ÷ 4, n ÷ 2, 3n ÷ 4
Box plots (H)Five numbers; box = middle half; outliers as crosses
Histograms (H)Frequency density = frequency ÷ class width; area = frequency

Can you…?

  • I can say whether data is categorical, discrete or continuous, and design a fair question.
  • I can complete tally charts, frequency tables, grouped frequency tables and two-way tables.
  • I can spot bias, take a stratified sample and use capture–recapture.
  • I can find the mean, median, mode and range from lists, frequency tables and grouped tables.
  • I can draw and read bar charts, pictograms, pie charts, line graphs and stem-and-leaf diagrams.
  • I can describe correlation, draw a line of best fit and use it to estimate, knowing the dangers of extrapolating.
  • I can compare two distributions with an average and a spread.
  • Higher: I can find quartiles and the IQR, and draw and use cumulative frequency graphs and box plots.
  • Higher: I can draw and read histograms using frequency density.

Key terms

Population
The whole group being studied.
Sample
A smaller group taken from the population.
Bias
A sample or question that unfairly favours some answers.
Stratified sample
A sample with the same proportions from each group as the population has.
Mean
The total of the values divided by how many there are.
Median
The middle value when the data is in order.
Mode
The most common value.
Range
The biggest value minus the smallest value.
Frequency
How many times a value occurs.
Modal class
The class with the highest frequency.
Correlation
How two variables are related: positive, negative or none.
Line of best fit
A straight line that follows the trend of a scatter graph.
Interpolation
Estimating inside the range of the data.
Extrapolation
Estimating outside the range of the data (less reliable).
Outlier
A value far from the rest of the data.
Quartiles
Values that split ordered data into four equal parts.
Interquartile range
Upper quartile minus lower quartile.
Cumulative frequency
A running total of the frequencies.
Frequency density
Frequency divided by class width.

Exercise 24Z  Check-up (non-calculator)

Non-calculator
1
Is the number of cars in a car park categorical, discrete or continuous data?
2
Work out the mean of 8, 5, 9, 12 and 6.
3
Work out the median of 11, 4, 9, 2, 7 and 15.
4
Here are some numbers: 3, 5, 3, 8, 5, 3. Write down (a) the mode, (b) the range.
(a)Mode
(b)Range
5
The table shows the number of pets owned by 10 students.
Pets1234
Students2521
Work out the mean number of pets.
6
A pie chart is drawn for 30 students who each chose one sport: football 12, netball 8, rugby 6, tennis 4. Work out the angle for each sport.
7
A school has 90 boys and 60 girls. A stratified sample of 30 students is taken. How many boys and how many girls are in the sample?
8
25 deer are caught, marked and released. Later 40 deer are caught and 5 are marked. Estimate the number of deer in the forest.
9
The scatter graph shows the height of a candle and the time it has been burning. As the time increases, the height decreases and the points are close to a straight line. Describe the correlation.
10
The table shows the ages of 20 people.
AgeFrequency
0 < a ≤ 10 years6
10 < a ≤ 20 years10
20 < a ≤ 30 years4
Work out an estimate for the mean age.

Total 62 marks. Allow about 62 minutes.

Exercise 24X  Exam-style questions

Non-calculator
1
FThe ages of six people are 24, 31, 19, 27, 35 and 22. Work out the mean age.[2 marks]
2
FThe two-way table shows how 65 students travel to school. Complete the table.
WalkBusCarTotal
Year 101812   ?   35
Year 11   ?   9730
Total32   ?   1265
[3 marks]
(a)Year 10, car
(b)Year 11, walk
(c)Total, bus
3
FForty people chose a favourite drink: tea 14, coffee 10, juice 12, water 4. Draw an accurate pie chart. (Show the angle for each drink.)[4 marks]
4
FThe table shows the number of goals scored by a team in 20 matches.
Goals01234
Matches47621
Work out the mean number of goals.[3 marks]
5
FIn a lake, 60 fish are caught, marked and returned. The next day 50 fish are caught and 10 of them are marked. (a) Estimate the number of fish in the lake. (b) State one assumption you made.[4 marks]
(a)Estimate
(b)Assumption
6
FA college has 240 students in Year 12 and 160 students in Year 13. A stratified sample of 50 students is taken. Work out the number of students from each year group in the sample.[3 marks]
7
FThe scatter graph shows the revision hours and the test marks of 12 students, with a line of best fit. (a) Describe the correlation. (b) Use the line of best fit to estimate the test mark of a student who revises for 6 hours. (c) Explain why the line should not be used to estimate the mark for 14 hours.[5 marks]
0102030405060708090100012345678910Hours of revisionTest mark
(a)Correlation
(b)Estimate for 6 hours
(c)Why not 14 hours
8
FThe table shows the heights of 40 students.
HeightFrequency
140 < h ≤ 150 cm6
150 < h ≤ 160 cm14
160 < h ≤ 170 cm15
170 < h ≤ 180 cm5
(a) Write down the modal class. (b) Work out an estimate for the mean height.[5 marks]
(a)Modal class
(b)Estimated mean
9
FThe numbers of text messages sent by two students in six days are: Hana: 12, 15, 13, 18, 16, 14. Zoe: 20, 11, 25, 9, 22, 23. Compare the number of messages sent using the mean and the range.[4 marks]
10
FThe marks of 15 students are: 34, 41, 28, 35, 47, 52, 38, 41, 29, 44, 33, 41, 56, 38, 49. Draw an ordered stem-and-leaf diagram and include a key.[3 marks]
11
FThe line graph shows the ice cream sales (hundreds) in each quarter of three years. (a) What were the sales in Q2 of Year 3? (b) Describe the seasonal pattern. (c) Estimate the sales in Q3 of Year 4, and say why your estimate may be wrong.[4 marks]
020406080QuarterSales (hundreds)149211712258191611
(a)Q2 of Year 3
(b)Seasonal pattern
(c)Estimate for Q3 of Year 4
12
FTom wants to find out how often people in his town go to the cinema. He asks the first 20 people who leave the cinema one evening. (a) Give one reason why his sample may be biased. (b) Suggest a better way to choose a sample.[2 marks]
(a)Reason for bias
(b)Better method
13
FThe cumulative frequency graph shows the times taken by 80 workers. (a) Estimate the median. (b) Estimate the interquartile range. (c) Estimate the number of workers who took more than 45 minutes.[5 marks]
0204060800204060Time taken (minutes)Cumulative frequency
(a)Median
(b)Interquartile range
(c)More than 45 minutes
14
FThe box plots show the times, in minutes, taken by two groups to solve a puzzle.
Group A: smallest 12, Q1 20, median 26, Q3 38, largest 55.
Group B: smallest 15, Q1 22, median 30, Q3 34, largest 48.
Compare the two groups.[4 marks]
102030405060Time (minutes)Group AGroup B
15
FThe table shows the times that 80 people spent reading.
TimeFrequency
0 < t ≤ 10 min8
10 < t ≤ 20 min14
20 < t ≤ 40 min30
40 < t ≤ 50 min16
50 < t ≤ 70 min12
(a) Work out the frequency density for each class. (b) Estimate the number of people who read for between 30 and 40 minutes. (c) Estimate the median time.[7 marks]
(a)Frequency densities
(b)Between 30 and 40 minutes
(c)Median
16
FHere are eight numbers: 12, 14, 15, 17, 18, 20, 22, 60. (a) Work out the interquartile range. (b) Use the rule “more than 1.5 × IQR from a quartile” to show that 60 is an outlier.[4 marks]
(a)IQR
(b)Outlier
← Chapter 23: ProbabilityChapter 25: Functions and advanced graphs →

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