A Pie Chart Drawn from a Travel Survey, and a Sector Read Back to a Count
A school asked its 240 Secondary 2 students how they travel to school. 96 take the bus, 72 take the train, 48 walk and 24 come by car. The results are to be drawn as a pie chart. (a) Find the angle of each sector. (b) A second school has 450 students and draws the same kind of pie chart. Its sector for the students who walk has an angle of 56°. How many students of the second school walk?
A pie chart shares the 360° of a circle among all the people in the survey, so every person has the same small angle. To draw a sector, multiply that angle by the number of people in the category. To read a sector back, the angle over 360° is the fraction of the total that the category holds.
- Find the angle for one student. The 240 students share 360° equally, so each student has 360 ÷ 240 = 1.5°.
- Multiply each count by 1.5°. Bus: 96 × 1.5 = 144°. Train: 72 × 1.5 = 108°. Walking: 48 × 1.5 = 72°. Car: 24 × 1.5 = 36°.
- (a) The angles are 144°, 108°, 72° and 36°. Check: 144 + 108 + 72 + 36 = 360, so the four sectors fill the circle exactly.
- For the second school, write the sector as a fraction of the circle: 56360 = 745. The students who walk are the same fraction of the 450 students.
- (b) 745 × 450 = 70 students walk. Check: in the second school each student has 360 ÷ 450 = 0.8°, and 70 × 0.8 = 56°.
answer(a) bus 144°, train 108°, walking 72°, car 36°, which add up to 360°; (b) 70 students
techniquePie Charts
examsPSLE · O-Level · SAT · GCSE Higher
Common pitfalls
- Using the counts as the angles, so that the bus sector is drawn as 96°. The four counts add up to 240, not 360, so the sectors would not fill the circle. Each count has to be scaled by 360240 first.
- Reading the 56° sector with the first school's 1.5° for each student. The angle for one student depends on the size of the survey. The second school has 450 students, so its sectors are fractions of 450.
Sprint Times Put into an Ordered Stem-and-Leaf Diagram: The Median and the Range
The times, in seconds, of the 15 runners in a 100 m heat were 13.2, 12.4, 14.1, 12.9, 13.5, 15.0, 13.8, 12.7, 14.6, 13.2, 14.3, 13.9, 12.5, 14.8 and 13.6. (a) Draw an ordered stem-and-leaf diagram of the times and use it to find the median time. (b) Find the range of the times.
In a stem-and-leaf diagram each value is split into a stem and a leaf. Here the stem is the whole number of seconds and the leaf is the tenths digit. When the leaves in every row are in order, the whole list is in order, so the median can be found by counting along the rows and the range can be read from the first and the last leaf.
- Use the whole seconds 12, 13, 14 and 15 as the stems. Go through the list once and write the tenths digit of each time beside its stem: 13.2 puts a 2 in the row of 13, and 12.4 puts a 4 in the row of 12.
- Put the leaves of each row in order and write the key, 12 | 4 means 12.4 s. The rows are 12 | 4 5 7 9, 13 | 2 2 5 6 8 9, 14 | 1 3 6 8 and 15 | 0. Check: 4 + 6 + 4 + 1 = 15 leaves, one for each runner.
- With 15 values the median is the 15 + 12 = 8th value. The first row holds 4 values, so the 8th value is the 4th leaf in the row of 13, which is 6.
- (a) The median time is 13.6 s. Seven runners were faster than this and seven were slower.
- (b) The range is the slowest time minus the fastest time, read from the last leaf and the first leaf: 15.0 − 12.4 = 2.6 s.
answer(a) the median time is 13.6 s; (b) the range is 2.6 s
techniqueStem and Leaf
examsO-Level · SAT · GCSE Higher
Common pitfalls
- Finding the median from the leaves in the order they were first written. In the row of 13 the leaves were first written as 2, 5, 8, 2, 9, 6, and counting along them gives a different value. The median is the middle value of the ordered list, so the leaves must be ordered first.
- Giving the range as 15 − 12 = 3 from the stems alone. The stems are only the whole seconds. The fastest time is 12.4 s, not 12 s, so the leaf has to be read with its stem.
Museum Visitors in a Two-Way Table with Five Missing Cells, and a Proportion Within a Row
A museum sorted the 150 visitors of one morning by where they live, local or tourist, and by the guide they took: an audio guide, a paper guide or no guide. 90 of the visitors were tourists. Of the local visitors, 18 took an audio guide and 30 took no guide. 24 tourists took a paper guide. In total, 63 visitors took an audio guide. (a) Complete the two-way table and find how many tourists took no guide. (b) What percentage of the local visitors took an audio guide? Compare it with the percentage of the tourists who did.
In a two-way table every row adds up to its row total and every column adds up to its column total. A row or a column with only one empty cell can be filled by subtraction, and each new number may leave another row or column with only one gap. A proportion of the local visitors is taken out of the total of the Local row, because only that row is being described.
- Begin with the Total column, which has one gap. The local visitors are 150 − 90 = 60.
- The Local row now has one gap: 60 − 18 − 30 = 12 local visitors took a paper guide. The Audio column has one gap: 63 − 18 = 45 tourists took an audio guide.
- The Tourist row now has one gap: 90 − 45 − 24 = 21. The column totals follow: 12 + 24 = 36 paper guides, and 30 + 21 = 51 visitors with no guide.
- (a) 21 tourists took no guide. Check with the Total row: 63 + 36 + 51 = 150.
- For the local visitors, the whole is the Local row total, 60. 1860 = 30% of the local visitors took an audio guide. For the tourists, the whole is 90, and 4590 = 50% took one.
- (b) 30% of the local visitors took an audio guide, compared with 50% of the tourists, so a tourist was more likely to take one.
answer(a) 21 tourists took no guide; (b) 30% of the local visitors, compared with 50% of the tourists
techniqueTwo-Way Tables
examsSAT · GCSE Higher
Common pitfalls
- Dividing the 18 local visitors by 150, which gives 12%. That is the percentage of all the visitors who were local and took an audio guide. The question asks about the local visitors only, so the whole is 60.
- Trying to fill the Tourist row first. That row starts with two empty cells, so it cannot be found by one subtraction. Fill the rows and columns that have one gap, and the Tourist row is left with one gap after the Audio column is done.
A Frequency Tree for a Screening Test, and the Positive Results That Are Correct
1000 people take a screening test for a condition. 40 of them have the condition, and 36 of those 40 test positive. Of the people who do not have the condition, 48 test positive. (a) Complete a frequency tree and find how many people test negative. (b) A person tests positive. What fraction of the people who test positive have the condition?
A frequency tree splits a group in stages and writes a count on every branch. The two counts that leave a point always add up to the count that arrives at it, so each missing count is found by subtraction. For the last part, the group in the question is the people who test positive, so they are the whole of the fraction.
- The first pair of branches splits the 1000 people into those who have the condition and those who do not: 1000 − 40 = 960 people do not have it.
- The second pairs split each group by the test result. Of the 40 people with the condition, 40 − 36 = 4 test negative. Of the 960 people without it, 960 − 48 = 912 test negative.
- (a) 4 + 912 = 916 people test negative. Check: 36 + 48 = 84 people test positive, and 84 + 916 = 1000.
- The people who test positive come from two branches, 36 who have the condition and 48 who do not, which is 84 people in all.
- (b) 3684 = 37 of the people who test positive have the condition. That is less than half, because the 48 positive results come from the very large group of people who do not have it.
answer(a) 916 people test negative; (b) 37 of the people who test positive have the condition
techniqueFrequency Trees
examsGCSE Higher
Common pitfalls
- Answering part (b) with 3640. That is the fraction of the people with the condition who test positive. The question starts from a positive result, so the whole is the 84 people who test positive.
- Adding only the two negative counts that are easy to see and forgetting to check. The four end branches must add up to the 1000 people at the start: 36 + 4 + 48 + 912 = 1000. A tree whose ends do not add up has a wrong subtraction in it.
Journey Times in a Histogram with Unequal Class Widths
A company drew a histogram of the journey times to work, in minutes, of its employees. The vertical axis shows frequency density, which is the frequency divided by the class width. The bar from 0 to 10 minutes has height 0.6, the bar from 10 to 20 has height 1.4, the bar from 20 to 40 has height 1.8, the bar from 40 to 60 has height 1.2 and the bar from 60 to 100 has height 0.5. (a) How many employees are shown in the histogram? (b) Estimate the number of employees whose journey takes more than 50 minutes.
When the classes have different widths, the height of a bar is the frequency density, and the frequency is the area of the bar: frequency density times class width. A wide, low bar can hold more people than a narrow, tall one. Part of a class is estimated by taking the same part of the bar's area.
- Write the width of each class from its two ends. The widths are 10, 10, 20, 20 and 40 minutes.
- Multiply each frequency density by its class width: 0.6 × 10 = 6, 1.4 × 10 = 14, 1.8 × 20 = 36, 1.2 × 20 = 24 and 0.5 × 40 = 20.
- (a) 6 + 14 + 36 + 24 + 20 = 100 employees.
- A journey of more than 50 minutes is in the part of the 40 to 60 bar that lies between 50 and 60, or anywhere in the 60 to 100 bar. The part bar has area 1.2 × 10 = 12. The whole last bar is 20.
- (b) About 12 + 20 = 32 employees. It is an estimate, because it takes the 24 employees of the 40 to 60 class to be spread evenly across the class.
answer(a) 100 employees; (b) about 32 employees
techniqueFrequency Density
examsO-Level · SAT · GCSE Higher
Common pitfalls
- Reading the heights as frequencies and saying that the 20 to 40 class has 1.8 employees, or that the 60 to 100 class is the smallest because its bar is the lowest. The last bar is four times as wide as the first, and it holds 20 employees, more than the first two classes together.
- Taking half of the last bar as well as half of the 40 to 60 bar. Only the class that contains 50 minutes is cut. Every journey in the 60 to 100 class is longer than 50 minutes, so the whole of that bar counts.
A Histogram of Fish Lengths Turned Back into a Grouped Frequency Table
A fisheries officer measured the length, x cm, of every fish in one catch and drew a histogram with equal class widths. The bars stand over 10 ≤ x < 20, 20 ≤ x < 30, 30 ≤ x < 40, 40 ≤ x < 50 and 50 ≤ x < 60, and their heights on the frequency axis are 4, 9, 14, 8 and 5. (a) Write the grouped frequency table and state the modal class. (b) A fish shorter than 30 cm must be put back. What percentage of the catch must be put back?
When every class has the same width, the vertical axis of a histogram can show the frequency itself, so each bar is read like a count. The grouped frequency table lists each class with its frequency. The modal class is the class with the greatest frequency, and it is given as an interval, because the histogram does not show the single lengths inside a class.
- Every class is 10 cm wide, so the height of each bar is its frequency. Read the five heights from the frequency axis: 4, 9, 14, 8 and 5.
- Write each class beside its frequency to make the grouped frequency table, and add the frequencies: 4 + 9 + 14 + 8 + 5 = 40 fish.
- (a) The frequencies are 4, 9, 14, 8 and 5. The tallest bar is over 30 ≤ x < 40, so that is the modal class.
- A fish shorter than 30 cm is in one of the first two classes, because 30 is the end of the second class: 4 + 9 = 13 fish.
- (b) 1340 = 0.325, so 32.5% of the catch must be put back.
answer(a) frequencies 4, 9, 14, 8, 5; the modal class is 30 ≤ x < 40; (b) 13 of the 40 fish, which is 32.5%
techniqueHistograms · Tally and Frequency Tables
examsO-Level · SAT · GCSE Higher
Common pitfalls
- Giving the modal class as 14. The number 14 is the frequency of the modal class. The modal class is the interval of lengths that has that frequency, 30 ≤ x < 40.
- Counting the 30 ≤ x < 40 class among the fish to put back. A fish of exactly 30 cm belongs to that class, and every fish in it is at least 30 cm long, so none of the 14 is shorter than 30 cm.
A Dot Diagram of Goals with One Outlier, and What It Does to the Mean
A dot diagram shows the goals scored by a hockey team in each of its 10 matches. There is 1 dot at 0, there are 2 dots at 1, 3 dots at 2, 2 dots at 3, 1 dot at 4 and 1 dot at 12. (a) Find the mean number of goals in a match. (b) The match with 12 goals was against a team that had only 7 players. Leave that match out, find the new mean, and say whether the mean or the median describes a usual match better.
In a dot diagram every dot is one value, so a column of dots is a value repeated. An outlier is a value far from all the others. The mean uses the size of every value, so one outlier pulls it toward itself. The median uses only the order of the values, so an outlier at the end hardly moves it.
- Find the total number of goals by multiplying each value by its number of dots: 0 × 1 + 1 × 2 + 2 × 3 + 3 × 2 + 4 × 1 + 12 × 1 = 0 + 2 + 6 + 6 + 4 + 12 = 30 goals.
- (a) The mean is 30 ÷ 10 = 3 goals in a match. Only 2 of the 10 matches had more than 3 goals, so the mean is higher than most of the dots.
- The dot at 12 is an outlier: it is far to the right of all the other dots. Without it there are 30 − 12 = 18 goals in 9 matches, and the mean is 18 ÷ 9 = 2 goals.
- Find the median both ways. With all 10 matches, the 5th and 6th values are both 2, so the median is 2. Without the outlier there are 9 values and the 5th is 2, so the median is still 2.
- (b) Without the outlier the mean falls from 3 goals to 2 goals, while the median stays at 2 goals. The median describes a usual match better, because one unusual match does not change it.
answer(a) the mean is 3 goals; (b) the mean falls to 2 goals and the median stays at 2 goals, so the median describes a usual match better
techniqueDot Diagrams
examsO-Level · SAT · GCSE Higher
Common pitfalls
- Adding the six values on the axis that have dots, 0 + 1 + 2 + 3 + 4 + 12 = 22, and dividing by 6. Each dot is a match, so a value with 3 dots must be counted 3 times, and the division is by the 10 matches.
- Dividing the 18 goals by 10 after the outlier is removed. Leaving out the match removes one value as well as its 12 goals, so there are only 9 matches to divide by.
A Vertical Line Chart of People in Each Car: The Mode and the Total
A student counted the people in each car that entered a car park and drew a vertical line chart. The line at 1 person has height 15, the line at 2 people has height 12, the line at 3 has height 7, the line at 4 has height 4 and the line at 5 has height 2. (a) State the mode, and find how many cars were counted. (b) Find the total number of people in the cars, and the mean number of people in a car.
A vertical line chart is used for a discrete count, such as the number of people in a car. Each line stands at one value, and its height is the frequency of that value. The heights add up to the number of cars. The number of people is found by multiplying each value by its frequency, because a line of height 12 at 2 stands for 12 cars with 2 people in each.
- Read the heights of the lines: 15, 12, 7, 4 and 2 cars. The tallest line stands at 1 person, so the mode is 1 person. The mode is the value under the tallest line, not the height 15.
- (a) The mode is 1 person in a car, and 15 + 12 + 7 + 4 + 2 = 40 cars were counted.
- Multiply each value by its frequency to count the people: 1 × 15 + 2 × 12 + 3 × 7 + 4 × 4 + 5 × 2 = 15 + 24 + 21 + 16 + 10 = 86.
- (b) There were 86 people, and the mean is 86 ÷ 40 = 2.15 people in a car. The mean need not be a whole number, although every single car holds a whole number of people.
answer(a) the mode is 1 person, and 40 cars were counted; (b) 86 people, a mean of 2.15 people in a car
techniqueVertical Line Charts
examsGCSE Higher
Common pitfalls
- Giving the mode as 15. The height 15 says how many cars had 1 person. The mode is the value that occurs most often, which is 1 person.
- Adding the heights to find the number of people. 15 + 12 + 7 + 4 + 2 = 40 counts the cars. A car at the line for 3 carries 3 people, so each height must be multiplied by its value before adding.
An Advertisement's Bar Chart Whose Axis Starts at 470
An advertisement shows a bar chart of the bottles of a drink sold, in thousands: 480 in 2024 and 504 in 2025. The vertical axis of the chart starts at 470, not at 0. (a) How many times as tall as the 2024 bar is the 2025 bar on this chart? (b) Find the true percentage increase in sales, and describe how the two bars compare when the axis starts at 0.
The height of a bar is measured from the bottom of the axis. When the axis starts at 0, the heights are in the same ratio as the values. When the axis starts at another number, each bar shows only the part of its value above that number, so a small difference between two large values can look very large. The honest comparison is the percentage change from the first value.
- On the advertisement each bar starts at 470. The 2024 bar has height 480 − 470 = 10 and the 2025 bar has height 504 − 470 = 34.
- (a) 34 ÷ 10 = 3.4, so the 2025 bar is drawn 3.4 times as tall as the 2024 bar. Sales did not become 3.4 times as large.
- Use the full values for the true change. The increase is 504 − 480 = 24 thousand bottles, and it is compared with the 2024 value, 480 thousand.
- (b) 24480 = 0.05, so sales rose by 5%. On an axis that starts at 0 the bars have heights 480 and 504, so the 2025 bar is only 1.05 times as tall and the two bars look almost the same.
answer(a) 3.4 times as tall; (b) a true increase of 5%, and from 0 the two bars are almost the same height
techniqueMisleading Graphs · Bar Charts
examsO-Level
Common pitfalls
- Saying that sales more than tripled because one bar is more than three times as tall as the other. The bars on this chart show only the amounts above 470. The values themselves are 480 and 504, which differ by 5%.
- Dividing the increase by the new value, 24504. A percentage increase is measured from where the change started, so the 24 thousand is divided by the 2024 value, 480 thousand.
Three Sets of Fitness Data and the Display That Suits Each
A PE teacher has three sets of data about one class: each student's favorite sport, the number of pull-ups each student can do, and each student's time for a 2.4 km run. (a) Choose a suitable display for each set of data, and give the reason from the type of data. (b) The run times, in minutes, of 12 students are 11.5, 12.0, 13.2, 13.9, 14.0, 14.6, 12.8, 15.1, 16.4, 13.5, 14.0 and 17.2. They are grouped into the classes 10 ≤ x < 12, 12 ≤ x < 14, 14 ≤ x < 16 and 16 ≤ x < 18 for a histogram. How many students are in the class 12 ≤ x < 14, and in which class does a time of 14.0 minutes belong?
The type of data decides the display. Categorical data are names with no number scale. Discrete data are counts, which take separate values. Continuous data are measurements, which can take any value in an interval. A display is suitable when its axis behaves like the data: separate bars for separate categories, a line or a column of dots at each whole number for a count, and bars that touch for a measurement.
- A favorite sport is a name, so the data are categorical. A bar chart suits them: there is one bar for each sport, and the gaps between the bars show that the categories are separate. A pie chart would also do.
- A number of pull-ups is a count, 0, 1, 2 and so on, so the data are discrete. A dot diagram or a vertical line chart suits them, because each whole number has its own column of dots or its own line, and nothing is drawn between the whole numbers.
- (a) A run time is a measurement and can be any value, such as 13.27 minutes, so the data are continuous. A histogram suits them: the times are grouped into classes, and the bars touch because each class starts where the class before it ends.
- Tally each time into its class. The class 12 ≤ x < 14 includes 12.0 but does not include 14.0, so a time of 14.0 goes into 14 ≤ x < 16. The tallies are 1, 5, 4 and 2. Check: 1 + 5 + 4 + 2 = 12 students.
- (b) 5 students are in the class 12 ≤ x < 14: their times are 12.0, 13.2, 13.9, 12.8 and 13.5. Each time of 14.0 minutes belongs to the class 14 ≤ x < 16.
answer(a) a bar chart for the sports, which are categorical; a dot diagram or a vertical line chart for the pull-ups, which are discrete; a histogram for the run times, which are continuous; (b) 5 students, and 14.0 belongs to 14 ≤ x < 16
techniqueChoosing a Chart · Categorical and Numerical Data · Tally and Frequency Tables
examsO-Level
Common pitfalls
- Drawing a histogram for the favorite sports. A histogram needs a number scale along its horizontal axis, and the sports have no order and no scale. Separate bars in a bar chart are correct for categories.
- Counting a time of 14.0 minutes in both classes, or in 12 ≤ x < 14. The sign < at the upper end means that 14 is left out of that class, and ≤ at the lower end of the next class means that it is included there. Every value belongs to exactly one class.
A Dual Bar Chart of Library Loans in Two Years, Compared as Percentage Changes
A dual bar chart shows the loans from a school library in 2023 and in 2024. Fiction went from 500 loans to 560, non-fiction from 200 to 250, and magazines from 100 to 110. (a) Which kind of loan rose by the greatest number, and which rose by the greatest percentage? (b) Find the percentage increase in the total number of loans.
A dual bar chart puts two bars side by side for each category, so that the two years can be compared within a category. The difference in height is the increase in number. The percentage increase divides that difference by the height of the first bar, so a short bar can have a small increase in number and a large percentage increase.
- Read each pair of bars and subtract. Fiction rose by 560 − 500 = 60 loans, non-fiction by 250 − 200 = 50 loans and magazines by 110 − 100 = 10 loans.
- Divide each increase by the 2023 value of its own category: fiction 60500 = 12%, non-fiction 50200 = 25%, magazines 10100 = 10%.
- (a) Fiction rose by the greatest number, 60 loans. Non-fiction rose by the greatest percentage, 25%, because its increase is measured against a much shorter 2023 bar.
- Add the bars of each year. In 2023 there were 500 + 200 + 100 = 800 loans, and in 2024 there were 560 + 250 + 110 = 920 loans. The increase is 920 − 800 = 120 loans.
- (b) 120800 = 15%. The total rose by 15%. Check: 800 × 1.15 = 920.
answer(a) fiction rose by the greatest number, 60 loans, and non-fiction by the greatest percentage, 25%; (b) the total rose by 15%
techniqueBar Charts
examsPSLE · O-Level · SAT · GCSE Higher
Common pitfalls
- Finding the change in the total by taking the mean of the three percentages, (12 + 25 + 10) ÷ 3, which is about 15.7%. The three categories are of different sizes, so their percentages do not count equally. The totals, 800 and 920, must be used.
- Choosing fiction for the greatest percentage increase because its two bars differ most in height. The difference in height is the increase in number. A percentage compares that difference with the first bar, and 60 out of 500 is a smaller part than 50 out of 200.