GCSE Maths: Histograms (Higher)
Part of the GCSE Maths statistics chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Histograms
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.
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.
| Time | Frequency |
|---|---|
| 0 < t ≤ 10 min | 8 |
| 10 < t ≤ 20 min | 14 |
| 20 < t ≤ 40 min | 30 |
| 40 < t ≤ 50 min | 16 |
| 50 < t ≤ 70 min | 12 |
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.
- 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 - 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 - Write the results in a table.
Time Frequency Class width Frequency density 0 < t ≤ 10 min 8 10 0.8 10 < t ≤ 20 min 14 10 1.4 20 < t ≤ 40 min 30 20 1.5 40 < t ≤ 50 min 16 10 1.6 50 < t ≤ 70 min 12 20 0.6 - 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.
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.
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.
- 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 - 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 - 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.
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.
- Find the frequency density of the bar. The bar 20 to 40 has frequency density 30 ÷ 20.30 ÷ 20 = 1.5
- Find the width of the part wanted. From 30 to 40.40 − 30 = 10
- 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.
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.
- 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 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Time Frequency (f) Midpoint f × midpoint 0 < t ≤ 10 min 8 5 40 10 < t ≤ 20 min 14 15 210 20 < t ≤ 40 min 30 30 900 40 < t ≤ 50 min 16 45 720 50 < t ≤ 70 min 12 60 720 Total 80 2590 - 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 - 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.
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.
- Work out the frequencies and the total. Frequency = frequency density × class width for each bar.Frequencies: 8, 14, 30, 16, 12; total n = 80
- Find the position of the median. It is half way through the data.80 ÷ 2 = 40
- Use a running total to find the median class. The running total first reaches 40 in the class 20 < x ≤ 40 min.
Class Frequency Running total 0 < x ≤ 10 min 8 8 10 < x ≤ 20 min 14 22 20 < x ≤ 40 min 30 52 40 < x ≤ 50 min 16 68 50 < x ≤ 70 min 12 80 - 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
- 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| Age | Frequency |
|---|---|
| 0 < a ≤ 20 years | 20 |
| 20 < a ≤ 30 years | 35 |
| 30 < a ≤ 40 years | 50 |
| 40 < a ≤ 60 years | 40 |
| 60 < a ≤ 100 years | 16 |
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.
- Find the width of each class. The bars have different widths, so the height cannot be the frequency. Width = upper end − lower end.20 − 0 = 20
30 − 20 = 10
40 − 30 = 10
60 − 40 = 20
100 − 60 = 40 - 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.20 ÷ 20 = 1
35 ÷ 10 = 3.5
50 ÷ 10 = 5
40 ÷ 20 = 2
16 ÷ 40 = 0.4 - Write the results in a table.
Age Frequency Class width Frequency density 0 < a ≤ 20 years 20 20 1 20 < a ≤ 30 years 35 10 3.5 30 < a ≤ 40 years 50 10 5 40 < a ≤ 60 years 40 20 2 60 < a ≤ 100 years 16 40 0.4
Check. Multiply back: 1 × 20 = 20, 3.5 × 10 = 35, 5 × 10 = 50, 2 × 20 = 40, 0.4 × 40 = 16. These are the frequencies. ✓
Watch out. The vertical axis is frequency density, not frequency.
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.
- Find the width of each class. The bars have different widths, so the height cannot be the frequency. Width = upper end − lower end.20 − 0 = 20
30 − 20 = 10
40 − 30 = 10
60 − 40 = 20
100 − 60 = 40 - 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.20 ÷ 20 = 1
35 ÷ 10 = 3.5
50 ÷ 10 = 5
40 ÷ 20 = 2
16 ÷ 40 = 0.4 - Write the results in a table.
Age Frequency Class width Frequency density 0 < a ≤ 20 years 20 20 1 20 < a ≤ 30 years 35 10 3.5 30 < a ≤ 40 years 50 10 5 40 < a ≤ 60 years 40 20 2 60 < a ≤ 100 years 16 40 0.4 - 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.
Check. Multiply back: 1 × 20 = 20, 3.5 × 10 = 35, 5 × 10 = 50, 2 × 20 = 40, 0.4 × 40 = 16. These are the frequencies. ✓
Watch out. The vertical axis is frequency density, not frequency.
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. In a histogram, area = frequency. So frequency = frequency density × class width.
- Find the class width. Upper end minus lower end.20 − 10 = 10
- Read the frequency density (the height of the bar). Frequency density = 3.2
- Multiply height by width. The area of the bar is the frequency.3.2 × 10 = 32
Check. Divide back: 32 ÷ 10 = 3.2. ✓
Watch out. The height of the bar is not the frequency. You must multiply by the class width.
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. In a histogram, area = frequency. So frequency = frequency density × class width.
- Find the class width. Upper end minus lower end.5 − 0 = 5
- Read the frequency density (the height of the bar). Frequency density = 2
- Multiply height by width. The area of the bar is the frequency.2 × 5 = 10
Check. Divide back: 10 ÷ 5 = 2. ✓
Watch out. The height of the bar is not the frequency. You must multiply by the class width.
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. Work out the frequency of every bar, then add.
- Frequency of each bar. Height × width for each bar.0 to 5: 2 × 5 = 10
5 to 10: 5 × 5 = 25
10 to 20: 3.2 × 10 = 32
20 to 30: 1.5 × 10 = 15
30 to 50: 0.6 × 20 = 12 - Add. 10 + 25 + 32 + 15 + 12 = 94
Check. Check the largest bar: 5 × 5 = 25 patients in 5 to 10, which is included. ✓
Watch out. Multiply each height by its width before adding.
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 bar from 20 to 30 has height 1.5. The part wanted is 6 wide.
- Height of the bar. Read the frequency density.Height = 1.5
- Width wanted. 30 − 24 = 6
- Area. 1.5 × 6 = 9
Check. The whole bar 20 to 30 has 1.5 × 10 = 15, and 6 out of 10 of it is 9. ✓
Watch out. Use only the width of the part, not of the whole bar.
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 bar from 30 to 50 has height 0.6. The part above 40 is 10 wide.
- Height of the bar. Height = 0.6
- Width wanted. 40 to 50.10
- Area. 0.6 × 10 = 6
Check. The whole bar has 0.6 × 20 = 12 patients, and 40 to 50 is half of it. ✓
Watch out. The class 30 to 50 is wider than the part you need.
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 bar from 10 to 20 has height 3.2. The part wanted is 5 wide.
- Height of the bar. Height = 3.2
- Width wanted. 15 to 20.20 − 15 = 5
- Area. 3.2 × 5 = 16
Check. The whole bar is 3.2 × 10 = 32; half of it is 16. ✓
Watch out. Take away the part you do not want, or multiply the part you want.
| Length | Frequency |
|---|---|
| 0 < t ≤ 10 min | 12 |
| 10 < t ≤ 20 min | 28 |
| 20 < t ≤ 40 min | 50 |
| 40 < t ≤ 60 min | 30 |
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.
- 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 + 60) ÷ 2 = 50 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Length Frequency (f) Midpoint f × midpoint 0 < t ≤ 10 min 12 5 60 10 < t ≤ 20 min 28 15 420 20 < t ≤ 40 min 50 30 1500 40 < t ≤ 60 min 30 50 1500 Total 120 3480 - 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 = 120
Total of f × midpoint = 3480 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.3480 ÷ 120 = 29.0 min
Check. The estimate should be inside the range of the classes, from 0 to 60, and 29.0 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.
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.
- Work out the frequencies and the total. Frequency = frequency density × class width for each bar.Frequencies: 12, 28, 50, 30; total n = 120
- Find the position of the median. It is half way through the data.120 ÷ 2 = 60
- Use a running total to find the median class. The running total first reaches 60 in the class 20 < x ≤ 40 min.
Class Frequency Running total 0 < x ≤ 10 min 12 12 10 < x ≤ 20 min 28 40 20 < x ≤ 40 min 50 90 40 < x ≤ 60 min 30 120 - Find how far into the class we need to go. Before this class there are 40 values, so we need 20 more from this class of 50.20 ÷ 50 = 0.4 of the class
- Find the estimate of the median. Start at the lower end and move that fraction of the class width.20 + 0.4 × 20 = 28 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.
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 area of a bar stands for the frequency. Work out how many people each square centimetre stands for, then use it for the second bar.
- Area of the first bar. Width times height.3 × 4 = 12 cm2
- Find the people per cm2. 24 people have an area of 12 cm2.24 ÷ 12 = 2 people per cm2
- Area of the second bar. 6 × 2 = 12 cm2
- People in the second bar. Multiply the area by 2.12 × 2 = 24
Check. Both bars have the same area, so they have the same frequency. ✓
Watch out. Area, not height, stands for the frequency.
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. Use frequency density = frequency ÷ class width for the second bar.
- Find the class width. Upper end minus lower end.30 − 10 = 20
- Divide the frequency by the width. 100 ÷ 20 = 5
- Compare. The first bar has height 4.5 > 4, so the second bar is taller, even though it has a bigger width.
Check. 5 × 20 = 100. ✓
Watch out. Divide by the class width of the class you are working on.
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. Compare frequency with frequency density.
- What does the height show? The height of a bar is the frequency density, not the frequency.frequency = height × width
- Use an example. Class A is 0 to 10 with height 3, and class B is 10 to 40 with height 2.A: 3 × 10 = 30 values. B: 2 × 30 = 60 values.
- Conclude. Class B has the shorter bar but twice as many values.Jon is wrong.
Check. B has more values because its bar is much wider. ✓
Watch out. Compare the areas, not the heights.
| Class | Frequency |
|---|---|
| 0 < x ≤ 10 | 15 |
| 10 < x ≤ 25 | 45 |
| 25 < x ≤ 30 | 10 |
| 30 < x ≤ 50 | 50 |
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.
- 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
25 − 10 = 15
30 − 25 = 5
50 − 30 = 20 - 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.15 ÷ 10 = 1.5
45 ÷ 15 = 3
10 ÷ 5 = 2
50 ÷ 20 = 2.5 - Write the results in a table.
Class Frequency Class width Frequency density 0 < x ≤ 10 15 10 1.5 10 < x ≤ 25 45 15 3 25 < x ≤ 30 10 5 2 30 < x ≤ 50 50 20 2.5
Check. Multiply back: 1.5 × 10 = 15, 3 × 15 = 45, 2 × 5 = 10, 2.5 × 20 = 50. These are the frequencies. ✓
Watch out. The vertical axis is frequency density, not frequency.
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. Probability = number in the class ÷ total number.
- Find the total. Add the frequencies.15 + 45 + 10 + 50 = 120
- Write the probability. The class 25 to 30 has 10 values.10120
- Simplify. 10120 = 112
Check. The probabilities of all four classes add up to 1. ✓
Watch out. Use the frequency, not the frequency density.
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. Work out the frequency of each known class, add them, and subtract from 150.
- Class 2 to 4. Frequency = density × width.15 × 2 = 30
- Class 4 to 10. 8 × 6 = 48
- Class 10 to 20. 4.6 × 10 = 46
- Add and subtract. The remaining parcels are in the first class.30 + 48 + 46 = 124 and 150 − 124 = 26
Check. 26 + 30 + 48 + 46 = 150. ✓
Watch out. Frequency = density × width.
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. Use the midpoints of the classes: 1, 3, 7 and 15.
- Multiply each frequency by its midpoint. 26 × 1 = 26
30 × 3 = 90
48 × 7 = 336
46 × 15 = 690 - Add and divide. Total of f × midpoint ÷ 150.(26 + 90 + 336 + 690) ÷ 150 = 1142 ÷ 150 = 7.6
Check. The estimate is between 0 and 20, as it must be. ✓
Watch out. Use the midpoint of each class.
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. Find the frequency of each bar from its area, and take the part of the bar from 30 to 40 minutes. Then add the bar above 40.
- Frequencies of the bars. Frequency = height × width.0 to 10: 2 × 10 = 20
10 to 20: 3 × 10 = 30
20 to 40: 1.5 × 20 = 30
40 to 60: 1 × 20 = 20 - Check the total. The question says there are 100 films.20 + 30 + 30 + 20 = 100
- Part of the bar from 20 to 40. The part above 30 is 10 wide, and the height is 1.5.1.5 × 10 = 15
- Add the bar from 40 to 60. All of it is above 30.15 + 20 = 35 films
- Write as a percentage. The total is 100 films.35%
Check. Films 30 or shorter: 20 + 30 + 15 = 65, and 65 + 35 = 100. ✓
Watch out. Use only the part of the bar above 30, and the same fraction of its area.