Histograms and comparing distributions
By the end of this chapter you will be able to
- Work out the frequency density of each class, and draw a histogram with equal or unequal class widths
- Read the frequency from a histogram by working out the area of a bar
- Estimate the mean of grouped data from the midpoints of the classes
- Compare two histograms using the shape, the averages and the range, in context
You should already know
- Grouped frequency tables and estimated means (Year 8, Chapter 13)
- Drawing bar charts and comparing data sets (Year 7 and Year 8)
- Class boundaries and the difference between discrete and continuous data (Year 8, Chapter 13)
Where this leads
- Cumulative frequency and box plots (Year 9, Chapter 12)
- Probability and sampling in statistics (GCSE)
- Histograms in GCSE statistics, including estimated median and mean
National Curriculum for England, Key Stage 3 mathematics, Year 9. Content: Construct and interpret histograms with equal and unequal class widths, using frequency density = frequency / class width, Read a frequency or a frequency density from a histogram, and find the total frequency, Estimate the mean from a grouped frequency table or histogram, using midpoints, Compare distributions drawn as histograms, using the shape, averages and range.
Exercise P Do you remember?
Non-calculatorMethod. Multiply first, then add.
- Multiply. 3 × 10 = 30 and 2 × 5 = 10.30 + 10
- Add. 40
Check. 30 + 10 = 40.
Method. Find the number halfway between them.
- Add. 10 + 20 = 30.30
- Halve. 30 ÷ 2 = 15.15
Check. 15 is 5 more than 10, and 5 less than 20.
Method. Width = upper boundary − lower boundary.
- Subtract. 35 − 20 = 15.15
- Write the answer. 15
Check. 20 + 15 = 35.
Method. Divide 6 by 2.
- Think of groups. How many groups of 2 are in 6?6 = 2 + 2 + 2
- Divide. Two groups of 3 make 6.6 ÷ 2 = 3
Check. 3 × 2 = 6.
Method. Add the frequencies.
- Add. 8 + 12 = 20, 20 + 15 = 35, 35 + 10 = 45, 45 + 5 = 50.50
- Write the answer. 50
Check. The total number of values is 50.
Histograms with equal class widths
A histogram has bars with no gaps, because the data are continuous. The area of each bar is proportional to the frequency. When all the class widths are equal, the height of each bar is the frequency itself.
Key ideas
- A histogram has no gaps between the bars.
- Equal class widths: height = frequency.
- The class boundaries are the edges of the bars.
- Total frequency = the total number of values.
Method. With equal widths, the height is the frequency.
- Read the height. The bar for 10 < x ≤ 15 has height 14.height = 14
- Write the frequency. With equal widths, the height is the frequency.frequency = 14
- Write the answer. 14
Check. Adding all frequencies gives the total, 40.
Method. Add the frequencies.
- Add the frequencies. The frequencies are 4, 10, 14 and 12.4 + 10 + 14 + 12
- Calculate. 4 + 10 = 14, 14 + 14 = 28, 28 + 12 = 40.40
- Write the answer. 40
Check. The survey has 40 people.
Method. With equal widths, the height is the frequency.
- Read the height. The bar for 0 to 5 has height 4.height = 4
- Check the widths. All the bars are 5 wide, so the heights are frequencies.width = 5 for each bar
- Write the answer. Equal widths mean height = frequency.4
Check. This is the first bar of the ages histogram.
Watch out
Watch out: with unequal class widths, the height of a bar is not the frequency. You must use frequency density (section 13.2).
Exam tip
Exam tip. Check the widths are equal before you read heights as frequencies.
Exercise 13A Histograms with equal widths
Non-calculatorMethod. Add the four frequencies.
- Add. 4 + 10 + 14 + 1240
- Write the answer. 40
Check. This matches the total number of people surveyed.
Method. Find the tallest bar.
- Compare heights. The heights are 4, 10, 14 and 12. The largest is 14.14
- Write the answer. The bar with height 14 is the class 10 to 15.10 to 15
Check. 14 is the largest of the four heights.
Method. Read the bar for 15 to 20.
- Read. The bar for the class 15 to 20 has height 12.12
- Write the answer. 12
Check. 12 people are in that class.
Method. Add the frequencies.
- Add. 2 + 6 + 9 + 3 = 20.20
- Write the answer. 20
Check. The width does not change the frequencies.
Method. Add the heights, which are the frequencies.
- Add. 5 + 9 + 6 = 20.20
- Write the answer. 20
Check. With equal widths, the heights are the frequencies.
Method. Think about the type of data.
- Histogram. The data can take any value in a range, so the classes join.continuous data
- Bar chart. The categories are separate, so there are gaps.discrete categories
- Write the answer. Continuous data has no gaps; discrete categories have gaps
Check. This is the difference between the two charts.
Method. Read the inequality signs.
- Read the first class. 0 < a ≤ 5 includes 5, because of the ≤ sign.a ≤ 5
- Read the next class. 5 < a means the next class starts just after 5.5 < a
- Write the answer. Each boundary belongs to the class on the left
Check. Classes never overlap.
Method. Add each set of frequencies.
- First. 4 + 6 + 8 = 18.18
- Second. 2 + 3 + 4 = 9.9
- Write the answer. 18 and 9
Check. The second total is half the first.
Method. Add the heights, because the widths are equal.
- Add. 11 + 3 = 14.14
- Write the answer. 14
Check. Equal widths mean the heights are the frequencies.
Histograms with unequal class widths
When the class widths are unequal, use frequency density for the height of each bar: frequency density = frequency ÷ class width. The area of each bar is then the frequency.
Key ideas
- Frequency density = frequency ÷ class width.
- Bar height = frequency density.
- Bar area = height × width = frequency.
- Use the class width, found from the boundaries, not the number of values in the class.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.50 − 30 = 20
- Divide the frequency by the width. Frequency density = frequency ÷ class width.10 ÷ 20 = 0.5
- Write the answer. 0.5
Check. 0.5 × 20 = 10, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.10 − 0 = 10
- Multiply. Frequency = frequency density × class width (the area of the bar).1.5 × 10 = 15
- Write the answer. 15
Check. Dividing 15 by 10 gives back 1.5.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Multiply each height by its width to find the frequency, then add the frequencies.
- Find each frequency. Frequency = height × width for each bar.8, 12, 15, 10, 5
- Add the frequencies. The total frequency is the sum of all the bar areas.8 + 12 + 15 + 10 + 5 = 50
- Write the answer. 50
Check. The total must match the number of values in the data.
Watch out. Do not add the heights. Add the areas.
Watch out
Watch out: the frequency density is frequency ÷ width. A bar twice as wide needs half the height for the same frequency.
Exam tip
Exam tip. Write the width of each class in a column before you work out any frequency density.
Exercise 13B Frequency density and unequal widths
Non-calculatorMethod. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.10 − 0 = 10
- Divide the frequency by the width. Frequency density = frequency ÷ class width.8 ÷ 10 = 0.8
- Write the answer. 0.8
Check. 0.8 × 10 = 8, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.20 − 10 = 10
- Divide the frequency by the width. Frequency density = frequency ÷ class width.12 ÷ 10 = 1.2
- Write the answer. 1.2
Check. 1.2 × 10 = 12, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.30 − 20 = 10
- Divide the frequency by the width. Frequency density = frequency ÷ class width.15 ÷ 10 = 1.5
- Write the answer. 1.5
Check. 1.5 × 10 = 15, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.50 − 30 = 20
- Divide the frequency by the width. Frequency density = frequency ÷ class width.10 ÷ 20 = 0.5
- Write the answer. 0.5
Check. 0.5 × 20 = 10, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.60 − 50 = 10
- Divide the frequency by the width. Frequency density = frequency ÷ class width.5 ÷ 10 = 0.5
- Write the answer. 0.5
Check. 0.5 × 10 = 5, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.50 − 30 = 20
- Divide the frequency by the width. Frequency density = frequency ÷ class width.10 ÷ 20 = 0.5
- Write the answer. 0.5
Check. 0.5 × 20 = 10, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Frequency = frequency density × width.
- Multiply. 2.5 × 4 = 10.10
- Write the answer. 10
Check. 10 ÷ 4 = 2.5.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.60 − 50 = 10
- Multiply. Frequency = frequency density × class width (the area of the bar).0.5 × 10 = 5
- Write the answer. 5
Check. Dividing 5 by 10 gives back 0.5.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Multiply each height by its width to find the frequency, then add the frequencies.
- Find each frequency. Frequency = height × width for each bar.8, 12, 15, 10, 5
- Add the frequencies. The total frequency is the sum of all the bar areas.8 + 12 + 15 + 10 + 5 = 50
- Write the answer. 50
Check. The total must match the number of values in the data.
Watch out. Do not add the heights. Add the areas.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.20 − 0 = 20
- Multiply. Frequency = frequency density × class width (the area of the bar).0.9 × 20 = 18
- Write the answer. 18
Check. Dividing 18 by 20 gives back 0.9.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Use frequency density = frequency ÷ width.
- Width. The class 30 to 50 has width 50 − 30 = 20.20
- Density. Frequency density = 10 ÷ 20 = 0.5.0.5
- Write the answer. A wider class needs a lower bar for the same frequency.Frequency density is 0.5
Check. Area = 0.5 × 20 = 10, the frequency.
Method. Multiply each density by its width.
- First class. 3 × 5 = 15.15
- Second class. 1.5 × 10 = 15.15
- Write the answer. They have the same frequency, 15
Check. Both bars have the same area.
Method. Multiply each density by its width, then add.
- First. 0.6 × 10 = 6.6
- Second. 1.2 × 10 = 12.12
- Add. 6 + 12 = 18.18
Check. 18 is the total frequency.
Method. Think about what the height means.
- Height. The height is frequency divided by width.height = frequency ÷ width
- Area. So the area of each bar is the frequency, not the height.area = frequency
- Write the answer. Height is frequency density, so the area shows the frequency
Check. This keeps the bars comparable.
Reading frequencies from a histogram
To read a frequency from a histogram with unequal widths, find the area of the bar: frequency = frequency density × width. Then add all the frequencies to find the total.
Key ideas
- Area of a bar = frequency.
- Frequency = frequency density × width.
- Total frequency = sum of all the bar areas.
- Read the frequency density from the vertical axis.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.25 − 10 = 15
- Multiply. Frequency = frequency density × class width (the area of the bar).2 × 15 = 30
- Write the answer. 30
Check. Dividing 30 by 15 gives back 2.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.20 − 0 = 20
- Multiply. Frequency = frequency density × class width (the area of the bar).1.5 × 20 = 30
- Write the answer. 30
Check. Dividing 30 by 20 gives back 1.5.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.10 − 0 = 10
- Multiply. Frequency = frequency density × class width (the area of the bar).0.6 × 10 = 6
- Write the answer. 6
Check. Dividing 6 by 10 gives back 0.6.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Watch out
Watch out: read the frequency density from the vertical axis, then multiply by the width. Do not read the area as the height.
Exam tip
Exam tip. Write the area calculation for each bar, so your method is clear even if the arithmetic is wrong.
Exercise 13C Reading frequencies from a histogram
Non-calculatorMethod. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.25 − 10 = 15
- Multiply. Frequency = frequency density × class width (the area of the bar).2 × 15 = 30
- Write the answer. 30
Check. Dividing 30 by 15 gives back 2.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.20 − 0 = 20
- Multiply. Frequency = frequency density × class width (the area of the bar).1.5 × 20 = 30
- Write the answer. 30
Check. Dividing 30 by 20 gives back 1.5.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.30 − 20 = 10
- Multiply. Frequency = frequency density × class width (the area of the bar).4 × 10 = 40
- Write the answer. 40
Check. Dividing 40 by 10 gives back 4.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.60 − 40 = 20
- Multiply. Frequency = frequency density × class width (the area of the bar).0.75 × 20 = 15
- Write the answer. 15
Check. Dividing 15 by 20 gives back 0.75.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Multiply each height by its width to find the frequency, then add the frequencies.
- Find each frequency. Frequency = height × width for each bar.30, 45, 20
- Add the frequencies. The total frequency is the sum of all the bar areas.30 + 45 + 20 = 95
- Write the answer. 95
Check. The total must match the number of values in the data.
Watch out. Do not add the heights. Add the areas.
Method. Multiply each height by its width to find the frequency, then add the frequencies.
- Find each frequency. Frequency = height × width for each bar.8, 12, 15, 10, 5
- Add the frequencies. The total frequency is the sum of all the bar areas.8 + 12 + 15 + 10 + 5 = 50
- Write the answer. 50
Check. The total must match the number of values in the data.
Watch out. Do not add the heights. Add the areas.
Method. Add the areas of the classes above 30 minutes.
- Class 30 to 50. Frequency = 0.5 × 20 = 10.10
- Class 50 to 60. Frequency = 0.5 × 10 = 5.5
- Add. 10 + 5 = 15.15
- Write the answer. 15
Check. The classes above 30 minutes are 30 to 50 and 50 to 60.
Method. Use area = height × width.
- Area. Frequency = frequency density × width = 0.4 × 10.4
- Write the answer. 4
Check. 4 ÷ 10 = 0.4.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.80 − 60 = 20
- Divide the frequency by the width. Frequency density = frequency ÷ class width.12 ÷ 20 = 0.6
- Write the answer. 0.6
Check. 0.6 × 20 = 12, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Compare the areas.
- Areas. Frequency = height × width, so a wider bar has a bigger frequency.frequency = height × width
- Write the answer. The wider bar has the larger area and frequency
Check. Area, not height, gives the frequency.
Estimating the mean from grouped data
To estimate the mean: find the midpoint of each class, multiply it by the frequency, add the products, then divide by the total frequency.
Key ideas
- Midpoint = (lower boundary + upper boundary) ÷ 2.
- Estimated mean = sum of (midpoint × frequency) ÷ total frequency.
- Round the estimate as the question asks.
- The estimate is not exact, because the values inside each class are unknown.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.2.5, 7.5, 12.5, 17.5
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.2.5 × 4, 7.5 × 10, 12.5 × 14, 17.5 × 12
- Add and divide by the total frequency. The total frequency is 40. Divide the total of the products by it.470 ÷ 40 = 11.7500
- Round the answer. Give the estimated mean to 1 decimal place.11.8
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.5, 15, 25, 40, 55
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.5 × 8, 15 × 12, 25 × 15, 40 × 10, 55 × 5
- Add and divide by the total frequency. The total frequency is 50. Divide the total of the products by it.1270 ÷ 50 = 25.4000
- Round the answer. Give the estimated mean to 1 decimal place.25.4
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.5, 15
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.5 × 3, 15 × 1
- Add and divide by the total frequency. The total frequency is 4. Divide the total of the products by it.30 ÷ 4 = 7.5000
- Round the answer. Give the estimated mean to 1 decimal place.7.5
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Watch out
Watch out: use the midpoint of each class, not the boundaries. A class 10 to 20 has midpoint 15.
Exam tip
Exam tip. Make a table with columns: class, frequency, midpoint, midpoint × frequency. Then add the last column.
Exercise 13D Estimating the mean (midpoints)
Non-calculatorMethod. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.2.5, 7.5, 12.5, 17.5
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.2.5 × 4, 7.5 × 10, 12.5 × 14, 17.5 × 12
- Add and divide by the total frequency. The total frequency is 40. Divide the total of the products by it.470 ÷ 40 = 11.7500
- Round the answer. Give the estimated mean to 1 decimal place.11.8
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.5, 15, 25, 40, 55
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.5 × 8, 15 × 12, 25 × 15, 40 × 10, 55 × 5
- Add and divide by the total frequency. The total frequency is 50. Divide the total of the products by it.1270 ÷ 50 = 25.4000
- Round the answer. Give the estimated mean to 1 decimal place.25.4
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Add the boundaries, then halve.
- Add. 30 + 50 = 80.80
- Halve. 80 ÷ 2 = 40.40
Check. 40 is halfway between 30 and 50.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.5, 15
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.5 × 2, 15 × 3
- Add and divide by the total frequency. The total frequency is 5. Divide the total of the products by it.55 ÷ 5 = 11.0000
- Round the answer. Give the estimated mean to 1 decimal place.11.0
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Think about what grouping loses.
- Grouping. The data are only known to lie within a class.within the class
- Write the answer. The exact values are not known, so the midpoints give an estimate
Check. This is why the answer is an estimate.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.2, 7
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.2 × 5, 7 × 5
- Add and divide by the total frequency. The total frequency is 10. Divide the total of the products by it.45 ÷ 10 = 4.5000
- Round the answer. Give the estimated mean to 1 decimal place.4.5
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.25, 35
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.25 × 6, 35 × 4
- Add and divide by the total frequency. The total frequency is 10. Divide the total of the products by it.290 ÷ 10 = 29.0000
- Round the answer. Give the estimated mean to 1 decimal place.29.0
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Multiply.
- Multiply. 25 × 8 = 200.200
- Write the answer. 200
Check. 200 ÷ 8 = 25.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.5, 15
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.5 × 4, 15 × 1
- Add and divide by the total frequency. The total frequency is 5. Divide the total of the products by it.35 ÷ 5 = 7.0000
- Round the answer. Give the estimated mean to 1 decimal place.7.0
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Comparing two distributions drawn as histograms
When comparing, always give an average (the estimated mean or the modal class) and a measure of spread (the range or the spread of the bars), then say what it means in the context of the test.
Key ideas
- Compare an average: the estimated mean, using midpoints.
- Compare spread: the range of the classes that have data.
- Describe the shape: where the bars are tallest, and whether they are symmetrical.
- Always compare the same measures for both distributions.
Method. Compare the estimated mean of the two histograms.
- Compare the values. Compare the estimated mean of each distribution.Class A: 21.5, Class B: 19
- Find the difference. The difference shows how far apart the two values are.2.5
- Say which is bigger. Class A has the higher estimated mean
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Method. Compare the range of the two histograms.
- Compare the values. Compare the range of each distribution.Class A: 40, Class B: 30
- Find the difference. The difference shows how far apart the two values are.10
- Say which is bigger. Class A has the higher range
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Method. Compare the range of the two histograms.
- Compare the values. Compare the range of each distribution.Class A: 25, Class B: 35
- Find the difference. The difference shows how far apart the two values are.10
- Say which is bigger. Class B has the higher range
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Watch out
Watch out: a comparison needs the same measure for both classes. Do not compare the mean of one with the range of the other.
Exam tip
Exam tip. Use phrases such as "on average, Class A scored higher" and "Class B had more spread" in your comparisons.
Exercise 13E Comparing distributions
Non-calculatorMethod. Compare the estimated mean of the two histograms.
- Compare the values. Compare the estimated mean of each distribution.Class A: 21.5, Class B: 19
- Find the difference. The difference shows how far apart the two values are.2.5
- Say which is bigger. Class A has the higher estimated mean
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Method. Compare the range of the two histograms.
- Compare the values. Compare the range of each distribution.Class A: 40, Class B: 30
- Find the difference. The difference shows how far apart the two values are.10
- Say which is bigger. Class A has the higher range
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Method. Compare the estimated mean of the two histograms.
- Compare the values. Compare the estimated mean of each distribution.Class A: 24, Class B: 27
- Find the difference. The difference shows how far apart the two values are.3
- Say which is bigger. Class B has the higher estimated mean
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Method. Compare the classes with the tallest bars.
- Class A. Its most common class is 20 to 30.20 to 30
- Class B. Its most common class is 30 to 40, which is higher.30 to 40
- Write the answer. Class B typically scored higher
Check. The modal class is a typical value.
Method. Think about outliers.
- Outlier. One very high or low value changes the range.range changes
- Write the answer. The spread of the bulk of the data may be the same.One outlier changes the range a lot
Check. The IQR or the bulk of the bars gives a fairer picture.
Method. Look at the spread of the bars.
- Class A. The bars are concentrated in a few classes.concentrated
- Class B. The bars are spread across many classes.spread
- Write the answer. Class A is more consistent; Class B is more spread out
Check. A concentrated histogram has less spread.
Method. Compare the estimated mean of the two histograms.
- Compare the values. Compare the estimated mean of each distribution.Class A: 35.5, Class B: 33.2
- Find the difference. The difference shows how far apart the two values are.2.3
- Say which is bigger. Class A has the higher estimated mean
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Chapter summary
| Skill | Method |
|---|---|
| Equal widths | Height = frequency. Area is proportional to frequency. |
| Unequal widths | Frequency density = frequency ÷ width. Height = frequency density. |
| Frequency from histogram | Frequency = frequency density × width (the area of the bar). |
| Estimated mean | Midpoint × frequency, summed, divided by the total frequency. |
| Comparing | Compare averages and spread, in context, using the same measures. |
Can you…?
- I can draw a histogram with equal or unequal class widths
- I can find frequency density and frequency from a histogram
- I can find the total frequency of a histogram
- I can estimate the mean from grouped data using midpoints
- I can compare two distributions using averages and spread
Key terms
- histogram
- a chart with bars whose areas show frequencies, used for continuous data
- frequency density
- frequency divided by class width, used for the height of a histogram bar
- class width
- the upper boundary minus the lower boundary of a class
- midpoint
- the value halfway through a class, found by averaging the boundaries
- continuous data
- data that can take any value in a range, such as height or time
Exercise 13Z Check-up
Non-calculatorMethod. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.20 − 0 = 20
- Multiply. Frequency = frequency density × class width (the area of the bar).2 × 20 = 40
- Write the answer. 40
Check. Dividing 40 by 20 gives back 2.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.50 − 20 = 30
- Divide the frequency by the width. Frequency density = frequency ÷ class width.15 ÷ 30 = 0.5
- Write the answer. 0.5
Check. 0.5 × 30 = 15, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Add the frequencies.
- Add. 6 + 9 + 5 = 20.20
- Write the answer. 20
Check. The width does not change the sum.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.5, 20
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.5 × 5, 20 × 5
- Add and divide by the total frequency. The total frequency is 10. Divide the total of the products by it.125 ÷ 10 = 12.5000
- Round the answer. Give the estimated mean to 1 decimal place.12.5
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Multiply each height by its width to find the frequency, then add the frequencies.
- Find each frequency. Frequency = height × width for each bar.8, 10
- Add the frequencies. The total frequency is the sum of all the bar areas.8 + 10 = 18
- Write the answer. 18
Check. The total must match the number of values in the data.
Watch out. Do not add the heights. Add the areas.
Method. Add and halve.
- Add. 40 + 60 = 100.100
- Halve. 100 ÷ 2 = 50.50
Check. 50 is halfway.
Method. Compare the estimated mean of the two histograms.
- Compare the values. Compare the estimated mean of each distribution.Class A: 42, Class B: 38
- Find the difference. The difference shows how far apart the two values are.4
- Say which is bigger. Class A has the higher estimated mean
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.5 − 0 = 5
- Multiply. Frequency = frequency density × class width (the area of the bar).6 × 5 = 30
- Write the answer. 30
Check. Dividing 30 by 5 gives back 6.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.15 − 10 = 5
- Divide the frequency by the width. Frequency density = frequency ÷ class width.12 ÷ 5 = 2.4
- Write the answer. 2.4
Check. 2.4 × 5 = 12, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. The area gives the frequency.
- Area. Area = height × width = frequency.area = frequency
- Write the answer. So the area is the frequency
Check. Frequency density keeps the areas correct.
Total 18 marks. Allow about 18 minutes.
Exercise 13X Practice test questions
Non-calculatorMethod. Add the frequencies.
- Add. 5 + 8 + 12 + 7 = 32.32
- Write the answer. 32
Check. The total is 32.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.30 − 10 = 20
- Multiply. Frequency = frequency density × class width (the area of the bar).1.5 × 20 = 30
- Write the answer. 30
Check. Dividing 30 by 20 gives back 1.5.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Frequency density is the frequency divided by the class width.
- Find the class width. Width = upper boundary − lower boundary.50 − 40 = 10
- Divide the frequency by the width. Frequency density = frequency ÷ class width.12 ÷ 10 = 1.2
- Write the answer. 1.2
Check. 1.2 × 10 = 12, the frequency.
Watch out. Use the frequency density, not the frequency, for the height of the bar when the class widths differ.
Method. Use the midpoints to estimate the mean of grouped data.
- Find the midpoint of each class. The midpoint is the average of the two boundaries.5, 15, 30
- Multiply each midpoint by its frequency. This gives an estimate of the total for each class.5 × 2, 15 × 6, 30 × 2
- Add and divide by the total frequency. The total frequency is 10. Divide the total of the products by it.160 ÷ 10 = 16.0000
- Round the answer. Give the estimated mean to 1 decimal place.16.0
Check. The estimate is only approximate, because the exact values in each class are not known.
Watch out. Use the midpoint of each class, not the boundaries. A class of 0 to 10 has midpoint 5.
Method. Multiply each height by its width to find the frequency, then add the frequencies.
- Find each frequency. Frequency = height × width for each bar.10, 10
- Add the frequencies. The total frequency is the sum of all the bar areas.10 + 10 = 20
- Write the answer. 20
Check. The total must match the number of values in the data.
Watch out. Do not add the heights. Add the areas.
Method. Compare the estimated mean of the two histograms.
- Compare the values. Compare the estimated mean of each distribution.Class A: 27.5, Class B: 25
- Find the difference. The difference shows how far apart the two values are.2.5
- Say which is bigger. Class A has the higher estimated mean
Check. Use the same measure for both distributions.
Watch out. Give the comparison in context, for example "on average".
Method. Think about continuous data.
- Type. The data are continuous, so there are no gaps between classes.continuous
- Write the answer. Continuous data has no gaps
Check. Bar charts have gaps because their categories are separate.
Method. Add and halve.
- Add. 15 + 25 = 40.40
- Halve. 40 ÷ 2 = 20.20
Check. 20 is halfway between 15 and 25.
Method. The area of a histogram bar is the frequency, so multiply the height by the width.
- Find the class width. Width = upper boundary − lower boundary.170 − 150 = 20
- Multiply. Frequency = frequency density × class width (the area of the bar).0.4 × 20 = 8
- Write the answer. 8
Check. Dividing 8 by 20 gives back 0.4.
Watch out. In a histogram, the area of each bar is proportional to the frequency.
Method. Multiply each height by its width to find the frequency, then add the frequencies.
- Find each frequency. Frequency = height × width for each bar.12, 12
- Add the frequencies. The total frequency is the sum of all the bar areas.12 + 12 = 24
- Write the answer. 24
Check. The total must match the number of values in the data.
Watch out. Do not add the heights. Add the areas.