Chapter 5 of 8

All 8 chapters
  1. 1Samples and bias, readunderstood
  2. 2Averages from frequency tables, readunderstood
  3. 3Cumulative frequency, readunderstood
  4. 4Box plots, readunderstood
  5. 5Histograms with unequal widths, readunderstood
  6. 6Scatter graphs and correlation, readunderstood
  7. 7Data over time, readunderstood
  8. 8Telling the truth with data, readunderstood

M4.7.5 Chapter 5

Histograms with unequal widths

The heights of plants are grouped in classes of different widths. How can one chart show them fairly?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Draw a histogram using frequency density
  • Read frequencies from a histogram

A gardener measures 85 bean plants. The heights h are grouped in classes of different widths, such as 40 ≤ h < 50 and 50 ≤ h < 80. A bar chart with equal widths would not fit this table.

The heights of 85 plants in five classes. The head is h (cm), f. Row 1: 0 ≤ h < 10, 8. Row 2: 10 ≤ h < 20, 14. Row 3: 20 ≤ h < 40, 36. Row 4: 40 ≤ h < 50, 12. Row 5: 50 ≤ h < 80, 15.h (cm)f0 ≤ h < 10810 ≤ h < 201420 ≤ h < 403640 ≤ h < 501250 ≤ h < 8015
The classes are 10, 10, 20, 10 and 30 cm wide.

Suppose we draw each bar as tall as its frequency. The picture below does that. Look at the last two bars. The class 50 ≤ h < 80 is three times as wide as the class 40 ≤ h < 50. Its frequency is only a little bigger. Yet its bar looks about four times as big.

A chart with the x axis height h (cm), with class edges 0, 10, 20, 40, 50 and 80, and the y axis plants. The bar heights are the frequencies 8, 14, 36, 12 and 15. The last bar is very wide.01020304081436121501020405080height h (cm)plants
A misleading chart. The wide last bar holds 15 plants, but looks much bigger than the bar before it, which holds 12.

The fix is to let the area of a bar show the frequency. The area of a bar is width × height. So the height must be frequency ÷ width. This height is called the frequency density. A chart drawn this way is a histogram.

Worked example

Find the frequency density of the classes 40 ≤ h < 50 and 50 ≤ h < 80.

  1. The width of 40 ≤ h < 50 is 50 − 40 = 10 cm.
  2. Frequency density = frequency ÷ width = 12 ÷ 10 = 1.2 plants per cm.
  3. The width of 50 ≤ h < 80 is 80 − 50 = 30 cm. So the density is 15 ÷ 30 = 0.5 plants per cm.

Answer: 1.2 plants per cm and 0.5 plants per cm

Draw it first

Draw the histogram

  1. The x axis is height h (cm) with class edges 0, 10, 20, 40, 50 and 80. The y axis is plants per cm. Step 1 shows the axes only. Step 2 draws the first two bars, 10 cm wide, with heights 0.8 and 1.4. Step 3 draws all five bars, with heights 0.8, 1.4, 1.8, 1.2 and 0.5.00.20.40.60.811.21.41.61.8201020405080height h (cm)plants per cm

    1 Draw the height axis with the true class edges: 0, 10, 20, 40, 50 and 80. Draw the frequency density axis up the side.

  2. The x axis is height h (cm) with class edges 0, 10, 20, 40, 50 and 80. The y axis is plants per cm. Step 1 shows the axes only. Step 2 draws the first two bars, 10 cm wide, with heights 0.8 and 1.4. Step 3 draws all five bars, with heights 0.8, 1.4, 1.8, 1.2 and 0.5.00.20.40.60.811.21.41.61.8201020405080height h (cm)plants per cm

    2 For each class, work out the density. The first two classes give 8 ÷ 10 = 0.8 and 14 ÷ 10 = 1.4. Draw their bars.

  3. The x axis is height h (cm) with class edges 0, 10, 20, 40, 50 and 80. The y axis is plants per cm. Step 1 shows the axes only. Step 2 draws the first two bars, 10 cm wide, with heights 0.8 and 1.4. Step 3 draws all five bars, with heights 0.8, 1.4, 1.8, 1.2 and 0.5.00.20.40.60.811.21.41.61.8201020405080height h (cm)plants per cm

    3 Draw the other three bars. 36 ÷ 20 = 1.8, 12 ÷ 10 = 1.2 and 15 ÷ 30 = 0.5.

The class, the frequency f and the frequency density of each class of plant heights. The head is h (cm), f, density. Row 1: 0 ≤ h < 10, 8, 0.8. Row 2: 10 ≤ h < 20, 14, 1.4. Row 3: 20 ≤ h < 40, 36, 1.8. Row 4: 40 ≤ h < 50, 12, 1.2. Row 5: 50 ≤ h < 80, 15, 0.5.h (cm)fdensity0 ≤ h < 1080.810 ≤ h < 20141.420 ≤ h < 40361.840 ≤ h < 50121.250 ≤ h < 80150.5
Frequency density = frequency ÷ class width.

On the histogram, the wide last bar is now low. Its area is 30 × 0.5 = 15 plants, which matches the table. The area of a bar is the frequency. This is how to read a frequency from a histogram.

The x axis is height h (cm) with class edges 0, 10, 20, 40, 50 and 80. The y axis is plants per cm. The five bars have heights 0.8, 1.4, 1.8, 1.2 and 0.5. A dashed line runs from the top of the third bar across to the y axis at 1.8.00.20.40.60.811.21.41.61.8201020405080height h (cm)plants per cm
The third bar reaches 1.8. A dashed line carries that height across to the axis.

Worked example

The bar for 20 ≤ h < 40 reaches a height of 1.8 plants per cm. How many plants are in this class?

  1. The class is from 20 to 40, so its width is 40 − 20 = 20 cm.
  2. The frequency is the area of the bar: width × height = 20 × 1.8.
  3. 20 × 1.8 = 36 plants.

Answer: 36 plants

We can also find part of a class. Suppose the 36 plants in 20 ≤ h < 40 are spread evenly along the class. Then the bar has the same height all along it. A piece of the class that is half as wide holds half as many plants.

See it three ways

With things

Cut the class 20 ≤ h < 40 into two pieces at 30 cm. Each piece is half the width, so it holds half of the 36 plants.

As a picture

A histogram with the x axis height h (cm), with edges 20, 30 and 40, and the y axis plants per cm. The two bars are both 1.8 high, and each holds 18 plants.00.30.60.91.21.51.81818203040height h (cm)plants per cm

In symbols

36 ÷ 20 = 1.8 plants per cm. A piece 10 cm wide: 10 × 1.8 = 18 plants.

Worked example

Estimate how many plants are between 20 cm and 30 cm tall.

  1. The plants in 20 ≤ h < 40 are spread evenly, at 36 ÷ 20 = 1.8 plants per cm.
  2. The part from 20 to 30 is 30 − 20 = 10 cm wide.
  3. Number of plants = width × density = 10 × 1.8 = 18.
  4. So there are about 18 plants between 20 cm and 30 cm tall.

Answer: About 18 plants

This is an estimate, because the plants are probably not spread perfectly evenly. We use it when we know nothing better. It is exact only if the spread is even.

Try it

Draw a histogram on squared paper

You need: squared paper, a ruler, a pencil.

  1. Use 85 walks. The classes of time t, in minutes, are 0 ≤ t < 5, 5 ≤ t < 10, 10 ≤ t < 20 and 20 ≤ t < 40.
  2. The frequencies of the classes are 10, 25, 30 and 20.
  3. Work out each density: frequency ÷ width. They are 2, 5, 3 and 1 walks per minute.
  4. Draw the bars. Make the widths true to the scale of the time axis. Check each area: width × height should give 10, 25, 30 and 20.

What you will see: The tallest bar is for 5 ≤ t < 10, but the biggest class is 10 ≤ t < 20, which has the largest area, 30.

If you cannot try it, this is what happens.
A histogram with the x axis time t (min) with edges 0, 5, 10, 20 and 40, and the y axis walks per min. The bars are 2, 5, 3 and 1 high. The numbers 10, 25, 30 and 20 are written above the bars.0123451025302005102040time t (min)walks per min

A wrong idea many people have

The height of a bar in a histogram is the frequency.

Why it feels right: In a bar chart the height of a bar is the frequency, so the same seems true here.

The test that breaks it: In the plant chart the class 50 ≤ h < 80 has frequency 15, but its bar is only 0.5 high. It is wide, so its area is 30 × 0.5 = 15.

What is true: In a histogram with unequal widths the area of a bar is the frequency. The height is the frequency density.

A wrong idea many people have

The tallest bar is always the class with the most values.

Why it feels right: We are used to reading the tallest bar as the biggest.

The test that breaks it: In the walks histogram the tallest bar is 5 high, but it is only 5 minutes wide, so it holds 25 walks. The bar 3 high and 10 wide holds 30.

What is true: The class with the most values has the largest area, not necessarily the largest height.

Key words

histogram
a chart for measured data in which the area of each bar is the frequency.
frequency density
frequency divided by class width. It is the height of a bar in a histogram.

Back to your guess

The answer is B: Make the area of each bar match the frequency of its class.

If the classes have different widths, a bar as tall as its frequency makes wide classes look too big. Letting the area of each bar show the frequency is fair. The height of a bar is then the frequency divided by the class width, called the frequency density.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A class 20 ≤ h < 40 has 36 plants. What is its frequency density?

    Tap your answer.

    1. Not this one. You divided the width by the frequency, 20 ÷ 36. Frequency density is the frequency divided by the width. Go to M4.7.5. Try another.

    2. Not this one. You multiplied 36 by 20. The area is width × height, so to get the height we divide the frequency by the width. Go to M4.7.5. Try another.

    3. Yes, that is right. The width is 40 − 20 = 20 cm. Frequency density = 36 ÷ 20 = 1.8.

    4. Not this one. 36 is the frequency itself. The density is the frequency shared out over the 20 cm of the class. Go to M4.7.5. Try another.

    See the answer

    1.8 plants per cm

    The width is 40 − 20 = 20 cm. Frequency density = 36 ÷ 20 = 1.8.

    • If you chose A: You divided the width by the frequency, 20 ÷ 36. Frequency density is the frequency divided by the width. Go to M4.7.5.
    • If you chose B: You multiplied 36 by 20. The area is width × height, so to get the height we divide the frequency by the width. Go to M4.7.5.
    • If you chose D: 36 is the frequency itself. The density is the frequency shared out over the 20 cm of the class. Go to M4.7.5.

  2. A bar of a histogram is 6 cm wide and 2.5 plants per cm high. How many plants does it stand for?

    Tap your answer.

    1. Not this one. You added the width and the height. The frequency is the area of the bar, which is width × height. Go to M4.7.5. Try another.

    2. Not this one. You divided the width by the height. The frequency is the area, so multiply. Go to M4.7.5. Try another.

    3. Not this one. You divided the height by the width, 2.5 ÷ 6. The frequency is width × height. Go to M4.7.5. Try another.

    4. Yes, that is right. The area of the bar is its frequency: 6 × 2.5 = 15.

    See the answer

    15

    The area of the bar is its frequency: 6 × 2.5 = 15.

    • If you chose A: You added the width and the height. The frequency is the area of the bar, which is width × height. Go to M4.7.5.
    • If you chose B: You divided the width by the height. The frequency is the area, so multiply. Go to M4.7.5.
    • If you chose C: You divided the height by the width, 2.5 ÷ 6. The frequency is width × height. Go to M4.7.5.

  3. This histogram shows the heights of some plants. Which class has the most plants?

    A histogram with the x axis height h (cm) with edges 0, 10, 30 and 40, and the y axis plants per cm. The first bar, from 0 to 10, is 1.5 high. The second bar, from 10 to 30, is 1.2 high. The third bar, from 30 to 40, is 0.6 high.00.30.60.91.21.51.80103040height h (cm)plants per cm
    Tap your answer.

    1. Not this one. Its bar is the tallest but only 10 cm wide. Area counts: 10 × 1.5 = 15 plants. Another bar has a bigger area. Go to M4.7.5. Try another.

    2. Yes, that is right. The areas are 10 × 1.5 = 15, 20 × 1.2 = 24 and 10 × 0.6 = 6. The middle bar has the biggest area, 24 plants.

    3. Not this one. The area of each bar is its frequency, so the chart does tell us: 15, 24 and 6. Go to M4.7.5. Try another.

    See the answer

    10 ≤ h < 30

    The areas are 10 × 1.5 = 15, 20 × 1.2 = 24 and 10 × 0.6 = 6. The middle bar has the biggest area, 24 plants.

    • If you chose A: Its bar is the tallest but only 10 cm wide. Area counts: 10 × 1.5 = 15 plants. Another bar has a bigger area. Go to M4.7.5.
    • If you chose C: The area of each bar is its frequency, so the chart does tell us: 15, 24 and 6. Go to M4.7.5.

  4. What does the vertical axis of a histogram with unequal class widths show?

    Tap your answer.

    1. Not this one. That is a bar chart. In a histogram with unequal widths, the frequency is the area of the bar, not its height. Go to M4.7.5. Try another.

    2. Not this one. The width of a class is shown along the horizontal axis, by the width of its bar. Go to M4.7.5. Try another.

    3. Yes, that is right. The height of a bar is frequency ÷ width, so the area of the bar is the frequency.

    See the answer

    The frequency density: the frequency divided by the class width.

    The height of a bar is frequency ÷ width, so the area of the bar is the frequency.

    • If you chose A: That is a bar chart. In a histogram with unequal widths, the frequency is the area of the bar, not its height. Go to M4.7.5.
    • If you chose B: The width of a class is shown along the horizontal axis, by the width of its bar. Go to M4.7.5.

  5. Plants in the class 50 ≤ h < 80 are spread evenly. The bar is 0.5 high. Estimate how many plants are between 50 cm and 60 cm.

    The x axis is height h (cm) with class edges 0, 10, 20, 40, 50 and 80. The y axis is plants per cm. The five bars have heights 0.8, 1.4, 1.8, 1.2 and 0.5. A dashed line runs from the top of the fifth bar across to the y axis at 0.5.00.20.40.60.811.21.41.61.8201020405080height h (cm)plants per cm
    Tap your answer.

    1. Not this one. 15 plants are in the whole class from 50 to 80. The question asks for the part from 50 to 60. Go to M4.7.5. Try another.

    2. Not this one. 0.5 is the height of the bar, the density. The number of plants is the area of the part. Go to M4.7.5. Try another.

    3. Not this one. You took half of the class. From 50 to 60 is 10 cm out of 30 cm, which is a third of it. Go to M4.7.5. Try another.

    4. Yes, that is right. The piece from 50 to 60 is 10 cm wide. The area is 10 × 0.5 = 5 plants.

    See the answer

    5

    The piece from 50 to 60 is 10 cm wide. The area is 10 × 0.5 = 5 plants.

    • If you chose A: 15 plants are in the whole class from 50 to 80. The question asks for the part from 50 to 60. Go to M4.7.5.
    • If you chose B: 0.5 is the height of the bar, the density. The number of plants is the area of the part. Go to M4.7.5.
    • If you chose C: You took half of the class. From 50 to 60 is 10 cm out of 30 cm, which is a third of it. Go to M4.7.5.

  6. When must the height of a bar in a histogram be the frequency density?

    Tap your answer.

    1. Yes, that is right. With unequal widths, bars as tall as the frequency would make wide classes look too big. Using the density makes the area show the frequency.

    2. Not this one. The number of values does not matter. What matters is whether the widths of the classes are all equal. Go to M4.7.5. Try another.

    3. Not this one. With equal widths the height can be the frequency, and the picture is still fair, because every bar has the same width. Go to M4.7.5. Try another.

    See the answer

    When the classes have different widths.

    With unequal widths, bars as tall as the frequency would make wide classes look too big. Using the density makes the area show the frequency.

    • If you chose B: The number of values does not matter. What matters is whether the widths of the classes are all equal. Go to M4.7.5.
    • If you chose C: With equal widths the height can be the frequency, and the picture is still fair, because every bar has the same width. Go to M4.7.5.

  7. From an earlier chapter

    The upper quartile of a list is the median of its upper half. Find the upper quartile of 9, 14, 16, 20, 23, 25, 31.

    See the answer

    25

    There are 7 values, so the median is 20, the 4th. Leave it out. The upper half is 23, 25, 31. Its middle value is 25.

  8. A class 10 cm wide has frequency 14. What is the frequency density?

    See the answer

    1.4

    Frequency density = frequency ÷ width = 14 ÷ 10 = 1.4.

  9. A bar is 20 cm wide and 1.5 plants per cm high. How many plants does it stand for?

    See the answer

    30

    The frequency is the area: 20 × 1.5 = 30.

Keep it

Copy this into your exercise book.

Frequency density = frequency ÷ class width. The height of a bar is the density. The area of a bar is the frequency: width × height.

Draw: A histogram with three bars of different widths, and the area of the widest bar worked out beside it.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (1.1 MB). It opens on any phone.