Chapter 3 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

M4.7.3 Chapter 3

Cumulative frequency

A bus company times 80 journeys and sorts the times into classes. How can one graph show the middle time, and how many journeys took under 25 minutes?

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 cumulative frequency graph
  • Read the median and quartiles from the graph
  • Find the interquartile range

The table shows the times of 80 bus journeys. A cumulative frequency is a running total. It tells how many values are below the top of a class.

The times of 80 bus journeys in classes, with the frequency f and the cumulative frequency cf of each class. The head is t (min), f, cf. Row 1: 0 ≤ t < 10, 4, 4. Row 2: 10 ≤ t < 20, 7, 11. Row 3: 20 ≤ t < 30, 18, 29. Row 4: 30 ≤ t < 40, 22, 51. Row 5: 40 ≤ t < 50, 18, 69. Row 6: 50 ≤ t < 60, 11, 80.t (min)fcf0 ≤ t < 104410 ≤ t < 2071120 ≤ t < 30182930 ≤ t < 40225140 ≤ t < 50186950 ≤ t < 601180
4 + 7 = 11, then 11 + 18 = 29, and so on. The last total is all 80 journeys.

Look at the class 10 ≤ t < 20. Every journey in it took under 20 minutes. So the running total of 11 is the number of journeys under 20 minutes. It belongs at 20, the upper boundary of the class, not at its middle.

Draw it first

Draw the cumulative frequency graph

  1. A graph with the x axis time t (min) and the y axis cumulative frequency. Step 1 plots dots at (0, 0), (10, 4), (20, 11), (30, 29), (40, 51), (50, 69) and (60, 80). Step 2 joins them with straight lines. Step 3 marks the median with a dashed line across from 40 and down to 35. The dot is named median.10203040506010203040506070800time t (min)cumulative frequency

    1 Plot (upper boundary, cumulative frequency) for each class. Start at (0, 0), because no journey took under 0 minutes.

  2. A graph with the x axis time t (min) and the y axis cumulative frequency. Step 1 plots dots at (0, 0), (10, 4), (20, 11), (30, 29), (40, 51), (50, 69) and (60, 80). Step 2 joins them with straight lines. Step 3 marks the median with a dashed line across from 40 and down to 35. The dot is named median.10203040506010203040506070800time t (min)cumulative frequency

    2 Join the points with straight lines. The graph never goes down, and ends at 80.

  3. A graph with the x axis time t (min) and the y axis cumulative frequency. Step 1 plots dots at (0, 0), (10, 4), (20, 11), (30, 29), (40, 51), (50, 69) and (60, 80). Step 2 joins them with straight lines. Step 3 marks the median with a dashed line across from 40 and down to 35. The dot is named median.median10203040506010203040506070800time t (min)cumulative frequency

    3 To find the median, go across from 40, half of 80, to the graph. Then go down to the time axis.

The graph lets us read other values. How many journeys took under 25 minutes? Go up from 25 on the time axis to the graph, then across to the cumulative frequency axis. The reading is 20.

Joining the points with straight lines treats the journeys in each class as spread evenly along it. So a reading between two plotted points is an estimate. Some books draw a smooth curve instead, and its readings differ a little.

The same graph, with the x axis time t (min) and the y axis cumulative frequency. A dot named A is at (25, 20). A dot named B is at (40, 51). Dashed lines run from each dot down to the time axis and across to the cumulative frequency axis.AB10203040506010203040506070800time t (min)cumulative frequency
A: about 20 journeys took under 25 minutes. B: 51 journeys took under 40 minutes.

Worked example

How many of the 80 journeys took 40 minutes or more?

  1. Read up from 40 minutes on the time axis. The graph shows 51 journeys.
  2. Check with the table of cumulative frequencies: 4 + 7 + 18 + 22 = 51.
  3. The others took 40 minutes or more. There are 80 journeys in all.
  4. 80 − 51 = 29 journeys.

Answer: 29 journeys

Now the median and the quartiles. The median is the middle value. The lower quartile, Q1, has a quarter of the values below it. The upper quartile, Q3, has three quarters of the values below it.

For 80 values we read the graph at 40 for the median, 20 for Q1 and 60 for Q3. Why 40, and not the 40.5th value of chapter 2? The graph does not show 80 separate journeys. Its height at a time is how many journeys took less than that time. The median is the time with half the journeys below it. Half of 80 is 40, so we read across from 40.

In the same way Q1 has a quarter below it, so we read at 80 ÷ 4 = 20. The graph is an estimate made from classes. Reading at 40.5 instead of 40 gives about 35.2 minutes, not 35, and that small gap does not matter here. For a short list of separate values we count instead, as the next chapter shows.

The same graph, with the x axis time t (min) and the y axis cumulative frequency. Three dots on it have dashed lines to both axes. Q1 is at (25, 20), the median at (35, 40) and Q3 at (45, 60).Q1medianQ310203040506010203040506070800time t (min)cumulative frequency
Q1 is read at 20, the median at 40 and Q3 at 60.

Worked example

Read the median and the two quartiles of the 80 journey times from the graph.

  1. The median is read at 80 ÷ 2 = 40. Across to the graph and down gives an estimate of 35 minutes.
  2. Q1 is read at 80 ÷ 4 = 20. Across to the graph and down gives an estimate of 25 minutes.
  3. Q3 is read at 3 × 80 ÷ 4 = 60. Across to the graph and down gives an estimate of 45 minutes.

Answer: The median is 35 minutes. The lower quartile is 25 minutes and the upper quartile is 45 minutes.

The interquartile range, IQR, is Q3 − Q1. It is the length of the middle half of the data. The range, highest minus lowest, depends on two extreme values. A single odd value can change it a lot. The IQR leaves out the lowest quarter and the highest quarter, so one odd value cannot move it much.

Worked example

Find the interquartile range of the journey times.

  1. From the graph, the upper quartile is 45 minutes.
  2. The lower quartile is 25 minutes.
  3. IQR = Q3 − Q1 = 45 − 25 = 20 minutes.
  4. The middle half of the journeys fit within a spread of 20 minutes.

Answer: 20 minutes

See it three ways

With things

Line up 80 journeys from the shortest to the longest. Split the line into four equal groups. Walk to the end of the first group, the second group, and the third group.

As a picture

Eighty journeys in a grid of four rows, in four groups of 20: lowest quarter, second quarter, third quarter and highest quarter. The 20th, 40th and 60th journeys are framed. They mark Q1, the median and Q3.lowest quartersecond quarterthird quarterhighest quarter

In symbols

80 ÷ 4 = 20. 80 ÷ 2 = 40. 3 × 80 ÷ 4 = 60. Read the graph at 20, 40 and 60.

Try it

Graph the lengths of 20 sticks

You need: 20 sticks between 10 cm and 60 cm long, a ruler, squared paper.

  1. Measure each stick. Sort them into classes: 10 ≤ L < 20, 20 ≤ L < 30 and so on up to 60.
  2. Write the frequencies, then the running totals.
  3. Plot (upper boundary, running total), starting at (10, 0). Join the points with straight lines. Read the median at 10.

What you will see: The line never goes down, because a running total never goes down. It ends at 20, the number of sticks. Here, the median is about 35 cm.

If you cannot try it, this is what happens.
A cumulative frequency graph for 20 sticks. The x axis is length L (cm) from 0 to 60. The y axis is cumulative frequency from 0 to 20. Straight lines join dots at (10, 0), (20, 2), (30, 6), (40, 14), (50, 18) and (60, 20). The dot named median is at (35, 10), with dashed lines to both axes.median102030405060481216200length L (cm)cumulative frequency

A wrong idea many people have

Plot each cumulative frequency at the middle of its class.

Why it feels right: Bar charts and mean calculations use the middle of a class, so it seems natural here too.

The test that breaks it: The running total of 11 for 10 ≤ t < 20 counts journeys that took under 20 minutes. If it were plotted at 15, the graph would say that 11 journeys took under 15 minutes. That is not known.

What is true: A running total up to the end of a class belongs at the upper boundary of that class.

Key words

cumulative frequency
a running total of the frequencies.
upper boundary
the top end of a class, where its cumulative frequency is plotted.
quartile
a value that splits ordered data into quarters. Q1 has a quarter below it, Q3 a quarter above.
interquartile range
the upper quartile minus the lower quartile, the length of the middle half of the data.

Back to your guess

The answer is C: Plot the running total of the frequencies, then read across from half of 80, which is 40.

A running total of the frequencies is plotted, and the graph is read at 40, half of 80. Reading across from 40 to the graph and down to the time axis gives the median. The middle class by position can be the wrong class. The class with the highest frequency is the modal class, not the median.

Check yourself

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

  1. Heights of 60 plants are in classes. Which point do you plot for the class 20 ≤ h < 30?

    The heights of 60 plants in classes. The head is h (cm), f. Row 1: 0 ≤ h < 10, 4. Row 2: 10 ≤ h < 20, 5. Row 3: 20 ≤ h < 30, 15. Row 4: 30 ≤ h < 40, 15. Row 5: 40 ≤ h < 50, 12. Row 6: 50 ≤ h < 60, 9.h (cm)f0 ≤ h < 10410 ≤ h < 20520 ≤ h < 301530 ≤ h < 401540 ≤ h < 501250 ≤ h < 609
    Tap your answer.

    1. Not this one. The running total belongs at the upper boundary of the class, not at its middle. All 24 plants are under 30 cm. Go to M4.7.3. Try another.

    2. Not this one. 15 is the frequency of this class. A cumulative frequency adds the classes before it: 4 + 5 + 15. Go to M4.7.3. Try another.

    3. Not this one. The running total of 24 counts plants under 30 cm, so it is plotted at 30, not at the lower boundary 20. Go to M4.7.3. Try another.

    4. Yes, that is right. The running total up to this class is 4 + 5 + 15 = 24 plants. They are all under 30 cm, so the point is at the upper boundary: (30, 24).

    See the answer

    (30, 24)

    The running total up to this class is 4 + 5 + 15 = 24 plants. They are all under 30 cm, so the point is at the upper boundary: (30, 24).

    • If you chose A: The running total belongs at the upper boundary of the class, not at its middle. All 24 plants are under 30 cm. Go to M4.7.3.
    • If you chose B: 15 is the frequency of this class. A cumulative frequency adds the classes before it: 4 + 5 + 15. Go to M4.7.3.
    • If you chose C: The running total of 24 counts plants under 30 cm, so it is plotted at 30, not at the lower boundary 20. Go to M4.7.3.

  2. This graph shows the heights of the 60 plants. What is the median height?

    A cumulative frequency graph of 60 plants. The x axis is height h (cm) from 0 to 60. The y axis is plants from 0 to 60. Straight lines join dots at (0, 0), (10, 4), (20, 9), (30, 24), (40, 39), (50, 51) and (60, 60).102030405060153045600height h (cm)plants
    Tap your answer.

    1. Yes, that is right. Half of 60 is 30. Read across from 30 to the graph, then down: the median is 34 cm.

    2. Not this one. 30 is the place to read at, half of 60. You must then go across to the graph and down to the height axis. Go to M4.7.3. Try another.

    3. Not this one. 24 is the cumulative frequency at a height of 30 cm. It is a number of plants, not a height. Read across from 30, then down. Go to M4.7.3. Try another.

    4. Not this one. 45 is the place where the upper quartile is read, 3 × 60 ÷ 4. The median is read at half of 60. Go to M4.7.3. Try another.

    See the answer

    34 cm

    Half of 60 is 30. Read across from 30 to the graph, then down: the median is 34 cm.

    • If you chose B: 30 is the place to read at, half of 60. You must then go across to the graph and down to the height axis. Go to M4.7.3.
    • If you chose C: 24 is the cumulative frequency at a height of 30 cm. It is a number of plants, not a height. Read across from 30, then down. Go to M4.7.3.
    • If you chose D: 45 is the place where the upper quartile is read, 3 × 60 ÷ 4. The median is read at half of 60. Go to M4.7.3.

  3. For some data, Q1 = 22 and Q3 = 41. What is the interquartile range?

    Tap your answer.

    1. Not this one. You added the two quartiles. The interquartile range is the difference, Q3 − Q1. Go to M4.7.3. Try another.

    2. Not this one. 41 is the upper quartile itself. The interquartile range is how far Q3 is above Q1. Go to M4.7.3. Try another.

    3. Yes, that is right. IQR = Q3 − Q1 = 41 − 22 = 19.

    See the answer

    19

    IQR = Q3 − Q1 = 41 − 22 = 19.

    • If you chose A: You added the two quartiles. The interquartile range is the difference, Q3 − Q1. Go to M4.7.3.
    • If you chose B: 41 is the upper quartile itself. The interquartile range is how far Q3 is above Q1. Go to M4.7.3.

  4. Why is the interquartile range often a better measure of spread than the range?

    Tap your answer.

    1. Not this one. The IQR uses only the middle half of the data. That is the reason one odd value cannot change it much. Go to M4.7.3. Try another.

    2. Yes, that is right. The range depends on the two most extreme values. The IQR looks at the middle half only.

    3. Not this one. The IQR is part of the range, so it is never bigger. The middle half cannot be wider than all of the data. Go to M4.7.3. Try another.

    See the answer

    It leaves out the lowest and highest quarters, so one odd value cannot change it much.

    The range depends on the two most extreme values. The IQR looks at the middle half only.

    • If you chose A: The IQR uses only the middle half of the data. That is the reason one odd value cannot change it much. Go to M4.7.3.
    • If you chose C: The IQR is part of the range, so it is never bigger. The middle half cannot be wider than all of the data. Go to M4.7.3.

  5. Use the graph of the 60 plants. How many plants are shorter than 40 cm?

    The cumulative frequency graph of 60 plants. The x axis is height h (cm) from 0 to 60. The y axis is plants from 0 to 60. Dots at (0, 0), (10, 4), (20, 9), (30, 24), (40, 39), (50, 51) and (60, 60) are joined by straight lines.102030405060153045600height h (cm)plants
    Tap your answer.

    1. Not this one. 15 is the frequency of the class 30 ≤ h < 40. Plants under 40 cm include the lower classes too. Go to M4.7.3. Try another.

    2. Not this one. 21 plants are 40 cm or more, since 60 − 39 = 21. The question asks how many are shorter. Go to M4.7.3. Try another.

    3. Yes, that is right. Read up from 40 on the height axis. The graph is at 39 plants.

    4. Not this one. 24 is the reading at 30 cm. The question asks about 40 cm. Go to M4.7.3. Try another.

    See the answer

    39

    Read up from 40 on the height axis. The graph is at 39 plants.

    • If you chose A: 15 is the frequency of the class 30 ≤ h < 40. Plants under 40 cm include the lower classes too. Go to M4.7.3.
    • If you chose B: 21 plants are 40 cm or more, since 60 − 39 = 21. The question asks how many are shorter. Go to M4.7.3.
    • If you chose D: 24 is the reading at 30 cm. The question asks about 40 cm. Go to M4.7.3.

  6. A graph shows 80 values. At which cumulative frequency do you read the upper quartile?

    Tap your answer.

    1. Not this one. You read the lower quartile, a quarter of 80. The upper quartile is three quarters of the way up. Go to M4.7.3. Try another.

    2. Not this one. You read the median, half of 80. The upper quartile is higher up. Go to M4.7.3. Try another.

    3. Not this one. You took 75% of 100. The quartile place depends on the number of values, so it is three quarters of 80. Go to M4.7.3. Try another.

    4. Yes, that is right. The upper quartile is three quarters of the way up: 3 × 80 ÷ 4 = 60.

    See the answer

    60

    The upper quartile is three quarters of the way up: 3 × 80 ÷ 4 = 60.

    • If you chose A: You read the lower quartile, a quarter of 80. The upper quartile is three quarters of the way up. Go to M4.7.3.
    • If you chose B: You read the median, half of 80. The upper quartile is higher up. Go to M4.7.3.
    • If you chose C: You took 75% of 100. The quartile place depends on the number of values, so it is three quarters of 80. Go to M4.7.3.

  7. From an earlier chapter

    Using the table of 80 bus journeys, estimate the mean journey time. Use the midpoints of the classes.

    See the answer

    34.5 minutes

    The midpoints are 5, 15, 25, 35, 45 and 55. 5 × 4 + 15 × 7 + 25 × 18 + 35 × 22 + 45 × 18 + 55 × 11 = 20 + 105 + 450 + 770 + 810 + 605 = 2 760. Then 2 760 ÷ 80 = 34.5.

  8. The lower quartile of some data is 31 and the upper quartile is 52. What is the interquartile range?

    See the answer

    21

    IQR = Q3 − Q1 = 52 − 31 = 21.

  9. A cumulative frequency graph of 60 values reads 42 at a height of 40 cm. How many values are 40 cm or more?

    See the answer

    18

    42 values are under 40 cm. The rest are 40 cm or more: 60 − 42 = 18.

Keep it

Copy this into your exercise book.

Cumulative frequency is a running total, plotted at the upper boundary. Read the median at n ÷ 2, Q1 at n ÷ 4, Q3 at 3n ÷ 4. IQR = Q3 − Q1.

Draw: A rising line through points, with dashed lines from 20, 40 and 60 on the upright axis down to the other axis.

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.