Chapter 2 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

M4.7.2 Chapter 2

Averages from frequency tables

Mei opens 25 pea pods and counts the peas in each. How can she find the mean number of peas without adding 25 numbers?

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

  • Find the mean from a frequency table
  • Estimate the mean from grouped data
  • Find the group that holds the median

Mei opens 25 pods. Instead of writing 25 numbers, she writes a frequency table. It lists each number of peas, and the frequency, f, which is how many pods have that number.

The number of peas in 25 pods. The head is peas x, pods f. Row 1: 3, 2. Row 2: 4, 5. Row 3: 5, 9. Row 4: 6, 6. Row 5: 7, 3.peas xpods f3245596673
Two pods have 3 peas. Five pods have 4 peas. Nine pods have 5 peas.

To find the mean we need two totals: the number of peas, and the number of pods. A row of the table is a group of equal values. Nine pods with 5 peas each hold 9 × 5 = 45 peas. So multiply x by f for every row.

Draw it first

Build the mean from the table

  1. Step 1: the table of x and f for 25 pods. The head is peas x, pods f. Row 1: 3, 2. Row 2: 4, 5. Row 3: 5, 9. Row 4: 6, 6. Row 5: 7, 3. Step 2: a third column f × x and a total row. The head is peas x, pods f, f × x. Row 1: 3, 2, 6. Row 2: 4, 5, 20. Row 3: 5, 9, 45. Row 4: 6, 6, 36. Row 5: 7, 3, 21. Row 6: total, 25, 128.peas xpods f3245596673

    1 Write each number of peas, x, and its frequency, f.

  2. Step 1: the table of x and f for 25 pods. The head is peas x, pods f. Row 1: 3, 2. Row 2: 4, 5. Row 3: 5, 9. Row 4: 6, 6. Row 5: 7, 3. Step 2: a third column f × x and a total row. The head is peas x, pods f, f × x. Row 1: 3, 2, 6. Row 2: 4, 5, 20. Row 3: 5, 9, 45. Row 4: 6, 6, 36. Row 5: 7, 3, 21. Row 6: total, 25, 128.peas xpods ff × x3264520594566367321total25128

    2 Add a column f × x. Add up the f column and the f × x column. The totals are 25 pods and 128 peas.

The mean is the total of all the values divided by how many values there are. In symbols, mean = ∑fx ÷ ∑f. The sign ∑ means "add up". So ∑fx = 128 is the total number of peas, and ∑f = 25 is the number of pods.

Worked example

Find the mean number of peas in a pod, from the table of 25 pods.

  1. Multiply each x by f: 3 × 2 = 6, 4 × 5 = 20, 5 × 9 = 45, 6 × 6 = 36 and 7 × 3 = 21.
  2. Add the f × x column: 6 + 20 + 45 + 36 + 21 = 128. That is the number of peas.
  3. Add the frequencies: 2 + 5 + 9 + 6 + 3 = 25. That is the number of pods.
  4. Mean = 128 ÷ 25 = 5.12 peas in a pod.

Answer: 5.12 peas in a pod

A wrong idea many people have

The mean of a frequency table is the mean of the x column.

Why it feels right: The x column lists the values, so it looks like the data.

The test that breaks it: The x values 3, 4, 5, 6 and 7 have mean 5. But 9 pods have more than 5 peas and only 7 have fewer, so the real mean is a little above 5. It is 5.12.

What is true: Each x counts f times. Multiply x by f first, then divide by the total frequency, not by the number of rows.

See it three ways

With things

A bowl holds 10 pods. Four pods have 2 peas each and six pods have 5 peas each.

As a picture

Ten pods in a grid in two groups: 2 peas, white, 4 pods; 5 peas, dotted, 6 pods.2 peas5 peas

In symbols

f × x: 4 × 2 = 8 and 6 × 5 = 30. ∑fx = 38 peas. ∑f = 10 pods. Mean = 38 ÷ 10 = 3.8.

Now measured data. Times and heights are often grouped in classes. The table of 40 journey times below says 4 journeys took 0 or more minutes but under 10. We no longer know each time, only its class.

The times of 40 journeys, in classes. The midpoint of a class is the number halfway between its ends. The head is t (min), f, midpoint. Row 1: 0 ≤ t < 10, 4, 5. Row 2: 10 ≤ t < 20, 11, 15. Row 3: 20 ≤ t < 30, 15, 25. Row 4: 30 ≤ t < 40, 7, 35. Row 5: 40 ≤ t < 50, 3, 45.t (min)fmidpoint0 ≤ t < 104510 ≤ t < 20111520 ≤ t < 30152530 ≤ t < 4073540 ≤ t < 50345
0 ≤ t < 10 means 0 or more, but under 10 minutes.
A bar chart of 40 journey times. The x axis is time t (min) with class edges 0, 10, 20, 30, 40 and 50. The y axis is journeys. The bars have heights 4, 11, 15, 7 and 3.0481216411157301020304050time t (min)journeys
The classes have equal widths, so the height of a bar is the frequency.

To find a mean we need one value for each class. We use the midpoint, the number halfway between the two ends of the class. For 10 ≤ t < 20 the midpoint is (10 + 20) ÷ 2 = 15. We pretend every journey in the class took 15 minutes.

This is a guess, so the answer is only an estimate. Some journeys took less than 15 minutes and some more. We hope the errors cancel.

Worked example

Estimate the mean journey time from the table of 40 journeys.

  1. The midpoints are 5, 15, 25, 35 and 45.
  2. Multiply each midpoint by f: 5 × 4 = 20, 15 × 11 = 165, 25 × 15 = 375, 35 × 7 = 245 and 45 × 3 = 135.
  3. Add them: 20 + 165 + 375 + 245 + 135 = 940.
  4. Estimate of the mean = 940 ÷ 40 = 23.5 minutes.

Answer: 23.5 minutes

The midpoint x, the frequency f and f × x for each class of the journey times. The head is x, f, f × x. Row 1: 5, 4, 20. Row 2: 15, 11, 165. Row 3: 25, 15, 375. Row 4: 35, 7, 245. Row 5: 45, 3, 135.xff × x542015111652515375357245453135
The f × x column adds up to 940. The f column adds up to 40.

How far can the estimate be wrong? If every journey took the least its class allows, the mean would be 740 ÷ 40 = 18.5. If every journey took almost the most, it would be about 1 140 ÷ 40 = 28.5. The true mean lies between 18.5 and 28.5, and 23.5 sits in the middle.

A number line of times in minutes, from 18 to 29. Three dots are labelled 18.5, 23.5 and 28.5. They mark the least the mean could be, the estimate, and the most it could be.18192021222324252627282918.523.528.5
Times in minutes. The dots are the least the mean could be, the estimate, and the most it could be.

Now the median. With n values in order, the median is the (n + 1) ÷ 2 th value. For 25 pods it is the 13th pod. From a table we cannot see the 13th value, but we can see which row holds it. Add up the frequencies until the running total reaches 13.

The 25 pods in order of size, numbered 1 to 25 in 3 rows. Pods 1 to 7 have 3 or 4 peas, white. Pods 8 to 16 have 5 peas, dotted. Pods 17 to 25 have 6 or 7 peas, hatched. Pod 13 is framed: it is the median pod.123456789101112131415161718192021222324253 or 4 peas5 peas6 or 7 peas
Pod 13 is the middle one of 25. It lies in the group with 5 peas.

Worked example

Find the class that holds the median of the 40 journey times.

  1. There are 40 values. The median is the (40 + 1) ÷ 2 = 20.5th value.
  2. The 20.5th value is not in the first two classes, which hold 4 + 11 = 15 values.
  3. It is in the third class, because the first three classes hold 15 + 15 = 30 values.
  4. So the median is in the class 20 ≤ t < 30.

Answer: The class 20 ≤ t < 30

We can name the class that holds the median, but not the median itself. The most common class is the modal class, the one with the greatest frequency. Here that is also 20 ≤ t < 30.

Try it

Throw a die 20 times

You need: a die from a game, paper and a pencil.

  1. Throw the die 20 times. After each throw, put a tally mark by that number.
  2. Write the frequency table: x is the face, f is the number of tallies.
  3. Work out f × x for each row, add them up, and divide by 20.

What you will see: The mean is close to 3.5, the mean of the six faces. Your 20 throws may not give exactly 3.5, and that is fine.

If you cannot try it, this is what happens.
One set of 20 throws of a die. The totals are 20 throws and 72 in f × x, so the mean is 3.6. The head is face x, throws f, f × x. Row 1: 1, 2, 2. Row 2: 2, 4, 8. Row 3: 3, 3, 9. Row 4: 4, 5, 20. Row 5: 5, 3, 15. Row 6: 6, 3, 18. Row 7: total, 20, 72.face xthrows ff × x122248339452053156318total2072

Key words

frequency table
a table that lists each value and how many times it happens.
midpoint
the number halfway between the two ends of a class.
estimate
an answer that is close, not exact, because some information is missing.
modal class
the class with the greatest frequency.

Back to your guess

The answer is C: Multiply each number of peas by how many pods have it, add these up, then divide by the number of pods.

The table says 2 pods have 3 peas, 5 pods have 4 peas, and so on. Multiply each number of peas by its number of pods, and add: 128 peas in all. Then divide by the 25 pods. The mean is 128 ÷ 25 = 5.12 peas in a pod.

Check yourself

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

  1. The table shows the seeds in 20 pods. What is the mean number of seeds in a pod?

    The seeds in 20 pods. The head is seeds x, pods f. Row 1: 1, 5. Row 2: 2, 8. Row 3: 3, 4. Row 4: 4, 3.seeds xpods f15283443
    Tap your answer.

    1. Not this one. You found the mean of the numbers 1, 2, 3 and 4, and ignored how many pods have each. Multiply each x by its f first. Go to M4.7.2. Try another.

    2. Not this one. 45 is the total number of seeds, ∑fx. The mean is the total divided by the number of pods. Go to M4.7.2. Try another.

    3. Not this one. You found the mean of the frequencies 5, 8, 4 and 3. The mean wanted is of the seeds in a pod, not of the numbers of pods. Go to M4.7.2. Try another.

    4. Yes, that is right. 1 × 5 + 2 × 8 + 3 × 4 + 4 × 3 = 5 + 16 + 12 + 12 = 45 seeds. There are 5 + 8 + 4 + 3 = 20 pods. 45 ÷ 20 = 2.25.

    See the answer

    2.25

    1 × 5 + 2 × 8 + 3 × 4 + 4 × 3 = 5 + 16 + 12 + 12 = 45 seeds. There are 5 + 8 + 4 + 3 = 20 pods. 45 ÷ 20 = 2.25.

    • If you chose A: You found the mean of the numbers 1, 2, 3 and 4, and ignored how many pods have each. Multiply each x by its f first. Go to M4.7.2.
    • If you chose B: 45 is the total number of seeds, ∑fx. The mean is the total divided by the number of pods. Go to M4.7.2.
    • If you chose C: You found the mean of the frequencies 5, 8, 4 and 3. The mean wanted is of the seeds in a pod, not of the numbers of pods. Go to M4.7.2.

  2. The table shows the heights of 30 plants in classes. Which is the best estimate of the mean height?

    The heights of 30 plants in classes. The head is height h (cm), f. Row 1: 0 ≤ h < 10, 3. Row 2: 10 ≤ h < 20, 9. Row 3: 20 ≤ h < 30, 12. Row 4: 30 ≤ h < 40, 6.height h (cm)f0 ≤ h < 10310 ≤ h < 20920 ≤ h < 301230 ≤ h < 406
    Tap your answer.

    1. Not this one. You found the mean of the midpoints 5, 15, 25 and 35, and ignored the frequencies. More plants are in the middle classes, so the mean is higher. Go to M4.7.2. Try another.

    2. Yes, that is right. The midpoints are 5, 15, 25 and 35. 5 × 3 + 15 × 9 + 25 × 12 + 35 × 6 = 15 + 135 + 300 + 210 = 660. Then 660 ÷ 30 = 22.

    3. Not this one. 660 is the total ∑fx. The mean is the total divided by the number of plants. Go to M4.7.2. Try another.

    4. Not this one. You used the lowest value of each class, not the midpoint. The midpoint stands for the whole class. Go to M4.7.2. Try another.

    See the answer

    22 cm

    The midpoints are 5, 15, 25 and 35. 5 × 3 + 15 × 9 + 25 × 12 + 35 × 6 = 15 + 135 + 300 + 210 = 660. Then 660 ÷ 30 = 22.

    • If you chose A: You found the mean of the midpoints 5, 15, 25 and 35, and ignored the frequencies. More plants are in the middle classes, so the mean is higher. Go to M4.7.2.
    • If you chose C: 660 is the total ∑fx. The mean is the total divided by the number of plants. Go to M4.7.2.
    • If you chose D: You used the lowest value of each class, not the midpoint. The midpoint stands for the whole class. Go to M4.7.2.

  3. Waiting times of 45 visitors are in classes. Which class holds the median?

    The waiting times of 45 visitors in classes. The head is wait t (min), f. Row 1: 0 ≤ t < 10, 12. Row 2: 10 ≤ t < 20, 20. Row 3: 20 ≤ t < 30, 6. Row 4: 30 ≤ t < 40, 4. Row 5: 40 ≤ t < 50, 3.wait t (min)f0 ≤ t < 101210 ≤ t < 202020 ≤ t < 30630 ≤ t < 40440 ≤ t < 503
    Tap your answer.

    1. Yes, that is right. There are 45 values, so the median is the (45 + 1) ÷ 2 = 23rd. The first class holds 12 values. Adding the second gives 12 + 20 = 32, so the 23rd is in the second class.

    2. Not this one. That is the middle class by position, the third of five. The median is the middle value, and 12 + 20 = 32 values already lie below 20. Go to M4.7.2. Try another.

    3. Not this one. This class holds only the first 12 values. The median is the 23rd, so we must go further up. Go to M4.7.2. Try another.

    See the answer

    10 ≤ t < 20

    There are 45 values, so the median is the (45 + 1) ÷ 2 = 23rd. The first class holds 12 values. Adding the second gives 12 + 20 = 32, so the 23rd is in the second class.

    • If you chose B: That is the middle class by position, the third of five. The median is the middle value, and 12 + 20 = 32 values already lie below 20. Go to M4.7.2.
    • If you chose C: This class holds only the first 12 values. The median is the 23rd, so we must go further up. Go to M4.7.2.

  4. What is the midpoint of the class 40 ≤ h < 60?

    Tap your answer.

    1. Not this one. That is the lower end of the class. The midpoint is halfway between the two ends. Go to M4.7.2. Try another.

    2. Not this one. That is the width of the class, 60 − 40. The midpoint is a place on the scale, halfway along it. Go to M4.7.2. Try another.

    3. Yes, that is right. The midpoint is (40 + 60) ÷ 2 = 50.

    4. Not this one. That is the upper end of the class. The midpoint is halfway between 40 and 60. Go to M4.7.2. Try another.

    See the answer

    50

    The midpoint is (40 + 60) ÷ 2 = 50.

    • If you chose A: That is the lower end of the class. The midpoint is halfway between the two ends. Go to M4.7.2.
    • If you chose B: That is the width of the class, 60 − 40. The midpoint is a place on the scale, halfway along it. Go to M4.7.2.
    • If you chose D: That is the upper end of the class. The midpoint is halfway between 40 and 60. Go to M4.7.2.

  5. Why is a mean worked out from grouped data only an estimate?

    Tap your answer.

    1. Not this one. Grouping still counts every value. It hides where each value lies inside its class. Go to M4.7.2. Try another.

    2. Yes, that is right. The exact values are hidden by the classes. The midpoint is our best single guess for each class.

    3. Not this one. Some values are below the midpoint and some above. The errors can go either way and often cancel. Go to M4.7.2. Try another.

    See the answer

    We know only which class each value is in, so we use the midpoint as its value.

    The exact values are hidden by the classes. The midpoint is our best single guess for each class.

    • If you chose A: Grouping still counts every value. It hides where each value lies inside its class. Go to M4.7.2.
    • If you chose C: Some values are below the midpoint and some above. The errors can go either way and often cancel. Go to M4.7.2.

  6. In a frequency table, what does f × x give for one row?

    Tap your answer.

    1. Not this one. The number of values in the row is f, the frequency. Multiplying by x gives their total. Go to M4.7.2. Try another.

    2. Not this one. Every value in the row is x, so the mean of the row is just x. f × x is their total. Go to M4.7.2. Try another.

    3. Yes, that is right. If f values are each equal to x, they add up to f × x. The mean is then the sum of these totals divided by ∑f.

    See the answer

    The total of all the values in that row.

    If f values are each equal to x, they add up to f × x. The mean is then the sum of these totals divided by ∑f.

    • If you chose A: The number of values in the row is f, the frequency. Multiplying by x gives their total. Go to M4.7.2.
    • If you chose B: Every value in the row is x, so the mean of the row is just x. f × x is their total. Go to M4.7.2.

  7. From an earlier chapter

    A garden has 80 plants and 30 of them are herbs. A stratified sample of 16 is wanted. How many herbs are in it?

    See the answer

    6

    Herbs are 30 ÷ 80 of the garden. 30 ÷ 80 × 16 = 6.

  8. Pods hold 1, 2, 3, 4 or 5 peas. The numbers of pods are 3, 5, 6, 4 and 2. What is the mean number of peas in a pod?

    See the answer

    2.85

    1 × 3 + 2 × 5 + 3 × 6 + 4 × 4 + 5 × 2 = 3 + 10 + 18 + 16 + 10 = 57 peas. There are 3 + 5 + 6 + 4 + 2 = 20 pods. 57 ÷ 20 = 2.85.

  9. Plant heights, in four classes of width 10 cm from 0 to 40 cm, have frequencies 2, 6, 8 and 4. Estimate the mean height.

    See the answer

    22 cm

    The classes are 0 ≤ h < 10 and so on, so the midpoints are 5, 15, 25 and 35. 5 × 2 + 15 × 6 + 25 × 8 + 35 × 4 = 10 + 90 + 200 + 140 = 440. Then 440 ÷ 20 = 22.

Keep it

Copy this into your exercise book.

Mean = ∑fx ÷ ∑f. For classes, use midpoints and say estimate. Median place = (n + 1) ÷ 2: add the frequencies until the running total reaches it.

Draw: A table with columns x, f and f × x, and the two totals under 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.