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Maths · Interpreting and representing data

Averages and range

Mean, median and mode each describe a "typical" value in a different way, and the range shows how spread out the data is. From a list, a frequency table or grouped data, the method changes - but the ideas stay the same.

  • 6 key terms
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Last Lesson and Before

Answer each one, then check.

  1. 1

    Work out the mean of 2, 4 and 6.

    Show answerHide answer

    \(\dfrac{2 + 4 + 6}{3} = 4\)

  2. 2

    Put 7, 3, 9, 1 in order.

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    1, 3, 7, 9

  3. 3

    What is the middle value of 1, 3, 7?

    Show answerHide answer

    3

  4. 4

    Last lesson: what is interpolation?

    Show answerHide answer

    Estimating inside the range of the data.

  5. 5

    Round 12.46 to 1 decimal place.

    Show answerHide answer

    \(12.5\)

Learning Objectives

  1. 1Find the mean, median, mode and range of a set of data.
  2. 2Choose the most suitable average.
  3. 3Find averages and the range from a frequency table.
  4. 4Estimate the mean of grouped data, and find the modal class and the class containing the median.
  5. 5Solve problems with missing values and combined means.

Three Averages and a Spread

  • Mean

    Add up all the values and divide by how many there are: \(\text{mean} = \dfrac{\text{total}}{\text{number of values}}\).

  • Median

    The middle value when the data is in order. With an even number of values, the mean of the middle two.

  • Mode

    The most common value. There can be more than one, or none.

  • Range

    Largest value \(-\) smallest value. It measures spread, not average.

All Four from a List

Find the mean, median, mode and range of 4, 8, 3, 9, 8, 6, 12.

Show the solutionHide the solution
  1. 1 Mean: add them and divide by 7 \(\dfrac{50}{7} = 7.14\) (2 d.p.)
  2. 2 Median: put them in order 3, 4, 6, 8, 8, 9, 12: the middle (4th) value is 8
  3. 3 Mode: the most common 8 appears twice
  4. 4 Range: largest − smallest \(12 - 3 = 9\)

AnswerMean 7.14, median 8, mode 8, range 9

Which Average?

Mean

  • Uses every value.
  • Pulled up or down by outliers.
  • Best when there are no extreme values.

Median and mode

  • The median is not affected by outliers - best for data like house prices or salaries.
  • The mode is the only average for qualitative data, such as favourite colour.
  • The mode can be unhelpful if every value is different.

Pets Owned by 30 Students

Add a column for frequency \(\times\) value: it gives the total number of pets.

  • 0

    Frequency (f): 8. f × x: 0

  • 1

    Frequency (f): 12. f × x: 12

  • 2

    Frequency (f): 6. f × x: 12

  • 3

    Frequency (f): 3. f × x: 9

  • 4

    Frequency (f): 1. f × x: 4

  • Total

    Frequency (f): 30. f × x: 37

Averages from a Frequency Table

Use the pets table to find the mean, median, mode and range.

Show the solutionHide the solution
  1. 1 Mean = total of \(f \times x\) \(\div\) total frequency \(\dfrac{37}{30} = 1.23\) (2 d.p.)
  2. 2 Median: 30 values, so it is halfway between the 15th and 16th The first 8 are 0; values 9 to 20 are 1
  3. 3 So the 15th and 16th values are both 1 Median = 1
  4. 4 Mode: the value with the highest frequency 1 (frequency 12)
  5. 5 Range: largest value − smallest value \(4 - 0 = 4\)

AnswerMean 1.23, median 1, mode 1, range 4

Journey Times of 30 Students

Each class is represented by its midpoint.

  • \(0 < t \le 10\)

    Frequency (f): 5. Midpoint: 5. f × midpoint: 25

  • \(10 < t \le 20\)

    Frequency (f): 12. Midpoint: 15. f × midpoint: 180

  • \(20 < t \le 30\)

    Frequency (f): 9. Midpoint: 25. f × midpoint: 225

  • \(30 < t \le 40\)

    Frequency (f): 4. Midpoint: 35. f × midpoint: 140

  • Total

    Frequency (f): 30. f × midpoint: 570

Estimating the Mean of Grouped Data

Use the journey times table to estimate the mean, and find the modal class and the class containing the median.

Show the solutionHide the solution
  1. 1 Use the midpoint of each class 5, 15, 25, 35
  2. 2 Multiply by the frequencies and add \(25 + 180 + 225 + 140 = 570\)
  3. 3 Divide by the total frequency \(\dfrac{570}{30} = 19\)
  4. 4 Modal class: the class with the highest frequency \(10 < t \le 20\)
  5. 5 Median: the 15.5th value. Running total: 5, then \(5 + 12 = 17\) so it is in \(10 < t \le 20\)

AnswerEstimated mean 19 minutes; modal class \(10 < t \le 20\); the median is in \(10 < t \le 20\).

Why Only an Estimate?

With grouped data, you do not know the actual values.

  • The midpoint stands in

    We assume every value in a class is at its midpoint.

  • Some are higher, some lower

    The errors tend to cancel out, so the estimate is usually close.

  • In an exam

    "Explain why your answer is an estimate" wants: "the exact values are not known, so the midpoints were used".

Missing Values and Combined Means

Mean \(\times\) number of values = total. That one fact solves most problems.

  • Missing value

    The mean of 5 numbers is 8, so they add up to \(5 \times 8 = 40\). If four of them add up to 31, the fifth is 9.

  • Adding a value

    The mean of 4 numbers is 9 (total 36). After adding one, the mean of 5 is 10 (total 50). The new number is \(50 - 36 = 14\).

  • Combined means

    Class A: 20 students, mean 64 (total 1280). Class B: 25 students, mean 73 (total 1825). Mean of all 45: \(\dfrac{3105}{45} = 69\).

  • Never average the averages

    \(\dfrac{64 + 73}{2} = 68.5\) is wrong: the classes are different sizes.

Choose Your Average

For each data set, find the mean, median and mode, then decide which is the best average and why. (a) Salaries at a small firm: £22 000, £24 000, £24 000, £26 000, £28 000, £150 000. (b) Shoe sizes sold by a shop in one hour: 5, 6, 6, 7, 7, 7, 8, 9. (c) Favourite fruit of 10 people: apple 4, banana 3, grape 2, pear 1.

1. Work out all three averages.

2. Look for outliers and the type of data.

3. Choose and justify.

A good answer shows: (a) Mean £45 667, median £25 000, mode £24 000 - the median, because the £150 000 outlier pulls the mean up. (b) Mean 6.875, median 7, mode 7 - the mode, because a shop orders whole shoe sizes. (c) Only the mode (apple) makes sense for qualitative data.

Can I...?

  1. 1Find the mean, median, mode and range of a list.
  2. 2Choose the best average.
  3. 3Find the mean from a frequency table.
  4. 4Find the median from a frequency table.
  5. 5Estimate the mean of grouped data.
  6. 6Find the modal class.
  7. 7Find the class containing the median.
  8. 8Solve missing value and combined mean problems.

Summary & Exam Focus

  • Mean = total \(\div\) number of values; median = middle; mode = most common; range = largest − smallest.
  • Frequency table: \(\text{mean} = \dfrac{\sum fx}{\sum f}\).
  • Grouped data: use midpoints - the mean is only an estimate.
  • Mean \(\times\) number of values = total: the key to missing values and combined means.

Exam focus

Work out an estimate for the mean weight of the 40 parcels. (4 marks) (4 marks)

For grouped data, add a midpoint column and an \(f \times \text{midpoint}\) column to the table. Then divide by the total FREQUENCY, not by the number of classes.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Mean
The total of the values divided by the number of values.
Median
The middle value when the data is in order.
Mode
The most common value.
Range
The difference between the largest and smallest values.
Modal class
The class with the highest frequency in grouped data.
Midpoint
The value halfway through a class, used to estimate the mean.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 2 marks

    Here are six numbers: 7, 3, 9, 4, 7, 12. (a) Find the median. (b) Find the range.

    Show answerHide answer

    Model answer

    (a) In order: 3, 4, 7, 7, 9, 12. The median is \(\dfrac{7 + 7}{2} = 7\). (b) \(12 - 3 = 9\).

    Mark scheme

    • (a) 7 — B1
    • (b) 9 — B1
  2. Question 2 Calculator 3 marks

    The table shows the number of goals scored by a football team in 20 matches. Goals 0: 4 matches. Goals 1: 7 matches. Goals 2: 5 matches. Goals 3: 3 matches. Goals 4: 1 match. Work out the mean number of goals per match.

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    Model answer

    Total goals: \(0 \times 4 + 1 \times 7 + 2 \times 5 + 3 \times 3 + 4 \times 1 = 30\). Mean \(= \dfrac{30}{20} = 1.5\).

    Mark scheme

    • At least three products of goals \(\times\) frequency correct — M1
    • \(30 \div 20\) — M1
    • \(1.5\) — A1
  3. Question 3 Calculator 4 marks

    The weights, \(w\) kg, of 40 parcels are recorded. \(0 < w \le 2\): 6 parcels. \(2 < w \le 4\): 15 parcels. \(4 < w \le 6\): 12 parcels. \(6 < w \le 8\): 7 parcels. Work out an estimate for the mean weight of the parcels.

    Show answerHide answer

    Model answer

    Midpoints 1, 3, 5, 7. \(6 \times 1 + 15 \times 3 + 12 \times 5 + 7 \times 7 = 6 + 45 + 60 + 49 = 160\). \(\dfrac{160}{40} = 4\) kg.

    Mark scheme

    • Midpoints used: 1, 3, 5, 7 — M1
    • At least three products of frequency \(\times\) midpoint correct — M1
    • \(160 \div 40\) — M1
    • 4 (kg) — A1
  4. Question 4 Calculator 1 mark

    For the parcels in the last question, write down the modal class.

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    Model answer

    \(2 < w \le 4\)

    Mark scheme

    • \(2 < w \le 4\) — B1
  5. Question 5 Non-calculator 2 marks

    The mean of five numbers is 8. Four of the numbers are 3, 7, 9 and 12. Find the fifth number.

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    Model answer

    Total \(= 5 \times 8 = 40\). \(3 + 7 + 9 + 12 = 31\). The fifth number is \(40 - 31 = 9\).

    Mark scheme

    • 40 or 31 seen — M1
    • 9 — A1
  6. Question 6 Calculator 3 marks

    Class A has 20 students. Their mean test score is 64. Class B has 25 students. Their mean test score is 73. Work out the mean score of all 45 students.

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    Model answer

    Class A total: \(20 \times 64 = 1280\). Class B total: \(25 \times 73 = 1825\). Mean \(= \dfrac{1280 + 1825}{45} = \dfrac{3105}{45} = 69\).

    Mark scheme

    • 1280 or 1825 — P1
    • \((1280 + 1825) \div 45\) — P1
    • 69 — A1

Quick check

  1. What is the median of 2, 9, 4, 7, 5?

    1. A4
    2. B5
    3. C7
    4. D5.4
    Show answerHide answer

    B: 5

    In order: 2, 4, 5, 7, 9. The middle value is 5.

  2. Which average is best for data with an extreme outlier, like salaries?

    1. AThe mean
    2. BThe range
    3. CThe median
    4. DThe mode
    Show answerHide answer

    C: The median

    The median is not pulled up or down by an outlier.

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

    1. AThe exact values are not known, so midpoints are used
    2. BThe frequencies might be wrong
    3. CThere are too many classes
    4. DThe range is too large
    Show answerHide answer

    A: The exact values are not known, so midpoints are used

    The exact values are not known, so the midpoint of each class is used instead.

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