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Averages and Range

Finding the mean, median, mode and range, using frequency tables and grouped data, and choosing the best average.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1Calculate the mean, median, mode and range of a small data set.
  2. 2Choose the most suitable average and explain the effect of an outlier.
  3. 3Find the mean, median and mode from a frequency table.
  4. 4Estimate the mean from grouped data using mid-points, and find the modal class and median class.

One number to describe the data

An average gives one value that stands for a whole set of data, and the range tells you how spread out the data is. Statistics questions on a non-calculator paper use small sets of numbers, so you need to be accurate rather than fast: write the numbers in order, show your adding up, and always say what the answer means. Frequency tables are where many marks are lost, because the mean needs the frequency multiplied by the value, not just the values added up.

Choosing an average

Each average has strengths and weaknesses, and exam questions often ask you to choose one.

  • Mean

    Uses every value, but one extreme value, an outlier, can pull it up or down.

  • Median

    Not affected by outliers, so it is best when there are some very large or very small values.

  • Mode

    The only average for non-numerical data, such as the most popular colour, but it may not exist or may not be unique.

  • Range

    A measure of spread, not an average. A smaller range means the data is more consistent.

Comparing with a mean and range

Sam's five test scores are 12, 15, 15, 18, 20. Tina's five test scores are 8, 14, 15, 20, 23. Compare their scores using the mean and the range.

Show the solutionHide the solution
  1. 1 Sam's mean \(\dfrac{12 + 15 + 15 + 18 + 20}{5} = \dfrac{80}{5} = 16\).
  2. 2 Tina's mean \(\dfrac{8 + 14 + 15 + 20 + 23}{5} = \dfrac{80}{5} = 16\).
  3. 3 Ranges Sam: \(20 - 12 = 8\). Tina: \(23 - 8 = 15\).
  4. 4 Conclusion The means are equal, so on average they score the same, but Sam's smaller range shows more consistent scores.

AnswerBoth have a mean of 16, but Sam's range of 8 is smaller than Tina's 15, so Sam is more consistent

Frequency tables

When values are repeated, a frequency table saves writing them all out, but the method changes.

  • Mean

    Multiply each value by its frequency, add up the products, and divide by the total frequency. \(\bar{x} = \dfrac{\sum fx}{\sum f}\).

  • Median

    The total frequency is \(n\), so the median is the \(\dfrac{n + 1}{2}\)th value. Use the cumulative frequencies to find which value that is.

  • Mode

    The value with the highest frequency.

  • Range

    Largest value minus smallest value, not the largest frequency.

Mean from a frequency table

The table shows the number of pets owned by 20 students. Pets 0, 1, 2, 3, 4 have frequencies 4, 4, 6, 4, 2. Work out the mean, median and mode.

Show the solutionHide the solution
  1. 1 Products \(0 \times 4 = 0\), \(1 \times 4 = 4\), \(2 \times 6 = 12\), \(3 \times 4 = 12\), \(4 \times 2 = 8\), which add up to 36.
  2. 2 Mean \(\dfrac{36}{20} = 1.8\).
  3. 3 Median There are 20 values, so the median is halfway between the 10th and 11th. Cumulative frequencies are 4, 8, 14, 18, 20, so the 10th and 11th are both 2, and the median is 2.
  4. 4 Mode The highest frequency is 6, for 2 pets, so the mode is 2.

AnswerMean 1.8, median 2, mode 2

Grouped data

When data is in groups, you do not know the exact values, so you estimate.

  • Mid-points

    Use the middle of each class as its value. For \(20 < w \leq 40\) the mid-point is 30.

  • Estimated mean

    Multiply each mid-point by its frequency, add up, and divide by the total frequency.

  • Modal class

    The class with the highest frequency.

  • Median class

    The class that contains the middle value, found with the cumulative frequency.

Estimating a mean from grouped data

The weights of 20 parcels are 5 in the class \(0 < w \leq 20\), 10 in \(20 < w \leq 40\) and 5 in \(40 < w \leq 60\). Estimate the mean weight.

Show the solutionHide the solution
  1. 1 Mid-points 10, 30 and 50.
  2. 2 Products \(10 \times 5 = 50\), \(30 \times 10 = 300\) and \(50 \times 5 = 250\).
  3. 3 Total \(50 + 300 + 250 = 600\).
  4. 4 Estimated mean \(\dfrac{600}{20} = 30\) kg. It is only an estimate, because the true values are not known.

Answer30 kg

Working backwards from a mean

The total is the mean times the number of values.

  • Total from a mean

    If 5 numbers have a mean of 8, they total \(5 \times 8 = 40\).

  • Adding a value

    Adding a sixth number to give a mean of 9 means the new total is \(6 \times 9 = 54\), so the new number is \(54 - 40 = 14\).

  • Combined means

    To combine groups, find each total, add the totals, and divide by all the values.

  • Missing value

    A set with a known mean and known values leaves the missing value as total minus the others.

Test yourself

  1. 1

    How do you find the median of 10 ordered values?

    Show answerHide answer

    The mean of the 5th and 6th.

  2. 2

    What is the range of 3, 8, 11, 20?

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    \(20 - 3 = 17\).

  3. 3

    How do you find the mean from a frequency table?

    Show answerHide answer

    Divide the total of frequency times value by the total frequency.

  4. 4

    Which average is least affected by an outlier?

    Show answerHide answer

    The median.

  5. 5

    What is the mid-point of the class \(10 < t \leq 20\)?

    Show answerHide answer

    15.

Exam technique: averages

Show the working, and say what the answer means.

  • Order the data

    Write the values in order before finding the median.

  • Show the products

    In a frequency table, show each \(f \times x\) separately.

  • Give a conclusion

    "On average" and "more consistent" are needed in comparison questions.

  • Compare like with like

    Compare an average with an average, and a range with a range.

Summary and exam focus

  • The mean, median and mode are averages, and the range is a measure of spread.
  • The median is not affected by outliers, and a smaller range means more consistent data.
  • With a frequency table, the mean is \(\dfrac{\sum fx}{\sum f}\), and the median is the \(\dfrac{n + 1}{2}\)th value.
  • For grouped data use mid-points to estimate the mean.

Exam focus

The mean of five numbers is 8. Four of the numbers are 5, 7, 9 and 10. Work out the fifth number. (2 marks) (2 marks)

The total of five numbers with a mean of 8 is \(5 \times 8 = 40\). The four known numbers add up to 31, so the fifth number is \(40 - 31 = 9\). Do not average the four known numbers.

Key terms

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

Mean
The sum of the values divided by how many there are.
Median
The middle value when the data is in order.
Mode
The value that occurs most often.
Range
The difference between the largest and smallest values.
Outlier
A value that is much larger or smaller than the others.
Frequency table
A table showing how often each value occurs.
Grouped data
Data sorted into classes, such as \(0 < t \leq 10\).
Mid-point
The middle value of a class, used to estimate the mean of grouped data.
Modal class
The class with the highest frequency.

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