Math Vale

📊 Statistics Steppe · Statistics

Mean, Median & Mode

Summarise a data set with a single number, and choose the summary that tells the truth about the data you actually have.

In short

  • Mean = total / count; it must land between the smallest and largest values, and it need not be one of them.
  • The median is the middle of the ORDERED data — one value if the count is odd, the average of two if it is even.
  • The mode is the most frequent value itself, not its frequency, and a data set may have several modes or none.
  • An outlier drags the mean but barely moves the median, so lopsided data is better summarised by the median.

Three ways to say "typical"

An average is a single number standing in for a whole data set. There are three in common use, and they answer three different questions.

  • The mean shares the total out equally. Add every value and divide by how many there are.
  • The median is the middle value once the data is in order.
  • The mode is the value that occurs most often.

For 4, 5, 5, 6, 30 the mean is 10, the median is 5 and the mode is 5. All three are correct; they simply describe different things about the same five numbers. Which one is *honest* is a separate question, and the last section is about that.

The mean

mean = total / number of values

For 12, 15, 18, 19: total = 64, count = 4, mean = 16.

Two checks are worth making every time. The mean must sit between the smallest and largest values, so a mean of 40 for that data is wrong before you look at your working. And a mean is not usually one of the values in the set — 16 is not in the list above, and that is fine.

The definition runs backwards too, which is how a missing value is recovered. If five days have a mean of 20, the five days must total 20 x 5 = 100. If the four known days come to 78, the fifth was 100 - 78 = 22. Work with totals, not with averages of averages.

The median and the mode

To find the median, put the data in order first — this is the step people skip, and skipping it makes the answer meaningless.

  • With an odd count, the median is the single middle value. With 7 values it is the 4th.
  • With an even count, there are two middle values, and the median is halfway between them: their mean. With 8 values it is the average of the 4th and the 5th, so it may end in .5 even when every value is a whole number.

The mode needs no ordering, only counting, but a tally table makes it obvious. A data set can have two modes, or none at all if every value appears once. And the mode is the *value*, not how often it appeared: if 13 turns up three times, the mode is 13, not 3.

Outliers, and choosing an average

An outlier is a value far from the rest. The mean and the median deal with one very differently.

The mean uses every value at full weight, so a single extreme value drags it. The median only cares which value sits in the middle, so it barely moves.

Take 32, 34, 34, 35, 38, 630. The mean is about 133 — a figure no member is anywhere near. The median is 34.5, which describes a typical member well.

So: when the data has an outlier or is strongly lopsided, the median is the honest summary. When the data is fairly symmetric with no extremes, the mean uses all the information and is the better choice. The mode is the only average available for categories that are not numbers at all — the most common crop, the most popular route.

Worked examples

Example 1

Find the median of 69, 49, 94, 133, 1, 72, 126, 40.

  1. Order the data first: 1, 40, 49, 69, 72, 94, 126, 133.
  2. Count the values: there are 8, which is even, so there are two middle ones.
  3. With 8 values the middles are the 4th and the 5th: 69 and 72.
  4. Median = (69 + 72) / 2 = 141 / 2 = 70.5.
  5. Sense check: four values lie below 70.5 and four above, which is exactly what a median should do.

Example 2

A porter worked 6 days. On five of them he carried 12, 44, 16, 13 and 65 sacks. The mean over all six days was exactly 31. How many did he carry on the sixth?

  1. Run the definition of the mean backwards: mean x count = total.
  2. All six days must total 31 x 6 = 186 sacks.
  3. The five known days total 12 + 44 + 16 + 13 + 65 = 150 sacks.
  4. The missing day makes up the shortfall: 186 - 150 = 36 sacks.
  5. Check: (150 + 36) / 6 = 186 / 6 = 31, which is the stated mean.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Over 4 days the tally of stones cleared was: 8, 6, 10, 4 What is the mean?

Answer: 7 stones

  1. Total = 8 + 6 + 10 + 4 = 28
  2. There are 4 values, so mean = 28 / 4
  3. Mean = 7 stones
  4. Sense check: the mean should sit between the smallest value (4) and the largest (10), and it does.

Problem 2

Difficulty 3 of 5

A digger recorded the stones cleared on 6 days: 57, 28, 34, 48, 27, 37 What is the median?

Answer: 35.5 stones

  1. In order: 27, 28, 34, 37, 48, 57
  2. 6 values, so the median is the mean of the 3rd and 4th: (34 + 37) / 2.
  3. Median = 35.5 stones

Problem 3

Difficulty 4 of 5

A digger recorded the stones cleared on 9 days: 16, 24, 24, 16, 3, 3, 24, 3, 24 What is the mode?

Answer: 24 stones

  1. In order: 3, 3, 3, 16, 16, 24, 24, 24, 24
  2. Tally: 3 appears 3x, 16 appears 2x, 24 appears 4x.
  3. The most frequent value is 24, appearing 4 times, so the mode is 24 stones.

Common mistakes

  • Finding the middle of the list as written instead of ordering the data first.
  • Giving the total instead of the mean, or forgetting to divide at all.
  • Reporting how often the mode occurred rather than the value that occurred that often.
  • Using the mean to describe data with a large outlier, and reporting a "typical" figure that no member is near.
  • Averaging the averages of two groups of different sizes instead of pooling the totals.

What you should be able to do

  • Compute the mean of a small data set.
  • Find the median of a data set with an odd or an even number of values.
  • Identify the mode, including when there is more than one or none at all.
  • Say which average describes a data set best when an outlier is present.

Where this fits in the curriculum

Common Core

  • 6.SP.A.2

    Grade 6 — Understand that a set of data has a distribution described by its centre, spread and overall shape.

  • 6.SP.A.3

    Grade 6 — Recognise that a measure of centre summarises all of the values with a single number, while a measure of variation describes how the values vary.

  • 6.SP.B.5

    Grade 6 — Summarise numerical data sets in relation to their context, reporting measures of centre and describing the shape of the distribution.

Ontario

  • G7.D1.5

    Grade 7 — Determine the impact of adding or removing data from a data set on a measure of central tendency, and describe how these changes alter the shape and distribution of the data.

  • G6.D1

    Grade 6 — Manage, analyse and use data to make convincing arguments and informed decisions, in various contexts drawn from real life.

    Ontario numbers its Data expectations per grade document and the specific numbering could not be verified line by line here, so this tag names the strand's OVERALL expectation rather than guessing at a specific one.

SAT

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