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Which average is fair?

Statistics and data · level 4, lesson 1 of 6

A question to start

A small bakery has eight people on its payroll. Seven bakers and helpers earn 30, 32, 34, 35, 35, 36 and 38 thousand dollars a year. The owner pays herself 300 thousand dollars a year.

A job advert says: "Join us! The average pay here is 67.5 thousand dollars a year."

That is true: it is the mean. But no worker earns anywhere near it. Every one of the seven earns 38 thousand or less. Is the advert honest? It used real numbers and did the arithmetic correctly, and still it gives a false picture. In this lesson you learn why that happens, and how to choose the average that tells the truth about your data.

The idea

A quick reminder

From level 3:

  • the mean is the total shared out equally (add, then divide by how many);
  • the median is the middle value once the data are in order;
  • the mode is the most common value.

All three are called measures of centre: each tries to give one "typical" value. They answer different questions, and they react differently to unusual values.

Outliers drag the mean

An outlier is a value far away from the rest of the data. The owner's 300 is an outlier.

Work out the bakery's averages:

  • Total: \(30 + 32 + 34 + 35 + 35 + 36 + 38 + 300 = 540\). Mean: \(540 \div 8 = 67.5\) thousand.
  • In order: 30, 32, 34, 35, 35, 36, 38, 300. Median: halfway between the 4th and 5th, \((35 + 35) \div 2 = 35\) thousand.
  • Mode: 35 thousand (twice).

Now take the owner out and look at only the seven workers. Their total is \(540 - 300 = 240\), and \(240 \div 7\) is about 34.3 thousand. The median of 30, 32, 34, 35, 35, 36, 38 is 35 thousand.

Mean Median
Seven workers about 34.3 35
Seven workers and the owner 67.5 35

One person nearly doubled the mean. The median did not move at all.

Why the mean moves so much

Think of the mean as a fair share. The owner's 300 thousand goes into the same pot as everybody else's pay, and then the pot is shared equally among 8 people. Her pay is 265 thousand more than a 35-thousand worker's, and sharing that extra among 8 people adds \(265 \div 8\), about 33 thousand, to everyone's share. One giant value raises the mean for everyone on paper, though nobody's real pay changed.

Or think of the balance point (level 3, lesson 2): the mean is where the number line balances. Put a heavy value far out at 300, and the balance point must slide a long way towards it.

050100150 200250300 median 35mean 67.5 owner

The median only asks "who is in the middle of the line?" The owner is at the end of the line whether she earns 40 thousand or 40 million, so the middle does not change. Statisticians say the median is resistant to outliers: it resists being pulled.

When the median is better

Use the median when the data have outliers or are lopsided (a long tail of large values on one side). Pay, house prices and the prices of things in general are classic examples: a few very large values pull the mean up, and the median is closer to what a typical person sees. This is why news reports about pay or house prices usually give the median.

When the mean is better

The mean is not a bad average. It uses every value, which is a strength when there are no wild values, and it has a special power: mean × number of values = total. So the mean is the right choice when the total matters:

  • Sharing a restaurant bill or the cost of a trip: each person's fair share is the mean.
  • Planning supplies: if a class of 25 eats a mean of 2 apples a week each, you need \(25 \times 2 = 50\) apples, whatever the median is.
  • Data that are roughly balanced around the middle, with no outliers, such as heights of children of the same age: the mean and median will be close, and the mean uses all the information.

When the mode is the only one that makes sense

Sometimes the mean and the median are useless.

  • Category data. Favourite colours, types of pet, the brand of a phone. There is no "mean colour" and no "middle colour", since colours have no order. Only the mode works: "blue was the most common choice".
  • Choosing what to stock. A shoe shop orders shoes in sizes. The mean shoe size of its customers might be 7.36, a size nobody can buy. The median might be 7.5, but the shop wants to know which size sells most, so it needs the mode. Shoe sizes are numbers, yet the useful question is "which is most common?".

A decision guide

Your data or question Use
Categories (colours, pets, brands) Mode
"Which is most popular / sells most?" Mode
Numbers with outliers or a long tail (pay, prices) Median
"What does the person in the middle have?" Median
Numbers with no outliers, roughly balanced Mean (and the median will be close)
"What is each person's fair share?" or the total matters Mean

The most honest habit of all: say which average you used, and when the mean and median are far apart, give both. The gap between them is itself a clue that something lopsided is going on. In the bakery, "the mean pay is 67.5 thousand but the median is 35 thousand" tells the whole story in one sentence.

Worked examples

1. House prices. Five houses on a street sold for 400, 420, 450, 480 and 2000 thousand dollars (the last one is a mansion).

  1. Mean: total \(400 + 420 + 450 + 480 + 2000 = 3750\), and \(3750 \div 5 = 750\) thousand.
  2. Median: 400, 420, 450, 480, 2000, so 450 thousand.
  3. Four of the five houses cost under 500 thousand, but the mean is 750 thousand. The median (450 thousand) describes a typical house on this street; the mean is dragged up by the mansion.

2. Test scores with no outlier. Scores: 68, 72, 75, 75, 79, 81.

  1. Mean: total 450, and \(450 \div 6 = 75\).
  2. Median: \((75 + 75) \div 2 = 75\).
  3. Mode: 75.

All three agree. With balanced data and no outliers, any of them is fair, and the mean is a good choice because it uses all six scores.

3. A low outlier. Seven children's times to run a lap, in seconds: 62, 64, 65, 66, 68, 70 and 140 (one child stopped to tie a shoelace).

  1. Mean: total \(62 + 64 + 65 + 66 + 68 + 70 + 140 = 535\), and \(535 \div 7\) is about 76.4 seconds.
  2. Median: 62, 64, 65, 66, 68, 70, 140, so 66 seconds.

Here the outlier is a slow time, and it drags the mean towards it (up, because slow means a big number). Six of the seven children ran faster than the mean. The median, 66 seconds, is the fair "typical time". Before choosing, also ask: does the outlier belong in the data at all? A time that includes tying a shoelace is not really a running time, and a careful report might leave it out and say so.

4. Shoe sizes. A shop sold these sizes on Saturday: 6, 7, 7, 7.5, 8, 8, 8, 8, 9, 10.

  1. Mean: total 78.5, and \(78.5 \div 10 = 7.85\). Not a size.
  2. Median: halfway between the 5th and 6th values, \((8 + 8) \div 2 = 8\).
  3. Mode: 8 (four times).

To decide what to reorder, the shop uses the mode: size 8 sold most.

Try it

  1. Find the mean and median of 2, 3, 3, 4, 38. Which is the fairer "typical value", and why?
  2. A cafe's daily customers for a week: 120, 115, 130, 125, 118, 122, 400 (a festival day). Find the mean and the median. Which would you tell someone thinking of opening a cafe nearby?
  3. Which average would you use for: (a) the most common eye colour in a class; (b) the typical price of a house in a city; (c) how much each of 6 friends pays towards a 90-dollar gift?
  4. The mean of five numbers is 10 and the median is 6. Make up five numbers that fit. What does the gap between mean and median suggest?
  5. Explain why adding one very large value changes the mean a lot but the median only a little.
  6. A list of 9 values has median 20. The largest value, 25, is changed to 2500. What is the new median?
  7. In a class, the mean number of hours of TV per week is 12 but the median is 8. Which of these is likely: (a) a few students watch a lot of TV; (b) a few students watch almost none? Explain.
  8. Can the mode be a fair "typical value" when it is very far from the mean and the median? Give an example.
Answers
  1. Mean \(50 \div 5 = 10\); median 3. The median: four of the five values are 4 or less, and the 38 is an outlier that pulls the mean up to 10, which is bigger than all the other values.
  2. Total \(120 + 115 + 130 + 125 + 118 + 122 + 400 = 1130\), mean \(1130 \div 7\), about 161.4. In order: 115, 118, 120, 122, 125, 130, 400, median 122. The median: the festival day is not typical. (Better still, mention the festival day separately.)
  3. (a) Mode (category data). (b) Median (prices have a long tail of expensive houses). (c) Mean: \(90 \div 6 = 15\) dollars each, the fair share.
  4. The total must be \(10 \times 5 = 50\) and the middle value 6. One answer: 1, 2, 6, 7, 34 (total 50). The mean is well above the median, which suggests one or more large values pulling the mean up.
  5. The mean shares the total: one huge value adds a lot to the total, and that extra is shared among everyone. The median depends only on which value is in the middle position; adding one value at the top moves the middle by only half a place.
  6. Still 20. The middle (5th) value has not changed; only the largest one has.
  7. (a). The mean is above the median, so some large values are pulling it up: a few students who watch a lot.
  8. Yes, sometimes. In a survey of shoe sizes 5, 5, 5, 5, 9, 10, 10 the mode is 5, mean is \(49 \div 7 = 7\) and median 5. For a question like "which size sells most?", the mode is exactly the right answer however far it is from the mean.

Common mistake

Thinking "average" always means the mean, and the mean is always right. The bakery advert was arithmetically correct and still misleading. Ask which question you are answering, check for outliers, and say which average you used.

Deleting outliers just because they are inconvenient. An outlier may be a mistake (a typing slip, a shoelace), and then it can be removed, saying so. But an outlier may also be real and important: the owner really does earn 300 thousand. Report it; do not hide it.

Where it leads