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Shapes of distributions

Statistics and data · level 7, lesson 1 of 7

A question to start

A news report says: "The average waiting time at the clinic is 5.5 minutes." Another report about the same clinic, using the same data, says: "The typical patient waits 4 minutes." Nobody lied and nobody made an arithmetic error. How can both be true at once?

The answer is not in the numbers alone. It is in the shape of the data: a few patients waited a very long time, and that long tail pulls one kind of average away from the other. In this lesson you learn to read the shape of a data set at a glance, and to know from the shape which average to trust.

The idea

A distribution is the whole picture

The distribution of a variable tells you which values it takes and how often each one happens. A list of 500 heights is data; the distribution is the pattern hidden in that list: where the values bunch up, where they thin out, how far they reach.

For numerical data the usual picture of a distribution is a histogram (level 4): cut the number line into equal intervals, called bins, count the values in each bin, and draw touching bars. The outline of the bars is the shape of the distribution.

When you describe a distribution, say four things. A handy memory aid is SOCS:

  • Shape: symmetric, skewed, uniform, one hump or two.
  • Outliers: values far from the rest (the 1.5 × IQR rule from level 5 is one test).
  • Centre: a typical value, the mean or the median.
  • Spread: how far the values reach, measured by the range, the IQR or the standard deviation.

Five shapes you will meet again and again

symmetric skewed right skewed left uniform bimodal
  • Symmetric. The left half is a mirror image of the right half. Heights of adults of one sex, the weights of eggs from one farm and errors in careful repeated measurements are roughly symmetric, usually with one hump in the middle.
  • Skewed right (also called positively skewed). Most values are bunched at the low end and a long, thin tail stretches to the right. The word "skewed" names the side of the tail, not the side of the hump. Waiting times, incomes, house prices and the sizes of cities are skewed right: they cannot go below zero, but a few can be huge.
  • Skewed left (negatively skewed). The tail stretches to the left. Scores on an easy test are like this: most people score high, a few score low, and nobody can score above 100%.
  • Uniform. Every bin holds about the same number of values. Rolling a fair die 600 times gives about 100 of each face: a flat histogram.
  • Bimodal. Two humps, each called a mode (the word mode here means a peak, not only the single most common value). Two humps are a clue that two different groups have been mixed together. The arrival times of customers at a restaurant have one hump at lunch and one at dinner. A famous real data set is the time between eruptions of the Old Faithful geyser in Yellowstone: it has one hump of short waits and one hump of long waits.

A distribution with one hump is called unimodal. Real data never make perfect shapes; you look for the overall pattern and ignore the small bumps that come from chance.

Why a tail pulls the mean but not the median

The mean is the balance point of the data. Imagine each value as a one-gram weight placed on a long ruler at that value. The ruler balances on a finger placed at the mean. The median is the middle value: half the weights are on each side of it.

Now take the largest value and slide it further to the right by some distance \(d\). What happens?

  • The total of the data goes up by \(d\), so the mean goes up by \(\dfrac{d}{n}\), where \(n\) is the number of values. The far weight has more leverage, so the balance point moves towards it.
  • The median does not move at all. The largest value was above the middle before, and it is still above the middle. Only the order of the values matters to the median, not how far away they are.

So a long tail of large values drags the mean out towards the tail, while the median stays with the bulk of the data. That gives the usual pattern:

Shape Where the mean sits
Symmetric mean ≈ median (both at the centre of symmetry)
Skewed right mean > median (the mean is pulled right, into the tail)
Skewed left mean < median (the mean is pulled left, into the tail)

In a smooth, single-humped skewed distribution the peak (the mode) is furthest from the tail, then the median, then the mean: for skewed right, mode < median < mean. This is a rule of thumb, not a theorem. With small data sets or with values that are whole numbers it can fail, so always check with the actual numbers.

The clinic, solved

Here are the 13 waiting times, in minutes, behind the two news reports:

1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 8, 12, 19

waiting time (minutes) median 4 mean 5.5 0 4 8 12 16 20
  • The mean is \(\dfrac{72}{13} \approx 5.5\) minutes.
  • The median is the 7th of 13 values: 4 minutes.
  • The mode is 3 minutes.

The histogram is skewed right: most patients waited 2 to 5 minutes, and the 12 and 19 minute waits form a long tail. Those two patients pull the mean above the median. "Average 5.5 minutes" and "typical wait 4 minutes" are both true; the median describes the typical patient better, and the mean is what the clinic needs if it wants the total time spent waiting (\(13 \times 5.5 \approx 72\) minutes).

Which summary goes with which shape

  • Roughly symmetric, no outliers: use the mean for the centre and the standard deviation for the spread. They use every value and work well with the normal curve (next lesson).
  • Skewed, or with outliers: use the median for the centre and the IQR for the spread. They resist the pull of the tail. A box plot shows them well.
  • Bimodal: no single centre describes the data honestly. Split the data into the two groups if you can, and describe each one.

Shape and bin width

The shape you see depends a little on the bin width you choose. Bins that are too wide (three bars for 500 values) hide the shape; bins that are too narrow (one bar for every value) make a spiky mess of chance bumps. A common start is about \(\sqrt{n}\) bins for \(n\) values, then adjust by eye. A real feature, such as two humps, survives a change of bin width; a chance bump usually does not.

Worked examples

1. Describe a distribution with SOCS. The waiting times above.

  • Shape: skewed right, single hump at 2 to 4 minutes.
  • Outliers: Q1 is 2.5 and Q3 is 7 (median of the lower six values 1, 2, 2, 3, 3, 3 and of the upper six 4, 5, 6, 8, 12, 19), so IQR \(= 7 - 2.5 = 4.5\). The upper fence is \(7 + 1.5 \times 4.5 = 13.75\). The 19-minute wait is an outlier; the 12-minute wait is not.
  • Centre: median 4 minutes (better than the mean 5.5 because of the skew).
  • Spread: IQR 4.5 minutes; the range is \(19 - 1 = 18\) minutes.

2. A left-skewed test. Fifteen students wrote an easy test. Their scores were

38, 55, 64, 70, 74, 78, 80, 82, 84, 85, 87, 88, 90, 92, 95

The total is 1162, so the mean is \(1162 \div 15 \approx 77.5\). The median is the 8th value, 82. The mean is below the median, which fits a tail to the left: the 38 and the 55 pull the mean down. A teacher reporting "the class average was 77" would make the class sound weaker than it is; most of the class scored above 80.

3. Shape from two numbers. In a city the mean house price is 720 000 dollars and the median is 610 000 dollars. What is the shape?

The mean is well above the median, so some very expensive houses pull the mean up: the distribution is skewed right. The median is the better "typical price": half the houses cost less than 610 000 dollars.

4. Sliding one value. A data set of 10 values has mean 50 and median 48. The largest value, 70, was typed wrongly and should be 170. Find the new mean and median.

The total goes up by \(170 - 70 = 100\), so the mean goes up by \(100 \div 10 = 10\), to 60. The median does not change: it is still 48, because the order of the middle values is the same. One wrong value moved the mean by 10 and the median by nothing.

5. Two humps. A survey records the ages of everyone at a children's swimming lesson, including the parents watching. The histogram has one hump at ages 6 to 9 and another at 32 to 42. The mean age is about 22, an age that almost nobody in the building has. Split the data: describe the children and the parents separately.

Try it

  1. Name the likely shape (symmetric, skewed right, skewed left, uniform or bimodal): (a) the number of days in each month for many years; (b) the age at which people retire; (c) the prices of all cars sold in a year; (d) the last digit of many phone numbers; (e) the weights of one brand of 500 g flour bags.
  2. A distribution has mean 30 and median 42. Which way is it skewed? Which side has the long tail?
  3. The data 2, 3, 3, 4, 5, 5, 6, 40. Find the mean and median. Which better describes a typical value?
  4. Six values have mean 10. One value is changed from 12 to 30. What is the new mean? Can you say what happens to the median without more information?
  5. Explain, using the idea of a balance point, why the mean of a right-skewed data set is usually bigger than the median.
  6. A histogram of the heights of all students in a school, from kindergarten to grade 12, has two humps. Suggest a reason.
  7. Make up a data set of 5 whole numbers that is skewed right, has median 10 and has mean 20.
  8. Shoe sizes of a class are 5, 6, 6, 7, 7, 7, 8, 8, 9. Draw a quick tally of the values. Is the distribution symmetric? What are the mean, median and mode?
Answers
  1. (a) Not one of the smooth shapes: 28, 29, 30 or 31 only, with 30 and 31 the most common; the closest of the five is left-skewed (most months are long, a few are short). (b) Roughly symmetric around the 60s, often slightly skewed left, since some people retire early and few retire very late. (c) Skewed right: most cars cost a moderate amount, a few cost a fortune. (d) Uniform: each digit 0 to 9 about equally often. (e) Symmetric, a narrow hump around 500 g.
  2. The mean is below the median, so it is skewed left: the long tail is on the left, at small values.
  3. Total \(= 2 + 3 + 3 + 4 + 5 + 5 + 6 + 40 = 68\), mean \(= 68 \div 8 = 8.5\). The median is the average of the 4th and 5th values: \((4 + 5) \div 2 = 4.5\). The median is better; the mean 8.5 is bigger than every value except one.
  4. The total was \(6 \times 10 = 60\); it rises by \(30 - 12 = 18\), to 78, so the new mean is \(78 \div 6 = 13\). The median might change or not: if 12 was one of the two middle values it changes, if 12 was already in the top half it does not. You need the other values to know.
  5. The mean is the point where the data would balance like weights on a ruler. Values far out in the right tail are far from the centre, so they have a lot of leverage, and the balance point has to move right to stay level. The median only counts values on each side, so far-out values do not move it.
  6. Two groups are mixed: young children (one hump of short heights) and teenagers (a hump of taller heights). More precisely, heights grow with age, and the mix of grades is not smooth.
  7. One answer: 1, 2, 10, 11, 76. Median 10 (the middle value). The total must be \(5 \times 20 = 100\), and \(1 + 2 + 10 + 11 + 76 = 100\) ✓. The one large value makes a long right tail.
  8. Tally: 5 (one), 6 (two), 7 (three), 8 (two), 9 (one). It is symmetric about 7. Mean \(= 63 \div 9 = 7\), median 7 (the 5th value), mode 7: in a symmetric single hump all three agree.

Common mistake

Naming the skew after the hump. "Skewed right" means the tail points right; the big hump of data is on the left. Think of the tail as the skew: where the tail goes, the mean follows.

Quoting the mean of skewed data as "typical". Incomes, house prices and waiting times are skewed right, so their means are above what a typical person sees. When a report gives an average of skewed data, ask for the median.

Where it leads