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Build: Spread, how far from the middle

Statistics and data · step 3 of 5

A question to start

Two basketball players each score an average of 10 points a game. Over their last five games:

  • Player A scored 9, 10, 10, 11, 10.
  • Player B scored 0, 20, 2, 18, 10.

Both means are \(50 \div 5 = 10\). Both medians are 10. Yet the two players are clearly not the same. If your team needs about 10 points tonight, you would pick Player A, because Player A is steady. Averages hide this. On this page you will learn to measure how spread out data is, and to draw it.

The idea

Why the range is not enough

The range (largest minus smallest) is quick, but it uses only the two most extreme values. One strange value can make it huge. We want measures of spread that look at the bulk of the data, or at every value.

Quartiles: cutting the data into four

The median cuts sorted data into two halves. The quartiles cut it into four quarters.

  • The lower quartile, \(Q_1\), is the median of the lower half of the data.
  • The median is sometimes called \(Q_2\).
  • The upper quartile, \(Q_3\), is the median of the upper half.

There are a few slightly different conventions for quartiles. On this page we use a common school one: when the number of values is odd, leave the median out of both halves.

The interquartile range (IQR) is the distance between the quartiles:

\[\text{IQR} = Q_3 - Q_1\]

It is the spread of the middle half of the data, so a few extreme values at either end do not affect it.

The box plot

A box plot (or box-and-whisker plot) draws five numbers: the smallest value, \(Q_1\), the median, \(Q_3\) and the largest value. The box goes from \(Q_1\) to \(Q_3\) with a line at the median; the whiskers reach out to the smallest and largest values that are not outliers. Outliers are drawn as separate dots.

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A box plot shows a lot at once: where the middle is, how wide the middle half is, whether one side is more stretched than the other, and whether there are outliers.

A rule for outliers

How far away must a value be to count as an outlier? A widely used rule says a value is an outlier if it is more than \(1.5 \times \text{IQR}\) beyond a quartile. In symbols, it is an outlier if it is

\[\text{below } Q_1 - 1.5 \times \text{IQR} \quad \text{or} \quad \text{above } Q_3 + 1.5 \times \text{IQR}.\]

These two cut-off values are called the fences. The rule is a convention, not a law of nature, but it is a sensible one.

Spread around the mean: deviations

Another way to measure spread uses every value. For each value, find how far it is from the mean. This distance is the value's deviation. Then ask: how big is a typical deviation?

  • The mean absolute deviation (MAD) is the mean of the distances, ignoring minus signs.
  • The standard deviation, written \(\sigma\) (the Greek letter sigma), squares the deviations first, takes their mean, and then takes the square root to get back to the original units:
\[\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}\]

Here \(x\) stands for each value in turn, \(\bar{x}\) (read "x bar") is the mean, \(n\) is the number of values, and \(\sum\) means "add them all up". The quantity under the square root, \(\sigma^2\), is called the variance.

Why square? Deviations above the mean are positive and below are negative, and they always add up to exactly zero, so their plain mean is useless. Squaring makes them all positive, and it gives extra weight to values far from the mean. It also leads to beautiful mathematics later on (the normal curve, in the next step).

When the data is a sample from a bigger group and you want to estimate the spread of the whole group, statisticians divide by \(n - 1\) instead of \(n\). Calculators often show both, as \(\sigma\) and \(s\). For now, dividing by \(n\) is fine.

Worked examples

1. Quartiles, box plot and outliers. The data is

3, 5, 6, 7, 8, 9, 10, 12, 13, 15, 30

  1. It is already sorted. There are 11 values, so the median is the 6th: 9.
  2. Lower half (the 5 values before the median): 3, 5, 6, 7, 8. Its median is \(Q_1 = 6\).
  3. Upper half (the 5 values after the median): 10, 12, 13, 15, 30. Its median is \(Q_3 = 13\).
  4. \(\text{IQR} = 13 - 6 = 7\), and \(1.5 \times 7 = 10.5\).
  5. Fences: \(6 - 10.5 = -4.5\) and \(13 + 10.5 = 23.5\). Nothing is below \(-4.5\), but 30 is above 23.5, so 30 is an outlier.
  6. The box plot (drawn above) has whiskers from 3 and to 15, the largest value that is not an outlier, and a dot at 30.

The range is \(30 - 3 = 27\), but the IQR is only 7. The range was blown up by one value; the IQR was not.

2. The two basketball players. Both means are 10.

Player A score Deviation Squared Player B score Deviation Squared
9 −1 1 0 −10 100
10 0 0 20 10 100
10 0 0 2 −8 64
11 1 1 18 8 64
10 0 0 10 0 0
Total 0 2 Total 0 328
  • Player A: variance \(= 2 \div 5 = 0.4\), so \(\sigma = \sqrt{0.4} \approx 0.63\) points.
  • Player B: variance \(= 328 \div 5 = 65.6\), so \(\sigma = \sqrt{65.6} \approx 8.1\) points.

The mean absolute deviations tell the same story: Player A's is \((1 + 0 + 0 + 1 + 0) \div 5 = 0.4\) and Player B's is \((10 + 10 + 8 + 8 + 0) \div 5 = 7.2\). Player B's scores are spread out more than ten times as widely. Notice that the deviations add to 0 in both columns, just as promised.

Try it

  1. Find the range, the quartiles and the IQR of 2, 4, 4, 5, 7, 8, 9, 12.
  2. Using your answer to question 1, would a new value of 20 be an outlier?
  3. Find the mean and the standard deviation of 2, 4, 4, 4, 5, 5, 7, 9.
  4. You add 10 to every value in a data set. What happens to the mean? To the standard deviation?
  5. You multiply every value in a data set by 3. What happens to the standard deviation?
  6. Why is the IQR often a better measure of spread than the range?
Answers
  1. Range: \(12 - 2 = 10\). There are 8 values (an even number), so the median is halfway between the 4th and 5th: \((5 + 7) \div 2 = 6\), and the halves are 2, 4, 4, 5 and 7, 8, 9, 12. \(Q_1 = (4 + 4) \div 2 = 4\) and \(Q_3 = (8 + 9) \div 2 = 8.5\). \(\text{IQR} = 8.5 - 4 = 4.5\).
  2. Upper fence: \(8.5 + 1.5 \times 4.5 = 8.5 + 6.75 = 15.25\). Since 20 is above 15.25, yes, it would be an outlier.
  3. Sum \(= 40\) and \(n = 8\), so the mean is 5. The deviations are \(-3, -1, -1, -1, 0, 0, 2, 4\); their squares are 9, 1, 1, 1, 0, 0, 4, 16, which add to 32. Variance \(= 32 \div 8 = 4\), so \(\sigma = 2\).
  4. The mean goes up by 10. The standard deviation does not change: every value moves together, so the distances from the mean stay the same.
  5. It is multiplied by 3. Every distance from the mean is stretched 3 times. (The variance is multiplied by 9.)
  6. The range depends only on the two most extreme values, so a single outlier can make it huge. The IQR measures the middle half of the data and ignores the extremes.

Common mistake

Forgetting the square root. After adding up the squared deviations and dividing by \(n\), you have the variance, not the standard deviation. The variance is in squared units (points squared, centimetres squared), which makes no sense as a "typical distance". Take the square root to get back to the units of the data. Quick check: the standard deviation should look like a typical distance from the mean. For Player B, 8.1 points does; 65.6 points does not.

Where it leads