Statistics · Lesson 1 of 6
Why Averages Need Dispersion
“Discover why a centre cannot describe a whole dataset, and measure spread with range and mean deviation.”
• Compare datasets with the same mean and median but different amounts of scatter. • Calculate range and explain the information it leaves out. • Interpret signed deviations and show why deviations from the mean sum to zero. • Calculate mean deviation about a mean or median for ungrouped observations. • Explain absolute distances, repeated values and the units of mean deviation.
The same average can hide different stories
Suppose two players each average 53 runs over ten matches. One may perform steadily near that average, while the other alternates between very low and very high results. Their means alone cannot tell us which pattern occurred. Statistics therefore needs a description of spread alongside a description of centre.
A dataset is a collection of observations recorded for a purpose. The mean, median and mode describe central tendency: roughly where the observations are centred. Dispersion describes how far apart the values are or how far they lie from a centre. These are complementary ideas. A useful interpretation asks both where the data sit and how widely they are scattered.
| Player | Runs in ten matches | Mean | Median |
|---|---|---|---|
| A | 30, 91, 0, 64, 42, 80, 30, 5, 117, 71 | 53 | 53 |
| B | 53, 46, 48, 50, 53, 53, 58, 60, 57, 52 | 53 | 53 |
Both totals are 530, so dividing by ten gives mean 53. For the median, arrange each list in order. A has middle values 42 and 64, whose average is 53; B has middle values 53 and 53. Yet A’s observations stretch from 0 to 117, while B’s are clustered between 46 and 60. A common-scale dot plot makes the contrast visible. Repeated values are stacked, so every observation is retained.
Dispersion is the spread or scatter of observations. A measure of dispersion uses one number to describe an aspect of that spread. Its meaning depends on how the number is calculated.
Recall the centre before measuring spread
The mean uses every observation and divides their total by how many observations there are. The median uses position after ordering the data. Before calculating deviations about either centre, find that centre correctly; using an unsorted list to locate a median changes the result.
For odd n, the median is the observation in position (n + 1)/2 in the ordered list. For even n, it is the average of the observations in positions n/2 and n/2 + 1. The median need not be one of the original values. The mode is a most frequently occurring value, but this chapter develops mean deviation about the mean and median rather than about the mode.
Range gives a quick first comparison
Range measures the distance from the smallest observation to the largest. It is easy to calculate and useful for an initial look at variability. It uses only two values, however, so it cannot describe what happens between the extremes or how values are positioned relative to the mean.
Problem
Find the range for each ten-match dataset.
- 1.Player A has minimum 0 and maximum 117. Its range is 117 − 0 = 117 runs.
- 2.Player B has minimum 46 and maximum 60. Its range is 60 − 46 = 14 runs.
- 3.This supports the visual observation that A’s results are more spread out. It does not by itself tell us whether either player is more successful in every relevant sense.
Problem
Compare 0, 5, 5, 5, 10 with 0, 0, 5, 10, 10 using range and mean.
- 1.Both lists have minimum 0 and maximum 10, so both ranges are 10.
- 2.Each list totals 25 across five values, giving the same mean 5.
- 3.The first list has three values at its centre; the second has four values at its extremes. Equal ranges and means do not guarantee equal spread around the centre.
- 4.A measure using every observation can distinguish them: their absolute distances from 5 total 10 and 20 respectively, giving mean deviations 2 and 4.
Range, mean deviation and standard deviation are measures of dispersion. Quartile deviation is another measure named in the source, but it is not developed in this chapter. Our next step is to use distances from a centre, so that all the observations contribute.
Why signed deviations fail to measure scatter
For a chosen centre a, the signed deviation of a value xᵢ is xᵢ − a. Values below a have negative deviations and those above a have positive deviations. Signs tell us direction, but spread requires distance. If we simply add signed deviations, opposite directions can cancel even when values are far apart.
The signed deviation from a is xᵢ − a. Its absolute value |xᵢ − a| is the nonnegative distance between xᵢ and a on a number line. Absolute deviation removes the direction while retaining the distance.
For the mean, cancellation is unavoidable. Adding all deviations gives Σ(xᵢ − x̄) = Σxᵢ − nx̄. But the definition of the mean says Σxᵢ = nx̄, so the result is zero. A dataset with huge deviations can therefore have the same signed-deviation total as a dataset with no variation at all.
Problem
Use the values 2, 5 and 8 to compare signed and absolute deviations about their mean.
- 1.The mean is (2 + 5 + 8)/3 = 5.
- 2.Signed deviations are −3, 0 and 3. Their total is zero, although two observations are 3 units from the mean.
- 3.Absolute deviations are 3, 0 and 3. Their mean is 6/3 = 2, which records the nonzero spread.
Mean deviation is an average distance
Once we replace each signed deviation by its absolute distance, we can average those distances. Choose the centre first, calculate each distance, add them and divide by the number of observations. Mean deviation about the mean and mean deviation about the median use the same procedure; only the centre changes.
Mean deviation about a chosen central value a is the arithmetic mean of the absolute deviations |xᵢ − a|. It is written MD(a). Common choices are the mean x̄ and the median M.
The result is nonnegative because every distance is nonnegative. It is zero only when every observation equals the chosen centre. Repeated values are separate observations: if 12 appears twice, its distance must be included twice. Do not take the absolute value only after adding signed deviations; the absolute value belongs to each deviation before summation.
Problem
Find the mean deviation about the mean for 6, 7, 10, 12, 13, 4, 8, 12.
- 1.There are eight observations, with total 72, so the mean is 72/8 = 9.
- 2.Subtract 9 from each observation: −3, −2, 1, 3, 4, −5, −1, 3.
- 3.The absolute deviations are 3, 2, 1, 3, 4, 5, 1, 3. Their total is 22.
- 4.Therefore MD(x̄) = 22/8 = 2.75. The two occurrences of 12 each contribute a distance of 3.
| Value xᵢ | Signed deviation xᵢ − 9 | Absolute deviation |
|---|---|---|
| 6 | -3 | 3 |
| 7 | -2 | 2 |
| 10 | 1 | 1 |
| 12 | 3 | 3 |
| 13 | 4 | 4 |
| 4 | -5 | 5 |
| 8 | -1 | 1 |
| 12 | 3 | 3 |
| Total | 0 | 22 |
Problem
Find mean deviation about the mean for 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5.
- 1.The twenty observations total 200, so the mean is 10.
- 2.Distances from 10 are 2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5.
- 3.Their sum is 124. A table or a careful row-by-row total helps prevent omissions in a long list.
- 4.The mean deviation is 124/20 = 6.2. This is an average distance from 10, not the range of the observations.
Mean deviation about the median
Finding a median requires ordering before selecting the middle position. After the median is known, distances may be calculated from the ordered or original list; rearrangement does not change their total. Keep the notation M for median so it is not confused with the mean.
Problem
Find mean deviation about the median for 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
- 1.Order the eleven values: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21.
- 2.The middle position is (11 + 1)/2 = 6, giving median M = 9.
- 3.Distances from 9 are 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12. Their sum is 58.
- 4.MD(M) = 58/11 ≈ 5.27. Keep 58/11 until the final rounding step.
Problem
Find mean deviation about the median for 1, 3, 6 and 10.
- 1.The list is ordered, and there are four observations. The median is the mean of the second and third: M = (3 + 6)/2 = 4.5.
- 2.Distances from 4.5 are 3.5, 1.5, 1.5 and 5.5. Their total is 12.
- 3.MD(M) = 12/4 = 3. The median does not need to be one of the observed values.
Σ|xᵢ − a| measures total distance; |Σ(xᵢ − a)| measures the magnitude of a signed total. They are different. About the mean, the second expression is always zero even when the data vary.
Check your understanding
Use the questions to separate centre, extreme span and average distance. For numerical items, find the centre first and use the number of observations as the denominator.
Quiz
Two datasets have the same mean. What can we conclude about their dispersion?
What is the range of 4, 4, 7, 10, 12?
What is Σ(xᵢ − x̄) for any finite dataset with its arithmetic mean?
What is MD about the mean for 2, 5, 8?
For ordered data 1, 3, 6, 10, what is the median?
If observations are lengths in cm, what is the unit of mean deviation?
Practice Problems
- Explain how two datasets can share a mean and median but have different spread. Give a numerical example.
- Find the range and mean deviation about the mean of 4, 7, 8, 9, 10, 12, 13, 17.
- Find mean deviation about the mean of 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
- Find mean deviation about the median of 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17. Show the two middle observations.
- Find mean deviation about the median of 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
- Show algebraically why deviations from the arithmetic mean sum to zero, then explain why this does not mean zero dispersion.
- Construct two lists with the same minimum and maximum but different mean deviations about their means.
Key Takeaways
• A centre and a measure of spread answer different questions about a dataset. • Range uses only the largest and smallest observations, so it leaves out the arrangement between them. • Signed deviations from the mean cancel; absolute distances preserve spread. • Mean deviation averages individual absolute deviations about the stated centre. • Order the data to find the median, include every repeated observation, and round only at the end.
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Next · Lesson 2
Mean Deviation in Frequency Distributions