Mean absolute deviation
Two punnets of strawberries have the same mean mass. Does that make them alike, and what one number says how spread out each one is?
Where you are 13 in Center and spread
Why this exists
Didier and Dilys each buy a punnet of six strawberries and weigh them one by one. Didier’s weigh 13, 14, 15, 15, 16 and 17 grams; Dilys’s weigh 8, 10, 15, 15, 20 and 22 grams. Both punnets hold 90 grams, so both have a Mean of \displaystyle \frac{90}{6} = 15 grams. By the mean the two punnets are the same. Tip them out, though, and the difference is plain: Didier’s strawberries are all much the same size, while Dilys’s run from tiny to large.
A center alone cannot tell these apart. What is missing is a second number that says how far the values tend to sit from the center. This lesson builds one, the mean absolute deviation, out of the distances from the mean that the mean as a balance point already measures.
The first day defines it, computes it, reads it off a dot plot and says what it means. The second compares data sets with the same mean, shows why the MAD adds the distances on both sides of the mean rather than taking one side from the other, and watches the MAD respond to a new value.
Definition
A value’s distance from the mean is how far apart the two sit on the number line: the larger minus the smaller. For a value x and a mean m it is written with absolute value bars, \lvert x - m \rvert, and read “the distance between x and m”; it is the same whichever of the two is larger. With x = 3 and m = 5 the distance is 5 - 3 = 2. As in the mean as a balance point, a value below the mean is on the low side, a value above it on the high side, and a value equal to the mean has distance 0 and sits on neither side.
The mean absolute deviation, or \text{MAD}, of a data set is the mean of these distances:
Each word of the name does a job. A deviation is how far a value strays from the mean; absolute says each one is taken as a distance, never negative; mean says the distances are averaged. The MAD carries the unit of the data: grams of strawberry, minutes of song.
Intuition
Figure 1 is Dimitra’s playlist of five songs, 3, 4, 4, 5 and 9 minutes long, drawn as a dot plot resting on its balance point, the mean 5. Under the plot a bar runs from each song to the mean; its length is that song’s distance from the mean.
The bars have lengths 2, 1, 1, 0 and 4, which total 8. Shared out equally among the five songs, that is 1.6 each. So the MAD is the fair share of the distances: the length every bar would have if the total length were spread evenly. Some songs sit closer to the mean than 1.6 minutes and the 9 minute song sits much farther, but 1.6 minutes is how far a song typically sits from the mean.
A small MAD means the dots huddle near the mean; a large MAD means they are scattered. A MAD of 0 means every bar has length 0: every value equals the mean.
Results
The MAD is never negative, and it is 0 only when all the values are equal.
High side minus low side is zero. For any data set, the distances on the high side of the mean total the same as those on the low side, so “how far above, minus how far below” is always 0 and says nothing about spread. That is why the MAD adds the two sides instead. (In Math 7, with negative numbers, the same fact reads: the values minus the mean, signs kept, add to 0.)
Twice one side. The distances on the low side of the mean total the same as those on the high side, so
Why it is true
High minus low. Write H for the total of the high-side distances and L for the low-side total; a value at the mean adds to neither. The mean is the balance point, so H = L, and H - L = 0 for every data set. A tightly packed list and a widely spread one both give 0, so this number cannot tell them apart.
Twice one side. The total of all the distances is L + H, and since H = L that is 2L.
Never negative. Each distance is 0 or more, so their total is 0 or more, and so is the total divided by the count. The total is 0 only when every distance is 0, which means every value equals the mean, so all the values are equal.
Worked examples
Examples 1 to 3 go with the first day, Examples 4 to 6 with the second.
Finding a MAD
Dimitra’s playlist has five songs lasting 3, 4, 4, 5 and 9 minutes. Find the MAD and say what it means.
SolutionThe mean comes first, because every distance is measured from it. The total is 25 minutes over 5 songs, so the mean is \displaystyle \frac{25}{5} = 5 minutes. Next, each song’s distance from 5:
| Song length (min) | 3 | 4 | 4 | 5 | 9 |
|---|---|---|---|---|---|
| Distance from 5 | 2 | 1 | 1 | 0 | 4 |
The distances total 2 + 1 + 1 + 0 + 4 = 8, and their mean is the MAD: \displaystyle \frac{8}{5} = 1.6 minutes.
In words: a song on this playlist typically lasts about 1.6 minutes more or less than the mean of 5 minutes. The 9 minute song supplies 4 of the 8 on its own, so one long track accounts for half the spread. The balance gives a check: the low side totals 2 + 1 + 1 = 4, the high side 4.
A mean that is a decimal
Dobromir throws five paper airplanes. They fly 3, 5, 6, 8 and 9 meters. Find the MAD.
SolutionThe mean is \displaystyle \frac{31}{5} = 6.2 meters. No flight landed at 6.2, so every distance is a decimal. On the low side, 6.2 - 3 = 3.2, 6.2 - 5 = 1.2 and 6.2 - 6 = 0.2, total 4.6. On the high side, 8 - 6.2 = 1.8 and 9 - 6.2 = 2.8, total 4.6. The sides agree, as the balance requires, which also checks the subtractions. By the twice-one-side result the distances total 2 \times 4.6 = 9.2, so the MAD is \displaystyle \frac{9.2}{5} = 1.84 meters.
A flight typically lands a little under 2 meters from the mean of 6.2 meters.
Reading distances off a dot plot
Domenico weighs ten strawberries from one punnet. Figure 2 shows the masses, with the mean, 15 grams, marked. Find the MAD from the plot.
The ticks are 1 gram apart, so a dot’s distance from the mean is the number of steps between its column and 15. Every dot in a column has the same distance, so work column by column: the distance times the number of dots.
| Mass (g) | Dots | Distance from 15 | Dots \times distance |
|---|---|---|---|
| 12 | 1 | 3 | 3 |
| 13 | 2 | 2 | 4 |
| 14 | 3 | 1 | 3 |
| 15 | 1 | 0 | 0 |
| 16 | 1 | 1 | 1 |
| 17 | 1 | 2 | 2 |
| 22 | 1 | 7 | 7 |
The distances total 20 over 10 strawberries, so the MAD is \displaystyle \frac{20}{10} = 2 grams. (Check: the low side totals 3 + 4 + 3 = 10 and the high side 1 + 2 + 7 = 10. The marked mean is right too: the masses total 150, and \displaystyle \frac{150}{10} = 15.)
Eight of the ten strawberries lie within 2 grams of the mean. The 22 gram strawberry lies 7 grams out and supplies 7 of the 20 by itself. The MAD is a typical distance, not a fence: a value can sit well beyond it.
Same mean, different spread
Return to the punnets from the start of the lesson. Didier’s strawberries weigh 13, 14, 15, 15, 16 and 17 grams and Dilys’s weigh 8, 10, 15, 15, 20 and 22 grams; both have mean 15 grams. Find each MAD and say what the two numbers show.
SolutionDidier’s distances from 15 are 2, 1, 0, 0, 1 and 2, total 6, so his MAD is \displaystyle \frac{6}{6} = 1 gram. Dilys’s are 7, 5, 0, 0, 5 and 7, total 24, so her MAD is \displaystyle \frac{24}{6} = 4 grams.
Figure 3 marks one MAD either side of each mean: 14 to 16 for Didier, 11 to 19 for Dilys. Dilys’s MAD is four times Didier’s. A strawberry from her punnet typically differs from 15 grams by about 4 grams; one from his, by about 1 gram. Anyone wanting strawberries of even size should take Didier’s punnet, and the means alone could never have said so.
Why add the two sides?
Dobromir throws four more airplanes: 2, 4, 7 and 11 meters. Find the total distance on each side of the mean, the high-side total minus the low-side total, and the MAD. Then total each side measured from 5 instead of from the mean.
SolutionThe mean is \displaystyle \frac{24}{4} = 6 meters. The low side holds 2 and 4, at distances 4 and 2, total 6. The high side holds 7 and 11, at distances 1 and 5, total 6. A measure of spread built as “how far above, minus how far below” would give 6 - 6 = 0, as if the flights did not spread at all, and by the balance it gives 0 for every data set.
The MAD adds the sides instead: the distances total 6 + 6 = 12, so the MAD is \displaystyle \frac{12}{4} = 3 meters.
From 5, the low side holds 2 and 4 at distances 3 and 1, total 4, while the high side holds 7 and 11 at distances 2 and 6, total 8. The sides do not match. Only the mean makes the two sides equal, because only the mean is the balance point.
A new value moves the MAD
Dimitra’s second playlist has songs of 2, 4, 6, 8 and 10 minutes. Find the MAD. Then find the new MAD if she adds (a) a 6 minute song, (b) an 18 minute song.
SolutionThe mean is \displaystyle \frac{30}{5} = 6. The distances are 4, 2, 0, 2 and 4, total 12, so the MAD is \displaystyle \frac{12}{5} = 2.4 minutes.
(a) The total becomes 36 over 6 songs, so the mean is still 6. The new song’s distance is 0: the distances still total 12, now shared among six songs, so the MAD falls to \displaystyle \frac{12}{6} = 2 minutes. A value at the mean adds a song but no distance.
(b) The total becomes 48 over 6 songs, so the mean moves to 8, and every distance has to be measured again from 8: 6, 4, 2, 0, 2 and 10, total 24. The MAD is \displaystyle \frac{24}{6} = 4 minutes. One far value raised the MAD from 2.4 to 4: like the mean, the MAD is pulled hard by a lone far value (see resistant and non resistant measures).
Boundaries
- The MAD needs numbers on a number line. Categorical data, such as favorite colors, has no mean and so no MAD.
- Every distance is measured from the mean of the same data. When a value changes or is added, the mean may move, and then every distance must be measured again (Example 6).
- The MAD is a typical distance, not a limit: a value may sit several MADs from the mean, as the 22 gram strawberry in Example 3 does.
- Two MADs compare spreads only when the data sets measure the same kind of thing in the same unit. A MAD of 2 grams and a MAD of 2 minutes are not equal spreads.
Where this goes next
With a MAD in hand, the gap between two groups’ means can be measured in MADs, which is how comparing two populations decides whether a difference is large. Adding the same amount to every value slides the whole plot along without spreading it, so the mean moves and the MAD stays put, as the spread does in shifting a data set. The interquartile range measures spread around the median instead of the mean, and a lone far value barely moves it. Later, the standard deviation also starts from the distances to the mean but squares them before averaging.
Exercises
Exercises 1 to 10 go with the first day (Examples 1 to 3), and Exercises 11 to 23 with the second (Examples 4 to 6).
Day 1: finding a MAD
- Five strawberries weigh 9, 12, 14, 15, 20 grams. Find the mean and the MAD.
- Five songs last 2, 3, 5, 6, 9 minutes. Find the mean and the MAD.
- Six paper airplanes fly 4, 5, 6, 8, 9, 10 meters. Find the mean and the MAD.
- Five strawberries weigh 11, 12, 15, 16, 18 grams. Find the MAD.
- Four songs last 3.5, 4, 4.5, 6 minutes. Find the MAD.
Day 1: reading a dot plot
- Figure 4 shows ten of Dobromir’s flights. Find the mean and the MAD.
- In Figure 4, how many flights land more than one MAD from the mean?
- Under each dot of a dot plot of eight song lengths, its distance from the mean is written: 0.5, 0.5, 1, 1, 1, 1.5, 2, 2.5 minutes. Find the MAD.
Day 1: what a MAD says
- A grower reports that the strawberries in a punnet have a mean of 18 grams and a MAD of 1.5 grams. Say in a sentence what the MAD tells a buyer.
- One playlist has a MAD of 0.4 minutes and another a MAD of 2.1 minutes. In which playlist are the songs closer in length to one another? Explain.
Day 2: same mean, different spread
- Dimitra’s five paper airplanes fly 5, 6, 7, 8, 9 meters and Domenico’s fly 1, 4, 7, 10, 13 meters. Find each mean and MAD. Whose flights are more alike, and how many times as large is the other MAD?
- Playlist A has four songs of 4, 4, 4, 4 minutes and playlist B four songs of 1, 3, 5, 7 minutes. Show that the means agree and find both MADs.
- Without working anything out, say which data set has the larger MAD: 11, 11, 12, 13, 13 or 2, 7, 12, 17, 22. Then find both MADs.
Day 2: the two sides
- For the data 3, 6, 8, 11, find the total distance on each side of the mean and the high-side total minus the low-side total. Then find the MAD.
- In a list of six values, the distances on the low side of the mean total 9. Find the MAD.
- For ten strawberries, the distances on the high side of the mean total 7 grams. Find the MAD.
Day 2: changing the data
- Three songs last 2, 5, 8 minutes. Find the MAD, and the new MAD if a fourth song is added lasting (a) 5 minutes, (b) 13 minutes.
- In a punnet of ten strawberries, one weighs 12 grams, three weigh 13 grams, three weigh 14 grams, two weigh 15 grams and one weighs 17 grams. Find the mean and the MAD, working column by column as in Example 3.
- Three paper airplane flights have a mean of 8 meters. Two of them are 5 and 8 meters. Find the third flight and the MAD.
- Give a data set of four values with mean 10 and MAD 0, and one with mean 10 and MAD 5.
Day 2: reasoning
- Explain why the high-side total minus the low-side total is 0 for every data set, so that it can never measure spread.
- Explain why the MAD can never be larger than the largest distance from the mean. When are the two equal?
- Explain why adding a value equal to the mean can never raise the MAD.
Answers
- Mean 14 grams: \displaystyle \frac{70}{5} = 14. Distances 5, 2, 0, 1, 6, total 14, so the MAD is \displaystyle \frac{14}{5} = 2.8 grams.
- Mean 5 minutes: \displaystyle \frac{25}{5} = 5. Distances 3, 2, 0, 1, 4, total 10, so the MAD is \displaystyle \frac{10}{5} = 2 minutes.
- Mean 7 meters: \displaystyle \frac{42}{6} = 7. Distances 3, 2, 1, 1, 2, 3, total 12, so the MAD is \displaystyle \frac{12}{6} = 2 meters.
- 2.32 grams. The mean is \displaystyle \frac{72}{5} = 14.4. Low side 3.4 + 2.4 = 5.8, high side 0.6 + 1.6 + 3.6 = 5.8, total 11.6, and \displaystyle \frac{11.6}{5} = 2.32.
- 0.75 minutes. The mean is \displaystyle \frac{18}{4} = 4.5. Distances 1, 0.5, 0, 1.5, total 3, and \displaystyle \frac{3}{4} = 0.75.
- Mean 5 meters: \displaystyle \frac{50}{10} = 5. Distances 3, 1, 1, 0, 0, 0, 1, 1, 1, 2, total 10, so the MAD is \displaystyle \frac{10}{10} = 1 meter.
- 2 flights: the 2 meter flight (3 from the mean) and the 7 meter flight (2 from it). The other eight are at most 1 meter away.
- 1.25 minutes. The distances total 10, and \displaystyle \frac{10}{8} = 1.25.
- A strawberry in this punnet typically weighs about 1.5 grams more or less than 18 grams, so the strawberries are fairly even in size.
- The playlist with MAD 0.4 minutes. Its songs typically sit only 0.4 minutes from their mean, against 2.1 minutes for the other, so its lengths are bunched more closely.
- Both means are 7 meters. Dimitra’s distances 2, 1, 0, 1, 2 total 6: MAD \displaystyle \frac{6}{5} = 1.2. Domenico’s distances 6, 3, 0, 3, 6 total 18: MAD \displaystyle \frac{18}{5} = 3.6. Dimitra’s flights are more alike, and Domenico’s MAD is 3 times hers.
- Both means are 4: \displaystyle \frac{16}{4} = 4. A’s MAD is 0, since every song sits at the mean. B’s distances 3, 1, 1, 3 total 8, so its MAD is \displaystyle \frac{8}{4} = 2 minutes.
- The second. Both have mean 12, and the second’s values sit 5 or 10 from it against at most 1. MADs: \displaystyle \frac{4}{5} = 0.8 and \displaystyle \frac{30}{5} = 6.
- MAD 2.5. The mean is \displaystyle \frac{28}{4} = 7. Low side 4 + 1 = 5, high side 1 + 4 = 5, and 5 - 5 = 0. The distances total 10, so the MAD is \displaystyle \frac{10}{4} = 2.5.
- 3. The high side also totals 9, so the distances total 18, and \displaystyle \frac{18}{6} = 3.
- 1.4 grams. The low side also totals 7, so the distances total 14, and \displaystyle \frac{14}{10} = 1.4.
- MAD 2 minutes: mean 5, distances 3, 0, 3, and \displaystyle \frac{6}{3} = 2. (a) 1.5 minutes: the mean stays 5 and \displaystyle \frac{6}{4} = 1.5. (b) 3.5 minutes: the mean becomes \displaystyle \frac{28}{4} = 7, distances 5, 2, 1, 6 total 14, and \displaystyle \frac{14}{4} = 3.5.
- Mean 14 grams: the masses total 12 + 39 + 42 + 30 + 17 = 140 over 10. Distances times counts: 2 + 3 + 0 + 2 + 3 = 10, so the MAD is \displaystyle \frac{10}{10} = 1 gram.
- Third flight 11 meters: 3 \times 8 = 24 and 24 - 13 = 11. Distances 3, 0, 3, so the MAD is \displaystyle \frac{6}{3} = 2 meters.
- MAD 0: 10, 10, 10, 10, the only choice. MAD 5: for example 5, 5, 15, 15, whose distances are all 5.
- The mean is the balance point, so the distances on the low side total exactly the same as those on the high side, and one total minus the other is 0. That happens for every data set, bunched or spread out, so the result cannot tell them apart.
- The MAD is a fair share of the distances, and a fair share can never exceed the largest share: if it did, every distance would be smaller than the mean of the distances, which is impossible. The two are equal only when every distance is the same, as in 5, 5, 15, 15.
- A value equal to the mean leaves the mean where it is, so no other distance changes, and it adds a distance of 0. The same total is shared among one more value, so the MAD stays the same (if it was 0) or falls.