The same average
Sam and Alex play three rounds of a game. Sam scores 4 points in every round: 4, 4 and 4. His mean score is (4 + 4 + 4) ÷ 3 = 12 ÷ 3 = 4 points, and every one of his scores is exactly the mean.
Sam’s three bars are all 4 points tall, and the dashed mean line at 4 runs along their tops.
The same average, spread wider
Alex scores 1, 4 and 7. His mean is (1 + 4 + 7) ÷ 3 = 12 ÷ 3 = 4 points as well, and his median, the middle score, is also 4. By the mean or the median, the two players are the same.
But their scores are nothing alike. Every one of Sam’s scores is the mean, while Alex’s lowest score is 3 below the mean and his highest is 3 above it. An average says where the data is centered. It says nothing about how spread out the values are, so an average alone does not describe a data set.
Alex’s bars are 1, 4 and 7 points tall. The mean line is at 4 again, with one bar 3 below it and one bar 3 above it.
Sam’s scores on a scale from 0 to 8: all three dots are stacked at 4.
Alex’s scores on the same scale: the dots at 1, 4 and 7 stretch across 6 points.
Measuring the spread with the range
A measure of spread is a single number that says how far apart the values are. The simplest is the range: the highest value minus the lowest. Alex’s range is 7 − 1 = 6 points. Sam’s range is 4 − 4 = 0 points, because his scores do not spread at all.
The range uses only the two extreme values, so one unusual value can make it large. For larger data sets the interquartile range, the upper quartile minus the lower quartile, measures the spread of the middle half and ignores the extremes.
Read the highest bar, 7, and the lowest bar, 1, against the scale: the range is 7 − 1 = 6 points.
Comparing with an average and a spread
Ana and Ben take five tests each. Ana scores 66, 69, 71, 73 and 76 points, and Ben scores 52, 60, 73, 84 and 86 points. Both totals are 355, so both means are 355 ÷ 5 = 71 points. Ana’s range is 76 − 66 = 10 points, and Ben’s is 86 − 52 = 34 points.
A comparison of two data sets gives one average and one measure of spread, and says what each means in the context: “Ana and Ben have the same mean score, 71 points, but Ana’s range is 10 points and Ben’s is 34 points, so Ana’s scores are much more consistent.” Both halves of the sentence are needed. The means alone would call the two students the same.
More consistent does not always mean better; the question decides. A teacher who needs a score above 60 every time would pick Ana: all five of her scores are above 60, and two of Ben’s, 52 and 60, are not. A teacher who needs a score above 80 would pick Ben, the only one of the two who has scored that high.
Ana’s five scores lie within 10 points of each other, from 66 to 76.
Ben’s five scores, on the same scale, spread over 34 points, from 52 to 86. Both sets have a mean of 71.
A spread that uses every value
The range uses only two values. Another measure of spread uses all of them: the mean absolute deviation, the mean distance of the values from their mean.
For each value, find its distance from the mean and drop the sign, since a value 3 below the mean is as far from it as a value 3 above. Then find the mean of those distances. Alex’s scores 1, 4 and 7 are 3, 0 and 3 points from the mean of 4, so their mean absolute deviation is (3 + 0 + 3) ÷ 3 = 2 points. Sam’s is 0.
The signs must come off first. Kept with their signs, the differences from the mean, each value minus 4, are −3, 0 and +3, and they add up to 0. That happens for every data set, because the mean is the balance point of the values.
Σ(x − x̄) = 0 for every data set, so the signs come off before the average measures anything
Try to make the two sides of the balance different lengths.
The values 1, 3, 6 and 10 have mean 5. Their distances from it are 4, 2, 1 and 5, so the mean absolute deviation is 12 ÷ 4 = 3. Drag the last value: the differences from the mean, with their signs, always add up to 0, while the mean absolute deviation changes.
Worked example: A Market Stall's Daily Takings, and the Average Distance from the Mean
Question A market stall took $190, $210, $230, $270, $290 and $310 on the six days it opened last week. (a) Find the mean of the takings and the mean absolute deviation. (b) The stall holder calls a day unusual when the takings are further from the mean than the mean absolute deviation is. How many of the six days were unusual?
1.Find the mean first, because every distance is measured from it: 190 + 210 + 230 + 270 + 290 + 310 = 1500, and 1500 ÷ 6 = $250.
The six days on one scale, with the mean drawn through them. 2.Take the distance of each day's takings from $250, ignoring the sign: 60, 40, 20, 20, 40 and 60 dollars.
The takings total $1500 over 6 days, so the mean is $250. 3.(a) Those distances total 60 + 40 + 20 + 20 + 40 + 60 = 240, so the mean absolute deviation is 240 ÷ 6 = $40. The mean is $250 and on a typical day the takings are about $40 from it. Check: the three days above the mean are 20 + 40 + 60 = $120 above it in total and the three below are $120 below it, so the distances must total 2 × 120 = 240.
(a) The distances from the mean are 60, 40, 20, 20, 40 and 60 dollars, which total 240, so the mean absolute deviation is 240 ÷ 6 = $40. 4.A day is unusual when its distance from $250 is more than $40. Comparing the six distances with 40: 60 is more, 40 is not more, 20 is not, 20 is not, 40 is not and 60 is more.
The dashed lines stand one mean absolute deviation each side of the mean, at $210 and $290. 5.(b) So 2 of the 6 days were unusual, the $190 day and the $310 day, each $60 from the mean. The other four days were $40 or less from it.
(b) Only the $190 day and the $310 day lie outside those lines, so 2 of the 6 days were unusual.
Answer: (a) the mean is $250 and the mean absolute deviation is $40; (b) 2 of the 6 days were unusual
Common mistakes
- Adding the distances with their signs: −60 − 40 − 20 + 20 + 40 + 60 = 0. That is true of every data set, because the mean sits exactly at the balance point of the values. The signs have to be dropped before the distances are added.
- Counting the two $40 days as unusual as well. Their distance from the mean equals the mean absolute deviation, and the stall holder's rule asks for a distance greater than it, so only the two $60 days qualify.