math-2017:7.SP.3 Describe and analyze distributions.
math-2017:7.SP.3a Summarize quantitative data sets in relation to their context by using mean absolute deviation (MAD), interpreting mean as a balance point.
math-2017:7.SP.3b Informally assess the degree of visual overlap of two numerical data distributions with roughly equal variabilities, measuring the difference between the centers by expressing it as a multiple of a measure of variability. For example, the mean height of players on the basketball team is 10 cm greater than the mean height of players on the soccer team, about twice the variability (mean absolute deviation) on either team; on a dot plot (line plot), the separation between the two distributions of heights is noticeable.
Provisional targets. Clause-by-clause educational review is still pending.
Simulated A: 2,3,3,4,4,5,5,6; B: 4,5,5,6,6,7,7,8. Means are four and six; each MAD is one. The centers differ by two MADs, while the observations still overlap from four to six.
Learning goal: Interpret a mean as a balance point, compute mean absolute deviation and compare distributions with roughly equal variability. Express a difference between centers as a multiple of a variability measure without claiming that the groups never overlap.
A mean summarizes a quantitative dataset by dividing its total by the number of observations. It is a balance point in the sense that signed deviations above and below it sum to zero. The reason is arithmetic: n observations with mean m total n × m, and subtracting m from every observation removes n × m altogether, leaving a total deviation of zero. This does not mean the mean must be one of the observed values or the most common value. Its role comes from the complete set's arithmetic balance.
Use two explicitly simulated groups completing the same task. Group A's times in minutes are 2, 3, 3, 4, 4, 5, 5 and 6. Group B's times are 4, 5, 5, 6, 6, 7, 7 and 8. Each group has eight observations. No random population sample or causal assignment is assumed; the records support a comparison of these observed groups, with broader conclusions requiring an appropriate design.
Group A's total is 32, so its mean is four minutes. Subtract four from each observation to obtain signed deviations −2, −1, −1, 0, 0, 1, 1 and 2. The negative deviations total −4 and the positive deviations total +4, balancing at zero. Taking the average of the signed deviations would therefore produce zero and fail to measure the spread.
Mean absolute deviation, or MAD, averages the absolute distances from the mean. For Group A, those distances are 2, 1, 1, 0, 0, 1, 1 and 2. They total eight, and eight divided by eight observations gives a MAD of one minute. The unit is the same as the original measurement because each deviation is a distance in minutes.
Group B is the same pattern shifted two minutes higher. Its total is 48 and mean is six. The absolute deviations from six are again 2, 1, 1, 0, 0, 1, 1 and 2, so its MAD is also one minute. Shifting all values by the same amount moves the center but preserves each distance from the shifted center. The groups therefore have equal variability by this measure.
The difference between means is six minus four, or two minutes. Dividing that difference by the shared MAD of one gives two. The centers are separated by two MADs. This describes the center difference relative to the within-group variability; it is not a probability or a statement that every Group B time is twice a Group A time.
The dot plots show overlap from four through six minutes. Some Group A observations are five or six, while some Group B observations are four or five. It would be false to say every member of B took longer than every member of A. A center comparison describes a tendency across groups while the individual values still vary.
If the distributions were much wider with the same two-minute center difference, the separation would be less noticeable relative to variability. For example, a shared MAD of two would make the same center gap one MAD rather than two. This is why reporting only the difference in means can leave out important information about how distinct the groups look.
Check the roughly-equal-variability condition before using a single variability value as a common reference, and use this ratio only when that MAD is greater than zero; if every value in a group is identical, that group's MAD is zero, and the group is better described by its constant value directly. If one group has MAD one and another has MAD five, selecting one merely because it makes the separation sound larger is misleading. Report both spreads, inspect the plots and qualify the comparison. At this level, use the center-gap multiple as an informal descriptive tool rather than a formal significance test.
MAD differs from range. Group A's range is six minus two, or four minutes, because range uses only the extremes. MAD uses every observation's distance from the mean. Both can be useful, but they answer different spread questions. Median and interquartile range provide other summaries learned earlier; choose measures in relation to the distribution and the question rather than treating them as interchangeable labels.
Preserve repeated observations in a graph. Two threes, two fours and two fives in Group A must appear as separate dots, not be collapsed into one dot at each distinct value. Losing repeats changes the dataset and can change its summaries: without its repeats, Group A becomes 2, 3, 4, 5, 6, whose mean happens to stay four but whose MAD rises from one to 6/5 = 1.2 minutes. The diagram displays all eight observations per group on the same scale so that its visual comparison matches the calculations.
✏️Section 2 · ModelWork through the examples
Worked small dataset: For 1, 3 and 5, the mean is three. Signed deviations are −2, 0 and +2, which sum to zero. Absolute deviations total four, so MAD is 4/3, not two. There are three observations in the average, including the zero deviation at the mean.
Worked relative separation: Two groups have means ten and thirteen and both have MAD 1.5. The difference is three, or 3 ÷ 1.5 = two MADs. That comparison describes the centers relative to their spreads. The actual plots are still needed to describe overlap and unusual observations.
Worked false claim: “The MAD is one, so every value is one minute from the mean” confuses an average distance with an identical distance. In Group A, some values are zero, some one and some two minutes from four. Their average distance is one.
🧩Section 3 · DoPractice, demonstrate, and explain
Computation workshop: Make a table for Group A with observation, signed deviation and absolute deviation. Verify the mean balance and compute MAD. Repeat for Group B without omitting duplicate values. Draw dot plots and mark the means, then calculate the center gap in minutes and in MAD units.
Independent performance: Construct two eight-value datasets with the same variability pattern, a MAD greater than zero and different centers. Compute mean and MAD, graph all observations and describe overlap. Then create a pair with noticeably different spreads and explain why a single shared-MAD comparison needs caution. Write a conclusion distinguishing group tendencies from statements about every individual and from causal or population claims the data do not support.
✅Section 4 · Check6-question lesson practice test
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Problem 1
Why do signed deviations from a mean sum to zero?
Show worked solution
For n observations with mean m, the observations total n × m. Subtracting m from each observation subtracts n × m altogether, so the signed deviations total n × m − n × m = 0. In Group A: 8 values total 32, and subtracting 4 eight times removes 32.
Problem 2
What are Group A’s mean and MAD?
Show worked solution
Mean four minutes and MAD one minute, using all eight observations and their absolute distances.
Problem 3
Why does Group B have the same MAD?
Show worked solution
Every value is shifted up two, moving the mean equally while preserving distances from that mean.
Problem 4
How far apart are the means in MAD units?
Show worked solution
Two MADs, because the two-minute mean difference is divided by the shared one-minute MAD.
Problem 5
Why can’t we say every Group B member took longer than every Group A member?
Show worked solution
The observed times overlap from four through six, so individual values cross despite different centers.
Problem 6
Why must repeated values appear as separate dots?
Show worked solution
They are separate observations; omitting duplicates changes the dataset and can change its numerical summaries; in Group A the mean stays four but the MAD changes from 1 to 1.2.