Choosing and Reporting a Measure of Spread

The two calculator values come from different denominators. The population standard deviation divides the sum of squared deviations by n: for Set A, the squared deviations are 16, 4, 0, 0, 4, 16, whose sum is 40, so σ = √(40/6) ≈ 2.58. The sample standard deviation divides by n − 1: s = √(40/5) = √8 ≈ 2.83. Set B gives σ = √(4/6) ≈ 0.82 and s = √(4/5) ≈ 0.89.

The choice depends on what the six values represent. If the six values are the entire population of interest-for example, all six students in a small group-then σ is appropriate. If the six values are a sample used to estimate the spread of a larger population, then s is appropriate, because dividing by n − 1 corrects for the fact that the sample mean is estimated from the same data. In a school statistics course, data are usually treated as a sample, so s is the safer default, and the class should say so explicitly. The teacher should not present one value as "the real one" without stating the denominator and the context. For comparison, consistency matters more than the choice: report both sets to the same precision, e.g., 2.83 and 0.89, or 2.6 and 0.8 if rounding to one decimal place, and state the rounding rule.

Sequence (about 25 minutes):

1. Dot plots (5 min). Students plot Set A and Set B on parallel dot plots with a shared axis from 5 to 15. Ask: "Which distribution is more spread out? How can you tell without computing anything?" Students should notice that Set A reaches farther from the center, while Set B is tightly clustered around 10.
2. Deviations and MAD (10 min). Build a table of deviations from the mean 10 for each set. For Set A: −4, −2, 0, 0, 2, 4. For Set B: −1, −1, 0, 0, 1, 1. Ask: "Why do the deviations sum to zero, and what can we do about that?" Students compute mean absolute deviation: Set A has absolute deviations 4, 2, 0, 0, 2, 4, sum 12, MAD = 2; Set B has absolute deviations 1, 1, 0, 0, 1, 1, sum 4, MAD ≈ 0.67. This confirms the visual comparison.
3. Why square? (7 min). Ask: "If we square the deviations before averaging, how does that change the comparison?" Students compute the mean squared deviation: Set A 40/6 ≈ 6.67; Set B 4/6 ≈ 0.67. Then take square roots to return to the original units. Connect the result to the calculator's σ and s values.
4. Exit question (3 min). "A class reports a standard deviation of 2.8. What must you know before comparing it with another class's standard deviation of 0.9?"

Key question throughout: "What does a measure of spread have to do to be useful?" The sequence uses the dot plot to anchor the idea, MAD to make absolute distance concrete, and squaring to explain why standard deviation is not just an arbitrary calculator command.

The claim is false as a general statement, even though it happens to be true that Set A has both a larger range (14 − 6 = 8, versus 11 − 9 = 2 for Set B) and a larger standard deviation. Range depends only on the two extreme values; standard deviation depends on every deviation from the mean.

A counterexample with equal ranges but different standard deviations makes this clear. Consider Set C = 0, 0, 0, 10, 10, 10 and Set D = 0, 0, 0, 0, 0, 10. Both have range 10. Set C has mean 5 and deviations −5, −5, −5, 5, 5, 5; the sum of squared deviations is 6(25) = 150, so σ = √(150/6) = 5. Set D has mean 10/6 ≈ 1.67 and squared deviations dominated by the single 10, giving a different standard deviation (about 3.73). Same range, different standard deviation. Conversely, two sets can have the same standard deviation but different ranges if one has a few extreme values and the other has many moderate deviations.

For the original data, the correct explanation is that Set A's values are consistently farther from 10: its squared deviations are 16, 4, 0, 0, 4, 16, while Set B's are 1, 1, 0, 0, 1, 1. The standard deviation reflects that entire pattern, not just the endpoints. The student has noticed a correlation between range and standard deviation but has mistaken it for a definition.