True/False: A high standard deviation value indicates there is low variation in a dataset.

Question

BlackTom AI · Accepted Answer

Standard deviation measures how spread out values are around the mean, so a larger standard deviation means greater variation, not lower.

Relevant concepts:
- Standard deviation (SD) is the square root of the average squared deviations from the mean (population SD uses mean of squared deviations; sample SD uses n−1 in denominator).
- SD has the same units as the data and increases when data points are more spread or when outliers are present.

Step-by-step reasoning with examples:
1) Consider dataset A: [5, 5, 5, 5]
   - Mean = (5+5+5+5)/4 = 5
   - Squared deviations = (5−5)² repeated 4 times = 0,0,0,0
   - Sum of squared deviations = 0; variance = 0/4 = 0; SD = sqrt(0) = 0
   - Interpretation: SD = 0 indicates no variation.

2) Consider dataset B: [1, 3, 5, 7, 9]
   - Mean = (1+3+5+7+9)/5 = 25/5 = 5
   - Squared deviations = (1−5)²=16, (3−5)²=4, (5−5)²=0, (7−5)²=4, (9−5)²=16
   - Sum = 16+4+0+4+16 = 40; variance = 40/5 = 8; SD = sqrt(8) ≈ 2.828
   - Interpretation: SD ≈ 2.828 shows moderate spread around the mean.

3) Consider dataset C with more spread: [0, 0, 0, 10, 10]
   - Mean = 20/5 = 4
   - Squared deviations = (0−4)²=16 (three times) → 48; (10−4)²=36 (two times) → 72; sum = 120
   - Variance = 120/5 = 24; SD = sqrt(24) ≈ 4.899
   - Interpretation: SD ≈ 4.899 is larger than for B, reflecting greater variation.

Why the correct answer is False:
- The statement "A high standard deviation value indicates there is low variation" contradicts the definition and examples: high SD indicates high variation. Therefore the statement is false.

Why the other option (True) is incorrect:
- Accepting True would invert the meaning of SD and misinterpret dispersion measures; SD increases with spread and outliers, so True is wrong.

Conclusion: The correct answer is False, which matches the provided answer.

True/False: A high standard deviation value indicates there is low variation in a dataset.True/False

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