DA0-002 Data Analysis Practice Question
A dataset contains employee salaries ranging from $30,000 to $200,000. An analyst wants to scale the salaries to a range of 0 to 1 for use in a distance-based clustering algorithm. Which method should they use?
⚠ Common exam trap
DA0-002 often tests normalization vs. standardization — candidates pick z-score because it is commonly used, missing that the question explicitly requires a 0-to-1 bounded range that only min-max normalization provides.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
Min-max normalization
Min-max normalization rescales values linearly to a fixed range, typically [0, 1], using the formula (x − min) / (max − min). This is exactly what the analyst needs for a distance-based clustering algorithm, where features on different scales would otherwise dominate the distance metric. It preserves the relative ordering and shape of the distribution while bounding all values between 0 and 1.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Log transformation
Why it's wrong here
Log transformation compresses skew and stabilises variance but leaves values on an unbounded scale, so salaries would not fall within 0 to 1. It would be correct for reducing right-skew in monetary data before linear modelling, not for bounded rescaling.
- ✗
Robust scaling
Why it's wrong here
Robust scaling centres on the median and divides by the interquartile range, producing values that are not bounded to 0 to 1. It would be correct when outliers must not distort the scaling, whereas the stem requires an explicit minimum-maximum bound.
- ✓
Min-max normalization
Why this is correct
Min-max normalization rescales each value using (x − min)/(max − min), mapping the $30,000–$200,000 salary range linearly onto 0–1. This preserves relative distances, which distance-based clustering requires, unlike z-score standardisation, which centres on the mean with unbounded output.
- ✗
Z-score standardization
Why it's wrong here
Z-score standardisation subtracts the mean and divides by the standard deviation, yielding a distribution centred on zero with no fixed upper or lower bound. It would be correct where a normal distribution and unit variance are wanted, not a 0 to 1 range.
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Senior Network & Security Engineer · founder of Courseiva
Last reviewed September 2026 · checked against the official CompTIA exam blueprint
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