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MLA-C01 Practice Question: A data engineer is preparing a dataset for a…
A data engineer is preparing a dataset for a k-means clustering algorithm. The features have different scales: age (18-100), income ($20k-$200k), and number of purchases (0-50). Without scaling, which feature will dominate the distance calculations?
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
✓
Income
Income has the largest range (180,000 compared to 82 and 50), so it will dominate Euclidean distance calculations. Standardization or normalization is needed before clustering.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
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All features will contribute equally
Why it's wrong here
Features with larger numeric ranges produce larger squared differences, so income ($20k-$200k) dominates the Euclidean distance and skews centroid assignment. It is tempting to assume k-means treats all features equally, but without normalisation the algorithm weights whichever feature has the widest raw scale most heavily.
- ✓
Income
Why this is correct
Euclidean distance sums squared differences, so the feature with the widest numeric range contributes most. Income spans roughly $180k against age's 82 and purchases' 50, making its squared deviations dominate the distance metric and skew cluster assignment.
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Number of purchases
Why it's wrong here
Purchases span 0–50, so its contribution to Euclidean distance is far smaller than income's 20,000–200,000 range. It is tempting because purchases are numeric and unscaled, but that suits a dataset where all features share comparable magnitudes, not this one.
- ✗
Age
Why it's wrong here
Age spans 18–100, giving a range of roughly 82, dwarfed by income's 180,000 spread, so it cannot dominate squared distances. It is tempting because age is unscaled, but that suits features with similar ranges, not this mix.
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