DA0-002 Data Analysis Practice Question
A data scientist is performing K-means clustering on customer data. She plots the within-cluster sum of squares (WCSS) for different values of k and observes an 'elbow' at k=4. What does this indicate?
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
✓
The optimal number of clusters is 4
The elbow method suggests that adding more clusters beyond k=4 yields diminishing returns, so k=4 is a suitable number of clusters.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
The optimal number of clusters is 4
Why this is correct
The elbow marks where adding clusters stops yielding meaningful WCSS reduction, so k=4 balances model complexity against fit. Beyond four, each extra cluster captures only marginal variance, indicating diminishing returns. This inflection point therefore identifies four as the optimal cluster count for the customer dataset.
- ✗
The algorithm should be run with k=3 to avoid overfitting
Why it's wrong here
The elbow at k=4 suggests four clusters balance fit against complexity; choosing k=3 discards a genuine grouping. Overfitting concerns apply when k approaches the sample size, not at the bend where marginal WCSS gain drops sharply.
- ✗
The data contains exactly 4 outliers
Why it's wrong here
The elbow reflects the point of diminishing returns in WCSS reduction as cluster count increases, not the number of outliers. Outliers are individual distant points, identified through distance measures or isolation techniques, and bear no fixed relationship to the chosen k.
- ✗
The WCSS is minimized at k=4, indicating perfect clustering
Why it's wrong here
The elbow marks where adding clusters yields diminishing WCSS reductions, not the minimum; WCSS keeps falling as k rises, reaching zero when k equals the number of points. Minimisation alone would always favour the largest k, so it cannot indicate perfect clustering.
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