DA0-002 Data Analysis Practice Question
A company wants to segment its customers into distinct groups based on purchasing behavior. Which algorithm is best suited for this task?
⚠ Common exam trap
Test-takers frequently confuse supervised learning algorithms (like decision trees or logistic regression) with unsupervised clustering, mistakenly thinking that any algorithm that 'groups' data can be used for segmentation without recognizing the need for unlabeled data.
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
✓
K-means clustering
K-means clustering is an unsupervised learning algorithm that partitions data into K distinct clusters based on feature similarity, making it ideal for segmenting customers by purchasing behavior without predefined labels. It groups customers who exhibit similar purchasing patterns, enabling the company to identify natural segments for targeted marketing.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Decision tree
Why it's wrong here
Decision trees can classify, but without labels, they cannot create segments.
- ✗
Logistic regression
Why it's wrong here
Logistic regression is for classification, but it requires labeled data, and segmentation is typically unsupervised.
- ✓
K-means clustering
Why this is correct
K-means clustering groups similar customers together based on features.
- ✗
Linear regression
Why it's wrong here
Linear regression predicts a continuous outcome, not groups.
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JA
Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
This DA0-002 practice question is part of Courseiva's free CompTIA certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the DA0-002 exam.