DA0-002 Data Acquisition and Preparation Practice Question
During EDA, an analyst calculates the Z-score for each data point in a dataset. A data point with a Z-score of 3.5 is identified. What does this indicate?
⚠ Common exam trap
DA0-002 often tests whether candidates confuse Z-score (standard deviations from mean) with IQR-based outlier detection or with frequency counts, luring them to pick 'high frequency' or 'within IQR' answers.
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 data point is likely an outlier
A Z-score of 3.5 means the data point lies 3.5 standard deviations above the mean. In most distributions, values beyond ±3 standard deviations are statistically rare (about 0.3% of a normal distribution) and are commonly flagged as outliers. This is the standard EDA heuristic for outlier detection.
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 data point has a high frequency
Why it's wrong here
Z-scores standardise distance from the mean in standard-deviation units; they say nothing about how often a value occurs. Frequency is measured by counts or density, not by deviation. The confusion arises because extreme values are often rare, but rarity is a consequence, not the definition, and a Z-score of 3.5 simply denotes 3.5 SD above the mean.
- ✗
The data point is exactly at the mean
Why it's wrong here
A Z-score of exactly 0 places a point at the mean; 3.5 sits 3.5 standard deviations above it, indicating an extreme outlier. The mean is the reference point from which Z-scores are measured, so confusion arises because the mean anchors the calculation. Z-scores near zero, not 3.5, would indicate proximity to the mean.
- ✓
The data point is likely an outlier
Why this is correct
A Z-score of 3.5 lies beyond three standard deviations from the mean, satisfying the stem's outlier criterion. Under a normal distribution roughly 99.7% of values fall within three standard deviations, so such an extreme standardised deviation is statistically improbable and warrants flagging as a likely outlier during EDA.
- ✗
The data point is within the interquartile range
Why it's wrong here
A Z-score of 3.5 lies 3.5 standard deviations from the mean, far outside any interquartile range, which spans only the middle 50% of values (roughly ±0.67 SD). The IQR is a positional measure, not a standardised deviation score, so it cannot express this value. It tempts because both describe spread, but IQR suits box-plot outlier fencing.
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Written and reviewed by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
Last reviewed September 2026 · checked against the official CompTIA exam blueprint
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