DA0-002 Data Analysis Practice Question
A data analyst wants to use a Z-score to standardize a dataset. The variable has a mean of 50 and a standard deviation of 10. What is the Z-score for a raw value of 70?
⚠ Common exam trap
Many candidates confuse the difference between the raw value and the mean (20) with the Z-score, or incorrectly reversing the numerator to get a negative Z-score, which would misrepresent the direction from the mean.
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
✓
2
The Z-score formula is Z = (X - μ) / σ, where X is the raw value, μ is the mean, and σ is the standard deviation. Plugging in the given values: Z = (70 - 50) / 10 = 20 / 10 = 2. Thus, the raw value of 70 is 2 standard deviations above the mean, corresponding to a Z-score of 2.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
0.5
Why it's wrong here
Dividing the raw value by the standard deviation ignores the mean entirely; the correct calculation subtracts 50 from 70 before dividing by 10, giving 2. Using the raw value directly is tempting when the mean is a round number, but Z-scores measure deviation from the mean.
- ✗
20
Why it's wrong here
Subtracting the mean from the raw value gives 20, which is the deviation itself, not the Z-score; dividing by the standard deviation of 10 yields 2. Reporting the raw difference is tempting when the mean is a round number, but standardisation always requires that division step.
- ✗
-2
Why it's wrong here
The negative sign reverses the direction: 70 sits above the mean of 50, so its Z-score must be positive. The magnitude 2 is right, but the sign is wrong. Negating is tempting when recalling that values below the mean produce negative Z-scores, yet this raw value lies above it.
- ✓
2
Why this is correct
Applying the Z-score formula (x minus mean, divided by standard deviation) gives (70−50)/10 = 2. This standardised value states the raw score sits two standard deviations above the mean, satisfying the stem's requirement to standardise using the given mean of 50 and standard deviation of 10.
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JA
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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