Questions: Standard Normal Distribution and Z-Scores

Question 1

Student A scores 75 on Exam 1 (μ = 60, σ = 10). Student B scores 82 on Exam 2 (μ = 70, σ = 15). Who performed better relative to their class?

• A. Student B, because 82 > 75 in absolute terms
• B. They performed equally — both scored above their class mean
• C. Student A: Z = (75−60)/10 = 1.5 vs. Student B: Z = (82−70)/15 = 0.8, so Student A is farther above her class average
• D. Cannot be determined without knowing the shape of each score distribution

Answer: (C)

Z-scores standardize scores to a common scale, enabling comparisons across different distributions. Student A is 1.5 standard deviations above her class mean; Student B is only 0.8 standard deviations above his. The absolute scores (75 vs. 82) are meaningless for comparison because the exams have different means and spreads. This is the defining use case for Z-scores: translating raw scores into relative positions.

Question 2

What do the two operations in Z = (X − μ)/σ each accomplish?

• A. Subtracting μ scales the distribution; dividing by σ centers it at zero
• B. Subtracting μ centers the distribution at zero; dividing by σ rescales it so one standard deviation equals one unit
• C. Subtracting μ removes outliers; dividing by σ converts the distribution from normal to uniform
• D. Both operations together convert any distribution to a normal distribution

Answer: (B)

Subtracting μ shifts the distribution so its mean is at 0 — centering. Dividing by σ rescales the spread so that one standard deviation now equals exactly 1 unit — rescaling. The shape remains unchanged (still bell-shaped and symmetric if it was normal). Option D is a common misconception: standardization preserves the shape; it does not make a non-normal distribution normal.

Question 3

A student with Z = 1.5 and another student with Z = 1.5 on two completely different exams with different means and standard deviations have the same relative standing within their respective distributions.

• T. True
• F. False

Answer: True

True. Z = 1.5 means the student scored 1.5 standard deviations above the mean, regardless of what the underlying scale is. Whether the exam was scored 0–100 or 200–800, Z = 1.5 encodes the same relative position. This is exactly why standardization is useful: it strips away the original scale and leaves only the relative position, making cross-distribution comparison valid.

Question 4

Z-scores are primarily meaningful when the original data follows a normal distribution — for non-normal data, computing Z = (X − μ)/σ is undefined.

• T. True
• F. False

Answer: False

False. The formula Z = (X − μ)/σ can be computed for any data with a finite mean and standard deviation, regardless of the shape of the distribution. The result always measures 'how many standard deviations from the mean.' What you cannot do with non-normal data is use the standard normal probability table to look up P(Z < z) — those probabilities assume a normal distribution. Z-scores remain well-defined and useful (for relative ranking, outlier detection, etc.) even for non-normal data.

Question 5

Explain why the standard normal distribution (μ = 0, σ = 1) serves as a universal reference for all normal distributions. What do the two steps of Z = (X − μ)/σ each accomplish?

Model answer: Every normal distribution differs only in its mean and standard deviation — they are all the same bell shape, just shifted and stretched. The transformation Z = (X − μ)/σ removes both: subtracting μ centers the distribution at 0 (eliminates the shift), and dividing by σ rescales the spread to 1 (eliminates the stretch). Every normal distribution, regardless of its original parameters, maps to the same N(0,1) after this transformation. This means a single table or function for N(0,1) covers all possible normal probability calculations.

This is the key insight: standardization is a change of units, not a change of distribution. Just as converting miles to kilometers doesn't change the distance, converting X to Z doesn't change the probability — it just expresses it on a canonical scale. The universality of the standard normal table follows from the fact that all normal distributions are related by linear transformations, which Z = (X − μ)/σ exactly reverses.