Comparing Normal Models on a Common Scale

Why Put Normal Models on a Common Scale?

A value that is large in one setting may be typical in another. For example, a score of 80 points could be unusually high in a group whose scores cluster near 60, but relatively ordinary in a group whose scores cluster near 80. Comparing the raw values alone does not tell us how each value stands in its own distribution.

Standardization addresses this by measuring a value’s distance from its model mean in standard deviation units. The resulting z-score allows us to compare relative positions across normal models, even when their means or standard deviations differ. It does not make the original measurements identical, nor does it by itself say which raw value is greater.

Definition: If a variable \(X\) is modeled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\), the standardized value associated with an observation \(x\) is \(z=(x-\mu)/\sigma\). It states how many standard deviations \(x\) lies above or below that model’s mean. A positive z-score is above the mean; a negative z-score is below it.

The Common Scale: Standard Deviation Units

For each value, subtract the mean of its own model, then divide by that model’s standard deviation. This puts the difference from the mean in standard deviation units. For a normal model, the standardized variable has the standard normal distribution, \(Z\sim N(0,1)\). Thus the same z-score represents the same relative location in different normal models.

A z-score is relative to its reference distribution. It is not a measurement in the original units, and it does not say how many units separate two observations. To interpret relative standing further, use the standard normal cumulative distribution: \(\Phi(z)=P(Z\le z)\). This gives the model-based proportion of values at or below that z-score.

Z=\frac{X-\mu}{\sigma}

This comparison is appropriate when each distribution is adequately represented by its stated normal model and its mean and standard deviation are known or suitably specified. The values should also measure the same meaningful outcome, or outcomes for which relative position is a sensible comparison. If a normal distribution is only an approximation to a bounded or nonnegative quantity, its impossible tail values should be recognized as an approximation limitation.

Why Standardization Works

The common scale follows directly from the transformation. If \(X\sim N(\mu,\sigma^2)\), subtracting \(\mu\) centers the distribution at zero, and dividing by the positive value \(\sigma\) expresses deviations in standard deviation units. A linear transformation of a normal variable is normal; its mean and variance can be found from the rules for linear transformations.

E\left(\frac{X-\mu}{\sigma}\right) =\frac{E(X)-\mu}{\sigma} =\frac{\mu-\mu}{\sigma} =0

\operatorname{Var}\left(\frac{X-\mu}{\sigma}\right) =\frac{\operatorname{Var}(X)}{\sigma^2} =\frac{\sigma^2}{\sigma^2} =1

Therefore, \(Z=(X-\mu)/\sigma\sim N(0,1)\). The same reasoning applies separately to each normal model: after standardizing, both are expressed relative to the standard normal scale. Because \(\sigma\) is positive, the transformation also preserves ordering within a model: values above its mean have positive z-scores, and values below it have negative z-scores.

Key result: When \(X\sim N(\mu,\sigma^2)\), its standardized value \(Z=(X-\mu)/\sigma\) follows \(N(0,1)\). Equal z-scores therefore correspond to equal relative positions in their respective normal models, though not necessarily to equal raw values.

Worked Comparisons

Worked Example: Comparing Scores on Two Assessment Scales

Suppose two continuous assessment scales are intended to measure the same general level of performance. For this hypothetical comparison, their score distributions are modeled as normal: Scale A has mean 72 points and standard deviation 8 points, while Scale B has mean 88 points and standard deviation 12 points. A student scores 84 points on Scale A and another scores 104 points on Scale B. Compare their relative standing.

Check the model and comparison. Treat the stated normal distributions as adequate models for the assessment scores, and assume both scales measure a sufficiently similar outcome for relative standing to be meaningful. The values are being compared within their own scales, not as interchangeable raw points.

For the student on Scale A,

z_A=\frac{84-72}{8}=\frac{12}{8}=1.50

The score is 1.50 standard deviations above that scale’s mean. For Scale B,

z_B=\frac{104-88}{12}=\frac{16}{12}\approx 1.33

This score is about 1.33 standard deviations above its scale’s mean. The Scale A student has the higher relative standing, despite having the lower raw score. Using standard normal cumulative areas, \(\Phi(1.50)\approx 0.9332\) and \(\Phi(1.33)\approx 0.9082\). The models thus place the Scale A score around the 93rd percentile and the Scale B score around the 91st percentile. These are model-based percentile positions, not guarantees about a particular set of students.

Worked Example: The Same Raw Value Can Have Different Standing

Suppose the same continuous performance measure has one normal model for Group A, \(X_A\sim N(50,5^2)\), and another for Group B, \(X_B\sim N(70,10^2)\). A value of 60 points is observed in each group. Find and compare the model-based relative positions.

Check the model and comparison. Assume each group’s distribution is adequately modeled by the stated normal distribution and that the measurement has the same meaning and units in both groups. We compare each 60-point value with its own group’s model.

For Group A, 60 is 10 points above the mean, or two standard deviations:

z_A=\frac{60-50}{5}=\frac{10}{5}=2.00

For Group B, 60 is 10 points below the mean, or one standard deviation:

z_B=\frac{60-70}{10}=\frac{-10}{10}=-1.00

In Group A, the value is above the mean and has model-based percentile \(\Phi(2.00)\approx 0.9772\), about the 98th percentile. In Group B, it is below the mean, with percentile \(\Phi(-1.00)\approx 0.1587\), about the 16th percentile. The identical raw value therefore occupies very different positions in the two distributions.

Worked Example: A Shared Cutoff Can Be More Unusual in One Model

Two hypothetical processes produce signed measurement errors in millimetres, where positive values indicate measurements above a reference value. Suppose Process A’s errors follow \(X_A\sim N(0,2^2)\) and Process B’s follow \(X_B\sim N(0,5^2)\). Compare the relative standing of an error of \(+4\) mm in each process.

Check the model and comparison. Signed errors can be positive or negative, and the stated normal models are assumed adequate for both processes. Both means and observations use millimetres, so the raw difference is interpretable as well as the standardized comparison.

For Process A,

z_A=\frac{4-0}{2}=2.00

For Process B,

z_B=\frac{4-0}{5}=0.800

The error is two standard deviations above the Process A mean, but only 0.800 standard deviations above the Process B mean. Their model-based cumulative areas are \(\Phi(2.00)\approx 0.9772\) and \(\Phi(0.800)\approx 0.7881\). Thus \(+4\) mm is much higher in relative position under Process A’s narrower distribution. This does not mean the raw error is larger in A: it is \(+4\) mm in both cases. It means that the same departure from the mean is more unusual relative to A’s variability.

Raw Values and Relative Standing Answer Different Questions

Raw values preserve the original measurement and its units. A z-score describes position relative to a model’s center and spread. Which comparison matters depends on the question. If the concern is the actual number of points, millimetres, or other units, compare raw values. If the concern is how high or low a value is relative to its own distribution, compare z-scores or the corresponding percentiles.

A higher z-score means a higher position in the distribution, not necessarily a better outcome. For a variable where smaller values are desirable, such as an error magnitude or a completion time, a lower value may be preferable even though its z-score is lower. Always connect “higher” or “lower” to the outcome’s meaning.

Common Mistake

Incorrect reasoning: “The score of 104 is larger than 84, so the person with 104 must have the higher standing.” This compares raw values without accounting for the different model means and standard deviations.

Correction: Calculate each z-score with the mean and standard deviation of its own model. In the assessment example, the z-scores are 1.50 and about 1.33, so the 84-point score has the higher relative position even though 84 is the smaller raw value. Conversely, equal z-scores do not imply equal raw values; they imply equal relative positions under the stated models.

Key takeaway: To compare relative standing across normal models, standardize each value using its own model’s mean and standard deviation. Interpret the resulting z-scores in context, and keep the distinction between relative position and raw magnitude clear.

Check Your Understanding

For each question, distinguish the model-based relative position from the original measurement.

A value is 6 units above a model mean with standard deviation 3. What is its z-score, and what does its sign indicate?
Two normal models have means 40 and 55 and standard deviations 5 and 10. Compare the relative positions of values 50 and 65, respectively.
In a normal model, what does a z-score of \(-1.2\) say about a value’s position relative to its mean and standard deviation?
Two observations have the same z-score but come from models with different means. Must their raw values be equal? Explain.
Why can a larger z-score fail to mean a more desirable outcome?