Z-Scores and Standardization

A distance measured in standard deviations

A value of 85 has a different meaning on a test centered at 70 than on one centered at 90. A z-score expresses the distance from a mean relative to a standard deviation:

$z=\frac{x-\mu}{\sigma}$

Here x is the value, μ the population mean, and σ the population SD, which must be positive. If you standardize relative to a sample, use (x − x̄)/s and identify it as a sample-standardized score.

• z = 0: at the mean.
• z = 1.5: 1.5 SD above the mean.
• z = −2: 2 SD below the mean.

The sign tells direction; the magnitude tells relative distance. A z-score has no measurement unit.

Worked example: compare two tests

Maya scores 85 on a test with mean 70 and SD 10. Leo scores 90 on another test with mean 80 and SD 5.

$z_{Maya}=\frac{85-70}{10}=1.5,\qquad z_{Leo}=\frac{90-80}{5}=2$

Leo’s score is farther above the mean relative to the variability of his test. This does not establish that he knows more: the tests may measure different skills or use different reference groups.

If both reference distributions are reasonably normal, the corresponding cumulative probabilities are Φ(1.5) = 0.9332 and Φ(2) = 0.9772. Under those models, the scores are around the 93rd and 98th percentiles. The z-scores themselves did not require normality; translating them to those percentiles did.

Standardization does not make a distribution normal

Subtracting the mean and dividing by SD shifts and rescales a distribution. A right-skewed distribution stays right-skewed. Two clusters remain two clusters.

The standard normal model, Z ~ N(0, 1), is a particular bell-shaped distribution. It is not the distribution of every standardized dataset. Use empirical percentiles when you want rank within data whose shape is unknown.

Convert a z-score to a normal probability

The site’s z-table reports left-tail probability Φ(z) = P(Z ≤ z).

Question	Calculation for z = 1.5	Result
Below 1.5	Φ(1.5)	0.9332
Above 1.5	1 − Φ(1.5)	0.0668
Below −1.5	1 − Φ(1.5)	0.0668
Between −1.5 and 1.5	Φ(1.5) − Φ(−1.5)	0.8664

For a continuous model, P(Z = 1.5) = 0, so strict and non-strict inequalities have the same probability. Area over a range is what carries probability.

Sketch or name the required region before looking up a number. A common mistake is reporting 0.9332 for “above 1.5,” which answers the opposite question.

Work backward from a percentile

Suppose order preparation time is modeled as normal with mean 10 minutes and SD 2. The 90th percentile has z ≈ 1.2816.

$x=\mu+z\sigma=10+1.2816(2)=12.5632\text{ minutes}$

Under this model, 90% of times are at most about 12.56 minutes. The remaining 10% are longer. This is an individual-outcome percentile, not a confidence interval for the mean.

One observation and a sample mean use different denominators

For an individual value, use σ. For a mean of n independent observations with known population SD, the standard error is σ/√n:

$z_{mean}=\frac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}$

With μ₀ = 10, σ = 2, and n = 25, a sample mean of 10.8 has z = 0.8/0.4 = 2. An individual observation of 10.8 has z = 0.8/2 = 0.4. Confusing the two makes routine individual values look like unusual sample means.

When σ is estimated using the sample SD, a mean test usually uses a t-statistic, even for a large sample.

Use z-scores cautiously for outliers

A rule such as |z| > 3 can flag observations for investigation under a suitable model. It is not a universal deletion rule. SD and mean are themselves influenced by outliers; heavy-tailed populations naturally generate more extreme standardized values than a normal model predicts.

Check units, transcription, collection conditions, and the distribution shape. Report how any exclusions were decided.

Practice

1. A normal model has μ = 50 and σ = 8. Find the z-score for x = 62 and the probability of exceeding 62.

Show the calculation

z = (62 − 50)/8 = 1.5. The upper-tail probability is 1 − Φ(1.5) ≈ 0.0668, or 6.68%. The cumulative table value 0.9332 is the probability of being below 62.

2. A positively skewed sample has standardized score z = 2. Is that automatically the 97.7th percentile?

Show the explanation

No. Standardization changes location and scale, not shape. The 97.7th-percentile conversion comes from the normal model. Estimate the percentile from the relevant data or use a justified distribution model.

Continue learning

Use the z-score calculator, then explore the normal distribution and sampling distributions. NIST’s normal-distribution reference defines the standard normal density and cumulative probability.