Finocturne
Statistics & Data Analysis

Z-score

Place one monthly return on a standard-deviation ruler to see its direction and distance from the .

Question, try, then observe

How far is 6% above the mean?

Assume a 2% mean monthly return and a 2-percentage-point SD. Move the observation to convert return distance into SD units. This is a ruler, not a normal-distribution plot.

Illustrative data, not live market data. A monthly return of 2 = 2%; dispersion and return differences use percentage points, with no annualization.

01 · Make a predictionFrom 2% to 6%, how many 2-percentage-point steps?
Observation 6%: Z=2.00; horizontal axis is monthly return in percent.-6%Z=-4-2%Z=-22%Z=06%Z=210%Z=4

Find the percentage-point gap, then divide by SD; Z is unitless

z=6−22=42≈2
Z-score2.00
Distance in SD units2.00
Observation6.00%
Mean / SD2.00% / 2.00 pp

The observation is 2.00 SDs above the mean. The sign preserves its direction.

A Z-score does not establish normality or directly give an event probability. This distance ratio is undefined when SD=0.

Divide (observation − mean) by SD in the same units. The percentage-point units cancel, leaving dimensionless Z.

What did you just see?

With a 2% and a two-percentage-point SD, 6% is two steps to the right and −2% is two steps to the left.

Standardizing is not normalizing a distribution

Computing a does not require normality, but it does not remove heavy tails, skewness or time dependence. The diagram's marks show SD distances, not particular probabilities.

Another example

  1. Enter an observed return of 6 and a mean of 2, both in %.
  2. Enter an SD of 2, meaning two percentage points.
  3. Calculate (6−2)/2=2 and read the position two SDs above the mean.
Show formal definition and formula

A Z-score measures the signed distance between an observation and a reference mean, using the SD as a ruler.

Standard score

Z is the observation minus its mean, divided by a positive SD in the same units.

xᵢ / yᵢ
Observations from the same period. A return input of 2 means 2%, calculated directly in percentage units.
μ / σ
Reference mean and a positive SD in the same units as the observation; Z is undefined at σ=0.
Z
A dimensionless relative position, not a normality assumption or event probability.

Common mistakes

A Z-score is a relative position, not proof of normality, an anomaly probability or an investment signal. Mismatched reference periods can also make it misleading.

Frequently asked questions

Does a Z-score have units?

No. Observation, mean and SD must use matching input units; division produces a unitless relative distance.

Does Z=2 give an event probability?

Not by itself. Standardizing does not make data normally distributed. Tail probabilities require an appropriate, validated distribution model.

What if the SD is zero?

Division by the SD is undefined, so there is no Z-score. Even an observation equal to the mean must not be reported as zero in this case.

You can now explore…

  • Measure signed distance in SD units
  • Standardizing does not establish normality or probabilities
Term reference3 terms

Open a term when you need a reminder. Underlined words in the article work too.