A z-score is not a distribution
You can standardise an observation using z = (x − μ) / σ, where σ is a positive standard deviation. This does not prove that the data are normally distributed. Using normal-distribution probabilities involves an additional model assumption. A test using the standard error of a sample mean has a different quantity in its denominator.
Read the table heading
In a table of cumulative left-hand areas, the value at z = 0 is 0.5. A table of areas between the mean and z shows 0 there instead. Identify the area being reported before matching the row and column.
| z | Left area Φ(z) | Right area |
|---|---|---|
| 0 | 0.5000 | 0.5000 |
| 1 | 0.8413 | 0.1587 |
| 1.96 | 0.9750 | 0.0250 |
| −1 | 0.1587 | 0.8413 |
A two-tailed example
For |z| = 1.96, the total area in both extremes is approximately 2 × 0.0250 = 0.0500. This is not the probability that a hypothesis is true. A z-score alone does not support a scientific verdict without the model and testing context.
Try the calculation
The tool below uses a numerical approximation to Φ(z). It displays left, right and two-tailed areas together. Very extreme values may round to 0 or 1. The tool does not replace specialist software for precise tail probabilities.
Normal-distribution areas
Select the button to calculate.
Numerical approximation to Φ(z); results are rounded to 6 decimal places.
Sources and provenance
Source check / web edition: 14 October 2026. Historical documents retain their own original dates.