Z-Score Calculator

Measure how far a value is from the mean in standard deviations, or work backward from a z-score or normal percentile.

Solve for

Z-score

1.5

85 is 1.5 SD above the mean · 93.32th percentile

  • Z-score1.5
  • Raw score85
  • P(X < x) — area to the left0.93319 (93.32%)
  • P(X > x) — area to the right0.06681 (6.68%)
  • Two-tailed P(|Z| > |z|)0.13361 (13.36%)
  • Percentile93.32th
Reference points for μ = 70, σ = 10
zRaw scorePercentile
-3400.13%
-2502.28%
-1.5556.68%
-16015.87%
-0.56530.85%
07050.00%
+0.57569.15%
+18084.13%
+1.58593.32%
+29097.72%
+310099.87%

How this was calculated

z = (x − μ) ÷ σ = (8570) ÷ 10 = 1.5

Probabilities from the standard normal cumulative distribution Φ(z).

Standardizing a value

Use z = (x − μ) ÷ σ, where x is the observation, μ the population mean, and σ a positive standard deviation. A positive z is above the mean; a negative z is below it. To reverse the operation, x = μ + zσ.

A z-score can describe distance from a mean without assuming a normal distribution. However, this page’s percentile and tail probabilities use the standard normal distribution, so interpreting them requires that model to be appropriate.

Example: score 85, mean 70, standard deviation 10

z = (85 − 70) ÷ 10 = 1.5. Under a normal model, about 93.3193% of observations are below this score and 6.6807% are above it. The probability outside ±1.5 standard deviations is about 13.3614%.

Percentiles and tails

The percentile mode inverts the cumulative normal distribution: a 90th percentile corresponds to z ≈ 1.281552. With mean 70 and standard deviation 10, that is a raw value near 82.815516.

Use percentiles strictly between 0 and 100. The endpoints correspond to infinite z-scores in an unbounded normal model. Standard deviation must be finite and greater than zero. Rounded probabilities near zero or one do not imply impossibility or certainty.

Frequently asked questions

Is a z-score a percentage?

No. It is a signed number of standard deviations. Converting it to a percentile requires a distributional model; this page uses the normal model.

Does a two-tailed area prove a result is significant?

No. A statistical test requires a justified null model, sampling assumptions and a defined test statistic. A normal tail area alone does not establish those conditions.

Last updated . Results are estimates for informational purposes only.