A score of 85 with mean 70 and SD 10
z = (85 − 70) ÷ 10 = 1.5
The difference is 15 original units. Dividing by 10 units per standard deviation gives 1.5 standard deviations above the mean.
Distance from the mean
Express a value in standard deviations from the mean, or work backward from a z-score to find the original value.
Z-score
Enter a score, mean, and standard deviation, then select Calculate.
A z-score describes relative distance. It does not by itself give a percentile, a probability, or an outlier decision.
≈ marks display rounding. Calculations run in your browser. Press Enter to calculate. Successful results appear in this tab's Recent calculations.
z = (85 − 70) ÷ 10 = 1.5
The difference is 15 original units. Dividing by 10 units per standard deviation gives 1.5 standard deviations above the mean.
z = (40 − 52) ÷ 8 = −1.5
The signed difference is −12. The z-score is −1.5, meaning the value is 1.5 standard deviations below the mean.
x = 100 + (−1.25 × 12) = 85
The value lies 15 units below the mean, so its raw score is 85. The result keeps the same units as the mean and standard deviation.
For a population mean μ and positive population standard deviation σ, standardize a raw value x with z = (x − μ)/σ. To reverse the calculation, use x = μ + zσ. For descriptive sample standardization, replace μ and σ with the sample mean and sample standard deviation. The supplied statistics determine the reference; this tool does not estimate them from observations. References: NIST — Z-scores and outlier detection and OpenStax — z-scores and the standard normal distribution.
Subtracting a mean and dividing by a positive standard deviation does not require a normal distribution. It does not turn an arbitrary distribution into a normal one. A standard normal probability table needs an appropriate normal model; a z-score alone does not establish a percentile or the probability of a score occurring. This calculator therefore returns distances and raw scores, without attaching probabilities.
The standard deviation must describe the same quantity and reference group as the mean. Enter the raw value, mean, and standard deviation in matching units; the resulting z-score has no physical unit. The supplied standard deviation is used directly. This is not a hypothesis-test z statistic for a sample mean and does not divide a standard deviation by the square root of a sample size to obtain a standard error.
Zero standard deviation cannot be used in the forward formula, even when the raw value equals the mean: that would require division by zero. Both modes require a positive standard deviation so that standardization and its inverse refer to the same valid scale. Negative or zero standard deviations are rejected.
A large absolute z-score can prompt further investigation, but it does not automatically prove that a value is erroneous or is an outlier. The reference data, sample size, distribution, and intended analysis matter. The calculator does not apply an outlier threshold or label a score as statistically significant.
Active numeric inputs accept up to 64 characters, 15 integer digits, and 10 decimal places. Expressions and scientific notation are not accepted. Hidden inputs are disabled and ignored. Exact fractions retain the entered decimals through subtraction, multiplication, and division. Display values round halfway away from zero, omit trailing zeros, and show ≈ when rounding occurs. A nonzero value that rounds to zero carries a notice, while the above/below interpretation uses the unrounded sign. Changing decimal places updates the display without adding a recent-calculation entry.
Yes. A negative z-score means the original value is below the supplied mean. It does not necessarily mean the original value itself is negative. For example, 40 is positive but lies below a mean of 52.
No. It means two standard deviations above the mean. Turning that distance into a percentile needs a distribution model or the underlying observations. The number is not a percentage.
Use the statistics that fit your reference. To describe position within a sample, supply its sample mean and sample standard deviation. To compare against a specified population, supply the population parameters. Do not mix statistics from different groups.
No. This calculation changes the center and scale. It preserves the distribution's shape rather than testing or correcting it. Normal probability calculations need a separate, justified model.
A very small negative z-score may round to zero at the selected precision. The ≈ marker and rounding notice distinguish it from an exact zero. The interpretation keeps the direction of the original unrounded result.