Z-score calculator
In a normal distribution about 68% of values fall within one standard deviation of the mean, 95% within two, and 99.7% within three. A z above 2 is therefore often treated as notable and above 3 as rare: roughly one value in 370 sits beyond three deviations in either direction. All three figures are properties of the normal curve alone, and they carry over to real data only as far as the data is actually normal.
The z-score itself needs no assumptions. Every percentile and tail figure below it comes from the standard normal curve and is only as good as that assumption.
A z-score is how many standard deviations a value sits from the mean: (x − μ) ÷ σ. A score of 115 against a mean of 100 and a standard deviation of 15 gives z = 1, the 84.13th percentile. The score is arithmetic; the percentile assumes a normal distribution.
How to calculate a z-score
Standardising strips out the units and the scale so that only position within a distribution remains. An IQ of 115 and a test score one deviation above its own mean are the same claim in different clothes, and the z-score is what lets you say so. The subtraction and the division are unconditional, which is exactly why the rows below the score deserve more suspicion than the score itself.
The percentile is borrowed from a curve, not from your data
A z of 2 reads as the 97.72nd percentile because 97.72 per cent of a standard normal distribution lies below 2. Take away normality and that number goes with it. What survives is Chebyshev’s inequality, which holds for every distribution with a finite variance and is correspondingly weak: at least 75 per cent of any data lies within two standard deviations of its mean, and at least 88.9 per cent within three. Against the normal figures of 95.45 and 99.73, the distribution-free guarantee is a much looser thing, and the gap between the two is the size of the assumption you are making.
A concrete case makes the size of it obvious. In an exponential distribution, the kind that describes waiting times, the mean and the standard deviation are equal, so a z of 2 lands at exactly 1 − e⁻³. That is the 95.02nd percentile rather than the 97.72nd, and the difference turns a top-2-per-cent reading into a top-5-per-cent one on the same number. Push the shape further and the one-sided form of Chebyshev’s bound allows a z of 2 as low as the 80th percentile. Where the shape is unknown, rank the values and read the percentile straight off the data rather than off the curve.
A small sample cannot produce a large z
The rule that anything beyond ±3 is an outlier quietly assumes you have enough data for a z of 3 to exist. In a sample of n values scored against their own mean and sample standard deviation, no z-score can exceed (n − 1) ÷ √n. Shiffler set that bound out in 1988, and the consequence is blunt: with ten observations the largest attainable z is 2.85, so the three-deviation rule cannot fire at all until the eleventh value arrives. With five observations the ceiling is 1.79.
That bound also explains a pattern that looks like coincidence in small data sets: the most extreme point often lands near the ceiling, because one wild value inflates the standard deviation it is being divided by. An outlier hides itself. This is the argument for the median and the median absolute deviation as a robust alternative when the data set is small and the question is which point does not belong.
Which mean, and whose deviation
Written out, the formula uses population parameters, μ and σ, and the panel takes whatever you type without asking where the numbers came from. If the mean and deviation are estimates from the same handful of values that contains x, the score inherits their uncertainty and comparing it against fixed normal cut-offs overstates the evidence. Feeding it a published population mean and deviation is the case the formula was written for.
What people use it for
- Comparing scores from tests with different scales
- Turning a raw score into a percentile
- Screening a data set for values that do not belong
- Checking a reported z-score against the sample it came from
Questions
The number of standard deviations a value sits from the mean. Positive is above, negative below, and zero is the mean itself.
Conventionally beyond ±3, roughly one value in 370 for normal data. Some fields use ±2, which flags about one in twenty and will therefore flag something in almost any data set.
Through the standard normal cumulative distribution, which is the percentile row here. A z of 1 is the 84.13th percentile and a z of 2 is the 97.72nd.
The score does; the percentile does not. Without normality all you can claim is Chebyshev’s bound: at least 75% within two deviations, at least 88.9% within three.
Because a z-score computed against a sample’s own mean and deviation is capped at (n − 1) ÷ √n. At n = 10 that is 2.85, and you need eleven values before 3 is reachable.
The share of a normal distribution within one, two and three standard deviations of the mean. It is a fact about the curve, not about your numbers.
The proportion of a normal distribution at least this far from the mean in either direction. At z = 1 that is 31.7%; at z = 2, 4.55%.
If the deviation was estimated from a small sample, yes. The t-distribution has heavier tails and gives the more cautious tail probability for the same statistic.