Z-Score from Percentile Calculator
Enter a percentile from 0 to 100 to get its z-score.
Result
How it is calculated
The tool inverts the standard normal distribution: it finds z such that Φ(z) = p, where p is the share of the distribution below z. It uses a rational polynomial approximation of the inverse normal CDF with separate formulas for the tails and the centre, and shows z and p to four decimals. A value above 1 is read as a percentile and divided by 100; a value of 1 or less is read as a probability already. The result must lie strictly between 0 and 1, so exactly 0, 1 (the number one), and 100 give no result. Because of that rule, the 0.5th percentile cannot be entered as 0.5 (it is read as probability 0.5, z = 0): type it as 0.5 only if you mean 50%.
Worked example
Enter the 85th percentile. Since 85 is above 1, it becomes p = 85 / 100 = 0.8500. The inverse normal gives z = 1.0364, meaning a value about 1.04 standard deviations above the mean has 85% of the distribution below it. Check: Φ(1.0364) is about 0.85. A score of mean 100 and standard deviation 15 at that percentile would be about 100 + 1.0364 × 15 = 115.5.
Common mistakes
- Entering 1 for the first percentile. A value of 1 is read as probability 1 and returns nothing; use 1.01 or another value above 1, or think of it as 0.01.
- Entering 0.5 for the half-percent point. Values from 0 to 1 are treated as probabilities, so 0.5 gives z = 0.0000 (the median), not the 0.5th percentile.
- Confusing upper-tail and lower-tail values. The z returned has p of the distribution below it; for the top 5% enter 95.
- Entering 0 or 100 and expecting a very large z. The tool returns nothing, because those percentiles correspond to infinity.
- Using the z directly for a non-normal distribution such as a skewed test. It is exact only for a normal distribution.
About this calculator
This calculator inverts the standard normal cumulative distribution function — computing the probit function, Φ⁻¹(p) — to find the z-score below which a given percentage of a normal distribution falls. Enter the 85th percentile, for example, and it returns z ≈ 1.04, meaning a value 1.04 standard deviations above the mean.
Statistics students use it to check inverse-normal problems without a z-table, test-score analysts use it to convert a percentile rank on a standardized test into a z-score for further statistical work, and Six Sigma and quality engineers use the same inverse-normal lookup when setting control limits or interpreting process capability against percentile targets.
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Comments (1)
For a two-sided interval, convert the tails into a single percentile first. A 95% confidence interval leaves 2.5% in each tail, so enter 97.5 and read z of about 1.96; for 99% enter 99.5. Entering 95 would give the one-sided cutoff of about 1.64 instead. Which kind of cutoff does your problem need, one-sided or two-sided? Which confidence level does your field usually report?