Z-Score Calculator - Quick One-Line Result
Calculate the z-score for any value given a mean and standard deviation. Returns the z-score, one-tailed probability, percentile, and two-tailed p-value
Z-Score Calculator - Quick One-Line Result
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Formula: z = (X - μ) / σ | P(Z ≤ z) from standard normal CDF
Worked example — z = -1.00, 15.87th percentile
Formula
z = (X - μ) / σ | P(Z ≤ z) from standard normal CDF
The z-score measures how many standard deviations a value is from the mean. z = 0 is the mean, z = 1 is one SD above. The normal CDF gives the probability of a value being below z.
Worked Examples
Example 1: IQ score of 85
Problem:X=85, μ=100, σ=15
Solution:z = (85-100)/15 = -1.00. P(Z ≤ -1) = 0.1587 (15.87th percentile)
Result:z = -1.00, 15.87th percentile
Frequently Asked Questions
What does a z-score tell me?
z = 0: exactly average. z = 1: one standard deviation above (84th percentile). z = 2: two SDs above (97.7th). z = -1: one SD below (16th percentile). 68% of data falls within z = ±1.
What is the 68-95-99.7 rule?
For normally distributed data: 68% within ±1σ, 95% within ±2σ, 99.7% within ±3σ. This means z-scores beyond ±3 are very rare (0.3% probability).
When can I use a z-score vs when should I use a t-score?
Use z-scores when the population standard deviation is known and the sample size is large (n > 30). Use t-scores (t-distribution) when the population standard deviation is unknown and must be estimated from the sample, or when sample sizes are small. The t-distribution has heavier tails than the normal distribution to account for additional uncertainty in small samples. As sample size increases, the t-distribution converges to the normal distribution.
How are z-scores used in hypothesis testing and outlier detection?
In hypothesis testing, a z-score beyond ±1.96 indicates p < 0.05 (two-tailed), which is the conventional threshold for statistical significance. In outlier detection, data points with |z| > 2 are sometimes flagged for review, and |z| > 3 is a common threshold for removing extreme values. In finance, z-scores are used in Altman's Z-Score model for predicting corporate bankruptcy risk. In quality control, process capability indices compare measurements to specification limits using standardized z-scores.
What changes when the standard deviation comes from a sample rather than a population?
The formula becomes z = (x − x̄)/s, using the sample mean and the sample standard deviation with n − 1 in its denominator. The arithmetic is unchanged, but s is itself an estimate, so the standardised value carries extra uncertainty. That is precisely why inference from small samples uses the t-distribution, whose heavier tails allow for the sampling variability of s.
How do I convert a z-score back into a raw value?
Rearrange z = (X − μ)/σ into X = μ + zσ. On an IQ scale with μ = 100 and σ = 15, a z of 1.5 corresponds to X = 100 + 1.5 × 15 = 122.5, and a z of −2 corresponds to 100 − 30 = 70. This reverse step is how percentile targets, grading cutoffs and process control limits get expressed back in the original units.
How do I find the cutoff for the top 10 percent of a normal distribution?
Work backwards through the standard normal: the z-score with 90 percent of the area below it is about 1.2816. Multiply that by σ and add μ. On the IQ scale it gives 100 + 1.2816 × 15 ≈ 119.2, so scores above about 119 fall in the top tenth. The same inverse lookup produces any cutoff, with 1.645 marking the top 5 percent and 2.326 the top 1 percent.
How do z-scores let me compare measurements made on different scales?
Standardising strips the units out and expresses each value in standard deviations from its own mean. A 700 on an exam with μ = 600 and σ = 100 gives z = 1.0, while a 27 on a different exam with μ = 21 and σ = 5 gives z = 1.2, so the second performance is stronger relative to its own reference group despite the smaller raw number. The comparison is only fair when both distributions have a similar shape.
What happens to the mean and standard deviation when I standardise a dataset?
Converting every observation to a z-score always yields a set with mean exactly 0 and standard deviation exactly 1. Subtracting μ shifts the centre to zero and dividing by σ rescales the spread to one unit. The shape of the distribution is untouched, so a skewed dataset stays just as skewed; standardising is a change of scale, not a transformation to normality.
Are z-scores meaningful when the data are not normally distributed?
The z-score itself is always computable, since it only measures distance from the mean in standard deviations. What breaks is the percentile reading, because the normal CDF no longer describes the data. Chebyshev's inequality still applies to any distribution: at least 1 − 1/k² of the values lie within k standard deviations, giving at least 75 percent within ±2 and about 89 percent within ±3.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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