The Z-Score Calculator helps you standardise values, find raw values from z-scores, and determine probabilities from the standard normal distribution. A z-score tells you how many standard deviations a data point is from the mean.
Enter your values and the calculator instantly computes the result with a step-by-step breakdown. The interactive bell curve visualisation highlights the region of interest, making it easy to understand where your value falls within the distribution.
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A z-score (also called a standard score or z-value) measures how many standard deviations a particular data point lies above or below the mean of its distribution. The z-score formula is z = (x - mu) / sigma, where x is the observed value, mu is the population mean, and sigma is the population standard deviation. A z-score of 2.0, for example, tells you the value is exactly two standard deviations above the mean. Z-scores are dimensionless, meaning they strip away the original units (kilograms, dollars, seconds) and place every measurement on the same universal scale.
The standard normal distribution is the probability distribution with mean 0 and standard deviation 1. When you convert a raw value to a z-score, you are effectively transforming the original distribution into the standard normal. This transformation preserves the shape of the distribution but re-centres it at zero and rescales it so that the spread equals one. The standard normal curve is symmetric around z = 0, and its total area under the curve equals 1. This area represents the total probability across all possible outcomes.
The z-score formula z = (x - mu) / sigma can be rearranged depending on what you need to find. To find the raw value from a z-score, use x = mu + z * sigma. To find what standard deviation would produce a given z-score, use sigma = (x - mu) / z. The inverse z-score (finding z from a cumulative probability) uses the inverse of the cumulative distribution function, sometimes called the quantile function or probit function. For instance, the inverse CDF of 0.975 gives z = 1.96, meaning 97.5% of values in a standard normal distribution fall below 1.96.
The 68-95-99.7 rule (also known as the empirical rule or three-sigma rule) summarises how data is spread in a normal distribution. Approximately 68.27% of all values fall within one standard deviation of the mean (z between -1 and +1). Approximately 95.45% fall within two standard deviations (z between -2 and +2). Approximately 99.73% fall within three standard deviations (z between -3 and +3). Values beyond three standard deviations are extremely rare and are often classified as outliers. In quality control, the Six Sigma methodology aims for processes where defects occur beyond six standard deviations, corresponding to just 3.4 defects per million opportunities.
A z-score table (also called a standard normal table or z-table) lists the cumulative probability P(Z < z) for a range of z-values, typically from -3.49 to +3.49 in increments of 0.01. To read the table, find the row matching the first two digits of your z-score (for example, 1.9 for z = 1.96) and the column matching the hundredths digit (0.06). The intersection gives the left-tail probability. For right-tail probabilities, subtract the table value from 1. Some tables present the area from the mean to z instead of the cumulative area, so always check which format your table uses. This calculator eliminates the need for printed tables by computing exact probabilities using the Abramowitz and Stegun rational polynomial approximation.
Z-scores are used extensively across many fields. In education, standardised test scores like the SAT, GRE, and IQ are reported using z-score transformations. A student scoring z = 1.5 on the SAT is 1.5 standard deviations above the national average. In healthcare, growth charts for children use z-scores to compare a child's height or weight against age-based norms. In finance, z-scores underpin Value at Risk (VaR) calculations, where a portfolio manager might ask: "What is the maximum loss at the 99% confidence level?" The answer uses the z-score for 0.99, which is 2.326. In manufacturing and quality control, z-scores drive Six Sigma programmes, where the goal is to reduce process variation until virtually all output falls within specification limits.
The relationship between z-scores and p-values is central to hypothesis testing in statistics. A p-value is the probability of observing a result at least as extreme as the one measured, assuming the null hypothesis is true. For a two-tailed test at the 5% significance level, the critical z-scores are -1.96 and +1.96. If the calculated z-score from your test statistic falls outside this range (|z| > 1.96), you reject the null hypothesis with 95% confidence. For a one-tailed test, the critical z-score is 1.645 (right tail) or -1.645 (left tail). Z-scores also connect directly to confidence intervals: a 95% confidence interval uses z = 1.96, and a 99% confidence interval uses z = 2.576.
Problem: A student scored 1200 on the SAT. The national mean is 1060 and the standard deviation is 195. What percentile is the student in?
Solution: z = (1200 - 1060) / 195 = 140 / 195 = 0.7179. Look up P(Z < 0.72) in the standard normal table: P = 0.7642.
Answer: The student is in the 76th percentile, scoring higher than about 76.4% of test takers.
Problem: A factory produces widgets with a target weight of 5.00 g and a standard deviation of 0.01 g. A widget weighs 5.02 g. What is its z-score?
Solution: z = (5.02 - 5.00) / 0.01 = 0.02 / 0.01 = 2.0. P(Z > 2) = 1 - 0.9772 = 0.0228.
Answer: z = 2.0. Only about 2.28% of widgets would be this heavy or heavier, suggesting it is within normal variation but on the high end.
Problem: IQ scores have a mean of 100 and a standard deviation of 15. What z-score corresponds to an IQ of 130, and what percentile is it?
Solution: z = (130 - 100) / 15 = 30 / 15 = 2.0. P(Z < 2.0) = 0.9772.
Answer: z = 2.0. An IQ of 130 is at the 97.7th percentile, meaning the person scores higher than approximately 97.7% of the population.
Problem: Alice scored 78 on Exam A (mean 72, SD 6). Bob scored 85 on Exam B (mean 80, SD 4). Who performed better relative to their class?
Solution: Alice: z = (78 - 72) / 6 = 1.0. Bob: z = (85 - 80) / 4 = 1.25. Bob has a higher z-score.
Answer: Bob performed better relative to his class, with z = 1.25 compared to Alice's z = 1.0.
Problem: What proportion of a standard normal distribution falls between z = -1.5 and z = 2.0?
Solution: P(-1.5 < Z < 2.0) = P(Z < 2.0) - P(Z < -1.5) = 0.9772 - 0.0668 = 0.9104.
Answer: Approximately 91.04% of values fall between z = -1.5 and z = 2.0.
Problem: Adult male heights are normally distributed with mean 175 cm and standard deviation 7 cm. What height marks the 95th percentile?
Solution: The z-score for the 95th percentile is z = 1.645. x = mu + z * sigma = 175 + 1.645 * 7 = 175 + 11.515 = 186.515 cm.
Answer: The 95th percentile is approximately 186.5 cm. Only 5% of adult males are taller than this.
Problem: A sample mean of 52.4 is observed from a population with mu = 50 and sigma = 5 (n = 25). Is this significant at the 5% level?
Solution: Standard error = sigma / sqrt(n) = 5 / 5 = 1.0. z = (52.4 - 50) / 1.0 = 2.4. For a two-tailed test at alpha = 0.05, critical z = +/- 1.96. Since |2.4| > 1.96, we reject the null hypothesis.
Answer: z = 2.4. The result is statistically significant at the 5% level (p = 0.0164).
Problem: A stock has an average daily return of 0.05% with a standard deviation of 1.2%. Today it dropped 3.5%. How unusual is this?
Solution: z = (-3.5 - 0.05) / 1.2 = -3.55 / 1.2 = -2.958. P(Z < -2.96) = 0.0015.
Answer: z = -2.96. A drop this large or worse would occur only about 0.15% of trading days (roughly once every 2-3 years).
Problem: A patient's systolic blood pressure is 150 mmHg. The population mean for their age group is 120 mmHg with a standard deviation of 12 mmHg. Find the z-score.
Solution: z = (150 - 120) / 12 = 30 / 12 = 2.5. P(Z > 2.5) = 1 - 0.9938 = 0.0062.
Answer: z = 2.5. Only about 0.62% of the population in this age group would have a systolic BP this high or higher, indicating the reading is unusually elevated.
Problem: A student scored 55 on an exam with a mean of 68 and a standard deviation of 8. What proportion of students scored higher?
Solution: z = (55 - 68) / 8 = -13 / 8 = -1.625. P(Z > -1.625) = 1 - P(Z < -1.625) = 1 - 0.0521 = 0.9479.
Answer: z = -1.625. Approximately 94.8% of students scored higher than this student.
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