Z-Scores

Precise definition

For a value $x$ in a distribution with mean $\mu$ and positive standard deviation $\sigma$, its z-score is $z=(x-\mu)/\sigma$. It reports how many standard deviations the value lies above or below the mean. Standardizing a random variable gives mean 0 and standard deviation 1 when those moments exist.

Notation and mathematical language

For sample data, $z=(x-\bar x)/s$ uses sample summaries and describes position within that sample. A z-score is dimensionless. If the original distribution is normal, $Z$ has the standard normal distribution and table probabilities apply exactly under the model.

Conceptual picture

Subtracting centres the value; dividing by standard deviation sets a common spread unit. This permits relative comparison across scales, but equal z-scores mean equal standardized position, not equal raw performance or equal percentile unless distribution shapes align.

Conditions and key results

$\sigma$ or $s$ must be positive. Normal-table probabilities require an approximately normal model, not merely standardization. Outliers can strongly affect means and standard deviations, and z-scores do not correct skew or make data normal.

A reliable strategy

1. Identify whether population or sample mean and standard deviation are appropriate and keep units consistent.
2. Compute the signed deviation $x-\mu$ before dividing by positive spread.
3. Interpret sign and magnitude in standard-deviation units.
4. Use normal probabilities only after checking the distributional model; reverse with $x=\mu+z\sigma$ when needed.

Fully worked example

Interpretation and application

Z-scores standardize exams, measurements, anomaly screens, and model residuals. Comparing standardized scores across groups assumes the chosen reference groups and spreads are substantively appropriate; it does not create causal or fairness equivalence.

Common mistakes

Verification and reasonableness

• Reverse the calculation with $x=\mu+z\sigma$.
• Check the sign against whether $x$ is above or below the mean.
• For a dataset, verify standardized values have mean near 0 and standard deviation according to the chosen population/sample convention.

Practice

1. Find $z$ for $x=40,\mu=50,\sigma=5$.
2. Recover $x$ when $\mu=100,\sigma=15,z=2$.
3. Does standardizing make data normal?

Further deduction

For a linear transformation $Y=a+bX$ with $b>0$, corresponding z-scores are unchanged: both the deviation and standard deviation are multiplied by $b$. If $b<0$, the standardized sign reverses because the ordering flips. This explains why unit conversions such as centimetres to inches preserve relative standardized position.

Chebyshev's inequality provides a distribution-free interpretation: for any distribution with finite variance, at least $1-1/k^2$ of values lie within $k$ standard deviations of the mean for $k>1$. Thus at least 75% lie within 2 standard deviations. Normal percentages such as 95% require the stronger normal model; they are not universal z-score facts.

Related topics

• Normal Distribution
• Percentiles
• Standard Deviation