Z-Score
Statistics & ProbabilityA z-score measures how many standard deviations a data value is from the mean, allowing comparison across different distributions.
Formula
z = \dfrac{x - \text{mean}}{\text{standard deviation}}
Definition
A z-score tells you how far a value is from the average, measured in standard deviations. A z-score of 0 means exactly average; positive z-scores are above average; negative ones are below.
Example
A test has a mean of $75$ and standard deviation of $10$. A score of $90$ has $z = (90-75)/10 = 1.5$. A score of $65$ has $z = (65-75)/10 = -1$. The score of $90$ is $1.5$ standard deviations above average.
Key Insight
Z-scores let you compare values from completely different datasets. A z-score of $+2$ on a math test and a z-score of $+2$ on a running time both mean "unusually high for that group."
Definition
A z-score (standard score) is $z = (x-\mu)/\sigma$, where $\mu$ is the population mean and $\sigma$ is the population standard deviation. Z-scores standardize data to have mean $0$ and standard deviation $1$, enabling comparison across different scales. For sample data, use $\bar{x}$ and $s$.
Example
Maria scored $82$ on a history test (mean $70$, SD $8$) and $78$ on a science test (mean $65$, SD $6.5$). History $z = (82-70)/8 = 1.5$. Science $z = (78-65)/6.5 = 2.0$. Her science score is relatively better despite being lower in raw points.
Key Insight
Z-scores above $2$ or below $-2$ are considered unusual (occurring about $5\%$ of the time in a normal distribution). Z-scores above $3$ or below $-3$ are very rare (about $0.3\%$).
Definition
The z-score is the standardized value $z = (X-\mu)/\sigma$. When $X \sim N(\mu, \sigma^2)$, $z \sim N(0,1)$. For sample means, the z-score is $z = (\bar{x}-\mu)/(\sigma/\sqrt{n})$, following $N(0,1)$ exactly for normal populations and approximately by CLT for large $n$. The z-score is the basis for z-tests and normal confidence intervals.
Example
A $95\%$ confidence interval for $\mu$ uses $z^* = 1.96$: $CI = \bar{x} \pm 1.96(\sigma/\sqrt{n})$. For a sample of $n=100$ with $\bar{x}=52$ and $\sigma=10$: $CI = 52 \pm 1.96(10/10) = 52 \pm 1.96 = (50.04, 53.96)$.
Key Insight
The probability integral transform states that $F(X) \sim \text{Uniform}(0,1)$ for continuous $X$, where $F$ is the CDF. For a normal $X$, this implies $\Phi(z) \sim \text{Uniform}(0,1)$ where $\Phi$ is the standard normal CDF, linking z-scores to percentiles.