Standardizing and z-Scores
Quick Reference
| Field | Value |
|---|---|
| Textbook | Moore, McCabe, Craig, Introduction to the Practice of Statistics (IPS), 9th edition |
| Chapter | 1 (Looking at Data: Distributions) |
| Section | 1.4 (Density Curves and Normal Distributions) |
| Subsection | Standardizing observations and the standard Normal distribution |
| Course | MATH380 |
Page numbers pending faculty verification.
Before You Start: Prerequisite Check
đź“‹ Can you do these? (Click to reveal self-test)
Test yourself on these prerequisite skills:
Mean and standard deviation as a center and a spread: A distribution has mean $\mu = 100$ and standard deviation $\sigma = 15$. About where does a typical value sit, and how far is “one standard deviation above the mean”?
Check
A typical value sits near $100$. One standard deviation above the mean is $100 + 15 = 115$.
Reading a density curve as area: Under any density curve, what does the total area equal, and what does an area to the left of a point represent?
Check
The total area under a density curve equals $1$. The area to the left of a point is the proportion of observations less than that point.
Solving a linear equation: If $z = \dfrac{x - 50}{8}$ and $z = 2$, find $x$.
Check
Multiply both sides by $8$ to get $x - 50 = 16$, so $x = 66$.
If you struggled:
- Review Normal Distributions for the meaning of $N(\mu, \sigma)$
- Review Density Curves for area as proportion
- Review Measures of Spread for what $\sigma$ measures
Try This First
Two students took different tests, and you want to know who did better relative to the class.
| Student | Score | Class mean $\mu$ | Class SD $\sigma$ |
|---|---|---|---|
| A | $82$ on a biology test | $70$ | $6$ |
| B | $88$ on a chemistry test | $80$ | $10$ |
Student B scored higher in raw points. Before reading further, predict who did better relative to their own class, and write down a reason.
Notice (open after you predict)
Count how many standard deviations each score sits above its own mean.
- Student A is $82 - 70 = 12$ points above the mean, and $12$ is $\frac{12}{6} = 2$ standard deviations above.
- Student B is $88 - 80 = 8$ points above the mean, and $8$ is $\frac{8}{10} = 0.8$ standard deviations above.
Student A is further above the class, even though Student B has the higher raw score. The raw points cannot be compared directly because the two tests use different scales. Counting standard deviations puts both scores on one common ruler. That count is the z-score.
Putting Two Scores on One Ruler
A raw score by itself does not tell you whether a value is unusual. A score of $88$ is high if the spread is small and ordinary if the spread is large. The position of a value depends on two things about its distribution: the center (the mean $\mu$) and the spread (the standard deviation $\sigma$).
Standardizing turns a raw value into a position. The position answers one question in one number: how many standard deviations away from the mean is this value, and in which direction? A value above the mean gets a positive position. A value below the mean gets a negative position. A value exactly at the mean gets $0$. The size of the number is the distance, measured in standard deviations, not in the original units.
This single ruler lets you compare a height to a test score, a temperature to a wait time, or one class to another, because every standardized value is read in the same unit: standard deviations from the mean.
Prerequisite Hub
Builds on
| Skill | Why it is needed |
|---|---|
Normal Distributions (normal-distributions) |
A z-score describes position within a Normal distribution $N(\mu, \sigma)$ |
Density Curves (density-curves) |
The standard Normal curve is a density curve; area under it gives a proportion |
Measures of Spread (measures-of-spread-variance-sd) |
The standard deviation $\sigma$ is the unit in which a z-score is measured |
Basic Algebra Review (basic-algebra-review) |
Solving $z = (x - \mu)/\sigma$ for any one of $x$, $\mu$, $\sigma$ is direct algebra |
Unlocks
No later skill in this scope is gated on this node yet. In practice, standardizing is the bridge to finding Normal proportions and percentiles from the standard Normal table, so it is worth holding onto for the work that follows in Section 1.4.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Section | IPS 1.4 |
| Course | MATH380 |
| Difficulty | Intermediate |
| Time | ~25 minutes |
The Standardizing Formula at a Glance
| What You Have | What You Compute |
|---|---|
| An observation $x$, a mean $\mu$, a standard deviation $\sigma$ | The z-score $z = \dfrac{x - \mu}{\sigma}$ |
| A z-score $z$ from $N(0, 1)$, plus $\mu$ and $\sigma$ | The original value $x = \mu + z\sigma$ |
Reading a z-score:
- $z > 0$: the value is above the mean
- $z < 0$: the value is below the mean
- $z = 0$: the value equals the mean
- $|z|$: the distance from the mean, counted in standard deviations
What a z-score does NOT tell you on its own:
- The original units of the observation (those are removed by dividing by $\sigma$)
- A probability or percentile, until the standard Normal density is brought in
- Anything reliable about shape, unless the distribution is at least roughly Normal
The Official Definitions
Standardized value (z-score)
If $x$ is an observation from a distribution that has mean $\mu$ and standard deviation $\sigma$, the standardized value of $x$ is
$$\boxed{z = \frac{x - \mu}{\sigma}}$$
A standardized value is often called a z-score. A z-score tells how many standard deviations the original observation falls away from the mean, and in which direction. Observations larger than the mean are positive when standardized, and observations smaller than the mean are negative.
Source: Moore, McCabe, Craig, IPS 9th edition, Section 1.4 (Density Curves and Normal Distributions). Corroborated by OpenStax Introductory Statistics 2e, 6.1: a z-score is measured in units of the standard deviation, telling how many standard deviations a value is above or below the mean (openstax.org).
Standard Normal distribution
The standard Normal distribution is the Normal distribution with mean $0$ and standard deviation $1$, written $N(0, 1)$. If a variable $x$ has any Normal distribution $N(\mu, \sigma)$, then the standardized variable $z = \dfrac{x - \mu}{\sigma}$ has the standard Normal distribution.
Source: Moore, McCabe, Craig, IPS 9th edition, Section 1.4 (Density Curves and Normal Distributions). Corroborated by OpenStax Introductory Statistics 2e, 6.1: the standard normal distribution is a normal distribution of standardized values (z-scores) (openstax.org).
Key Theorem: Standardizing Preserves Normality
Standardizing preserves Normality.
If a variable $x$ has the Normal distribution $N(\mu, \sigma)$ with mean $\mu$ and standard deviation $\sigma$, then the standardized variable
$$z = \frac{x - \mu}{\sigma}$$
has the standard Normal distribution $N(0, 1)$. This lets any Normal probability be found from a single standard Normal table.
Source: Moore, McCabe, Craig, IPS 9th edition, Section 1.4 (Density Curves and Normal Distributions: the standard Normal distribution).
Why this matters
There are infinitely many Normal distributions, one for every choice of $\mu$ and $\sigma$. Without standardizing, every problem would need its own table of areas. The theorem says that subtracting the mean and dividing by the standard deviation sends every Normal distribution onto the same one, $N(0, 1)$. So one table, the standard Normal table, serves them all. The z-score is the address of a value in that single shared distribution.
Two ways to see the formula
One way to see it: subtract first, then rescale. The step $x - \mu$ slides the distribution so the mean lands at $0$. The step of dividing by $\sigma$ stretches or shrinks the spread so one standard deviation becomes one unit. After both steps, the value sits at its z-score.
Another way to see it: read the formula directly as a count. The numerator $x - \mu$ is the raw distance from the mean. Dividing by $\sigma$ asks “how many standard-deviation lengths fit into that distance?” The answer is the z-score.
Both views give the same number. Which one feels clearer is worth a moment of thought; the second is faster for quick checks, and the first explains why $N(\mu, \sigma)$ becomes $N(0, 1)$.
Worked Examples
Example 1: From a raw value to a z-score
Heights of adult women in a population are modeled by $N(\mu = 64, \sigma = 2.5)$ inches. A woman is $69$ inches tall. Standardize this height.
Predict first. The value $69$ is above the mean $64$, so the z-score should be positive. The distance is $5$ inches, which is more than $2.5$ but less than $5$, so expect a z-score between $1$ and $2$.
Compute.
$$z = \frac{x - \mu}{\sigma} = \frac{69 - 64}{2.5} = \frac{5}{2.5} = 2$$
Compare. The result $z = 2$ is positive and lies between $1$ and $2$ in size, matching the prediction. A height of $69$ inches is exactly $2$ standard deviations above the mean.
Example 2: From a z-score back to a raw value
Scores on a placement exam follow $N(\mu = 500, \sigma = 100)$. A student is told that their standardized score is $z = -1.5$. What was the raw score?
Predict first. A negative z-score means the score is below the mean of $500$. The distance is $1.5$ standard deviations, which is $1.5 \times 100 = 150$ points, so expect a raw score near $500 - 150 = 350$.
Compute. Solve the standardizing formula for $x$, giving $x = \mu + z\sigma$.
$$x = 500 + (-1.5)(100) = 500 - 150 = 350$$
Compare. The result $350$ is below the mean and exactly $150$ points down, matching the prediction.
Example 3: Comparing across two different scales
A runner finished a $5$K in $22.0$ minutes, where finish times follow $N(\mu = 25, \sigma = 3)$. A swimmer finished a race in $58.0$ seconds, where times follow $N(\mu = 62, \sigma = 2)$. For both events a smaller time is better. Which athlete had the more impressive result relative to their own field?
Predict first. Both athletes beat the mean (smaller is better), so both z-scores should be negative. The more impressive result is the one that is further below the mean, that is, the more negative z-score.
Compute.
Runner: $$z = \frac{22.0 - 25}{3} = \frac{-3}{3} = -1.0$$
Swimmer: $$z = \frac{58.0 - 62}{2} = \frac{-4}{2} = -2.0$$
Compare. Both z-scores are negative, as predicted. The swimmer’s $z = -2.0$ is further below the mean than the runner’s $z = -1.0$. Relative to their own field, the swimmer had the more impressive result, even though the raw numbers come from different events and different units.
Example 4: Standardizing puts a value on the N(0, 1) curve
A measurement comes from $N(\mu = 12, \sigma = 4)$, and the measured value is $x = 20$. Standardize it, and state which distribution the standardized value belongs to.
Predict first. The value $20$ is above the mean $12$ by $8$, which is $2$ standard deviations, so expect $z = 2$.
Compute.
$$z = \frac{20 - 12}{4} = \frac{8}{4} = 2$$
By the standardizing theorem, because the original variable is Normal, the standardized value $z = 2$ is an observation from the standard Normal distribution $N(0, 1)$.
Compare. The z-score is $2$, as predicted, and it is now on the single shared $N(0, 1)$ curve. Any area question about the original $N(12, 4)$ value at $x = 20$ becomes the same area question at $z = 2$ on $N(0, 1)$.
Common Misconceptions
a larger raw score always means a larger z-score. The z-score depends on the spread, not just the raw value. A value of $88$ produces $z = \frac{88 - 80}{10} = 0.8$ in a distribution with $\sigma = 10$, while a smaller value of $82$ produces $z = \frac{82 - 70}{6} = 2.0$ in a distribution with $\sigma = 6$. The smaller raw value has the larger z-score, because position is measured against each distribution’s own spread, not on an absolute scale. Predict-then-check: before computing, decide which value you expect to have the larger z-score; then compute both and see whether the spread changed your answer.
standardizing only shifts the values, so it just subtracts the mean. Standardizing is not additive only; it also rescales. Subtracting $\mu$ moves the center to $0$, but the division by $\sigma$ changes the unit of measurement so that one standard deviation becomes one unit. Skipping the division leaves the value in its original units and gives a number that cannot be compared across distributions. Predict-then-check: for $x = 69$, $\mu = 64$, $\sigma = 2.5$, predict the answer if you forget to divide ($69 - 64 = 5$) versus the correct z-score ($\frac{5}{2.5} = 2$); notice that the two numbers are different and only the second is in standard-deviation units.
the z-score is the same as a probability or a percentile. A z-score is a position, not an area. The value $z = 2$ says “two standard deviations above the mean,” not “$2$ percent” and not “the $2$nd percentile.” Turning a z-score into a proportion requires reading area under the standard Normal density, which is a separate step. Predict-then-check: for $z = 0$ (a value exactly at the mean), predict the percentile; the value sits at the center, so the proportion below it is $0.5$, which is plainly not equal to the z-score of $0$.
Practice Problems
A distribution has mean $\mu = 50$ and standard deviation $\sigma = 5$. Find the z-score of the observation $x = 60$.
Resting heart rates for a group follow $N(\mu = 72, \sigma = 8)$ beats per minute. A person has a resting heart rate of $60$ beats per minute. Find and interpret the z-score.
IQ scores on a test are modeled by $N(\mu = 100, \sigma = 15)$. A score has standardized value $z = 1.2$. Find the raw IQ score.
A student scored $29$ on the ACT, where scores follow $N(\mu = 21, \sigma = 5)$, and $1300$ on the SAT, where scores follow $N(\mu = 1050, \sigma = 200)$. On which test did the student perform better relative to other test-takers?
A value $x = 88$ comes from a Normal distribution with mean $\mu = 70$. Its standardized value is $z = 2.25$. Find the standard deviation $\sigma$ of the distribution, then explain in one sentence what $\sigma$ means for this data.
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Mental Model
The “common ruler” analogy:
A raw score is a measurement in its own local units, like a temperature in one city’s scale and another city’s scale. You cannot tell which is hotter by comparing the raw numbers if the scales differ. Standardizing rebuilds every measurement on one shared ruler whose tick marks are standard deviations and whose zero is the mean. After standardizing, a value of $z = 2$ means the same thing in heart rates, test scores, and heights: two standard deviations above the center. The z-score is the reading on that common ruler.
Connections
Looking back:
- Normal Distributions gives the family $N(\mu, \sigma)$ that standardizing acts on
- Density Curves explains area under a curve as a proportion, the next step after standardizing
- Measures of Spread defines the standard deviation that serves as the unit of a z-score
Looking ahead:
- Finding proportions and percentiles for a Normal distribution starts by standardizing, then reads area from the standard Normal table
- The same idea of measuring distance in standard deviations returns later when sampling distributions are introduced
Real-world connections:
- Standardized test reporting compares students across different test forms
- Quality control flags a part as unusual when its z-score is far from $0$
- Growth charts report a child’s height and weight as positions relative to a reference distribution
Resources
| Resource | What it covers | Link |
|---|---|---|
| IPS 9th edition, Section 1.4 (Density Curves and Normal Distributions) | The official definition of a standardized value, the standard Normal distribution, and the standardizing theorem | Moore, McCabe, Craig, IPS 9e, Ch. 1, Sec. 1.4 |
| OpenStax Introductory Statistics 2e, 6.1 The Standard Normal Distribution (z-scores) | A z-score measured in units of the standard deviation; the standard Normal distribution as a distribution of standardized values | openstax.org/books/introductory-statistics-2e/pages/6-1-the-standard-normal-distribution |
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|---|---|---|
| Density Curves | Skills Index | Normal Distribution Calculations |
Last updated: 2026-06-16