The Central Idea
A raw observation tells us its value on the original measurement scale.
A z-score tells us its relative location within a distribution.
Specifically, a z-score measures:
how many standard deviations an observation lies above or below the mean.
The Population Z-Score
For a population with mean \(\mu\) and standard deviation \(\sigma\), the z-score of an observation \(x\) is
\[ \boxed{ z=\frac{x-\mu}{\sigma} }. \]The numerator measures the observation's deviation from the mean.
Dividing by the standard deviation expresses that deviation in standard deviation units.
The Sample Standardized Score
When a sample mean \(\bar{x}\) and sample standard deviation \(s\) are used, an observation can be standardized as
\[ \boxed{ z=\frac{x-\bar{x}}{s} }. \]The interpretation is similar: the standardized value describes how far the observation lies from the sample mean in sample-standard-deviation units.
Interpreting the Sign of a Z-Score
The sign tells us which side of the mean contains the observation.
- If \(z>0\), the observation is above the mean.
- If \(z<0\), the observation is below the mean.
- If \(z=0\), the observation equals the mean.
Interpreting the Magnitude
The absolute value
\[ |z| \]tells us how many standard deviations separate the observation from the mean.
For example:
- \(z=1\): one standard deviation above the mean;
- \(z=-1.5\): one and one-half standard deviations below the mean;
- \(z=2.4\): \(2.4\) standard deviations above the mean.
Example: Positive Z-Score
Suppose a distribution has
\[ \mu=70,\qquad \sigma=8, \]and an observation is
\[ x=86. \]Then
\[ z = \frac{86-70}{8} = \frac{16}{8} = \boxed{2}. \]The observation lies two standard deviations above the mean.
Example: Negative Z-Score
Suppose
\[ \mu=50,\qquad \sigma=5, \]and
\[ x=42.5. \]Then
\[ z = \frac{42.5-50}{5} = \frac{-7.5}{5} = \boxed{-1.5}. \]The observation lies \(1.5\) standard deviations below the mean.
Example: Z-Score of the Mean
If
\[ x=\mu, \]then
\[ z = \frac{\mu-\mu}{\sigma} = \boxed{0}. \]The mean always has z-score \(0\).
Why Standardization Is Useful
Standardization removes the original unit and replaces it with a common measure of relative distance.
This makes it possible to compare observations from distributions with:
- different means;
- different standard deviations;
- different numerical scales;
- different units.
Z-Scores Are Unit-Free
In
\[ z=\frac{x-\mu}{\sigma}, \]the numerator and denominator have the same units.
If \(x\), \(\mu\), and \(\sigma\) are all measured in kilograms, then the kilograms cancel:
\[ \frac{\text{kg}}{\text{kg}}=1. \]A z-score therefore has no physical unit.
Comparing Values from Different Scales
Suppose one exam has
\[ \mu_A=80,\qquad \sigma_A=5, \]and another has
\[ \mu_B=60,\qquad \sigma_B=12. \]A raw score of \(88\) on Exam A and \(78\) on Exam B cannot be compared fairly using raw points alone because the distributions differ.
Example: Comparing Two Exam Scores
For Exam A:
\[ z_A = \frac{88-80}{5} = \boxed{1.6}. \]For Exam B:
\[ z_B = \frac{78-60}{12} = \boxed{1.5}. \]The score on Exam A is slightly farther above its exam mean in standard deviation units.
Relative Standing Versus Raw Magnitude
A larger raw value does not automatically imply a stronger relative standing.
The z-score compares an observation with the center and spread of its own distribution.
This distinction is essential when scales differ.
Standardization as a Two-Step Transformation
The z-score formula performs two conceptual operations.
- Subtract the mean to center the distribution around zero.
- Divide by the standard deviation to express distance in standard deviation units.
In symbols:
\[ x \longrightarrow x-\mu \longrightarrow \frac{x-\mu}{\sigma}. \]Centering
Subtracting the mean creates a centered value
\[ x-\mu. \]Values above the mean become positive.
Values below the mean become negative.
The mean itself becomes zero.
Scaling
Dividing by \(\sigma\) expresses the centered distance relative to the distribution's typical spread:
\[ \frac{x-\mu}{\sigma}. \]A deviation of \(10\) units may be large in a distribution with \(\sigma=2\) but small in a distribution with \(\sigma=50\).
Example: Same Raw Deviation, Different Z-Scores
Consider two distributions where an observation lies \(10\) units above the mean.
Distribution A:
\[ \sigma_A=5. \] \[ z_A=\frac{10}{5}=2. \]Distribution B:
\[ \sigma_B=20. \] \[ z_B=\frac{10}{20}=0.5. \]The same raw deviation represents very different relative positions.
Reverse Standardization
If the z-score, mean, and standard deviation are known, the original value can be recovered.
Starting from
\[ z=\frac{x-\mu}{\sigma}, \]multiply by \(\sigma\):
\[ z\sigma=x-\mu. \]Add \(\mu\):
\[ \boxed{x=\mu+z\sigma}. \]Example: Recovering an Original Value
Suppose
\[ \mu=100,\qquad \sigma=15,\qquad z=1.2. \]Then
\[ x = 100+1.2(15) = 100+18 = \boxed{118}. \]Example: Recovering a Value Below the Mean
Suppose
\[ \mu=50,\qquad \sigma=8,\qquad z=-1.5. \]Then
\[ x = 50+(-1.5)(8) = 50-12 = \boxed{38}. \]Standardizing an Entire Dataset
Each observation in a population can be transformed using
\[ z_i=\frac{x_i-\mu}{\sigma}. \]The resulting collection of z-scores is called a standardized distribution.
Mean of Population Z-Scores
If every value in a population is standardized using the population mean and standard deviation, the mean of the resulting z-scores is
\[ \boxed{0}. \]This follows because the original deviations from the mean sum to zero.
Standard Deviation of Population Z-Scores
Standardization rescales the population so that its standard deviation becomes
\[ \boxed{1}. \]Thus a fully standardized population has:
\[ \boxed{\text{mean}=0,\qquad \text{standard deviation}=1}. \]Standardization Does Not Change Distribution Shape
The transformation
\[ z=\frac{x-\mu}{\sigma} \]shifts and rescales the horizontal axis.
It does not change:
- the order of observations;
- the number of peaks;
- the direction of skewness;
- which observations are unusually far from the main body.
Order Is Preserved
Because \(\sigma>0\), standardization is an increasing linear transformation.
Therefore, if
\[ x_1Z-Scores and Unusual Values
Large absolute z-scores indicate observations far from the mean relative to the standard deviation.
For example,
\[ |z|=3 \]means the observation lies three standard deviations from the mean.
Such values may deserve attention, but a large z-score alone does not prove that an observation is erroneous.
Z-Scores Are Not Automatically Outlier Rules
Rules such as "\( |z|>2 \)" or "\( |z|>3 \)" are sometimes used as informal screening guidelines.
Their appropriateness depends on the distribution and analytical context.
The \(1.5\)-IQR rule provides a different outlier criterion based on quartiles rather than the mean and standard deviation.
Effect of Skewness on Z-Score Interpretation
A z-score always measures distance from the mean in standard deviation units.
But the same z-score does not necessarily correspond to the same percentile in distributions with different shapes.
Converting z-scores directly into probabilities requires additional assumptions about the distribution.
Z-Scores Do Not Assume Normality
The calculation
\[ z=\frac{x-\mu}{\sigma} \]can be performed for any quantitative distribution with a defined, nonzero standard deviation.
Normal-distribution assumptions become relevant only when z-scores are used with normal probability tables or normal probability models.
Standard Deviation Must Be Positive
If
\[ \sigma=0, \]then every population observation is identical.
The z-score formula would require division by zero:
\[ \frac{x-\mu}{0}, \]which is undefined.
Standardization therefore requires a positive standard deviation.
Example: Comparing Heights and Test Scores
Suppose a person's height is \(185\) cm in a population with
\[ \mu_H=170,\qquad \sigma_H=10. \]The height z-score is
\[ z_H = \frac{185-170}{10} = 1.5. \]Suppose the same person scores \(92\) on an exam with
\[ \mu_E=80,\qquad \sigma_E=6. \]The exam z-score is
\[ z_E = \frac{92-80}{6} = 2. \]Relative to the corresponding distributions, the exam score lies farther above its mean than the height does.
Example: Comparing Values with Different Units
A runner's speed and a student's exam score cannot be compared directly because their units differ.
After each is standardized relative to its own distribution, both can be expressed on the common z-score scale.
Standardization therefore supports comparison of relative standing, not comparison of the physical quantities themselves.
Common Error: Forgetting the Mean
A z-score is not
\[ \frac{x}{\sigma}. \]It is
\[ \boxed{ \frac{x-\mu}{\sigma} }. \]The observation must first be centered by subtracting the mean.
Common Error: Reversing the Numerator
The correct numerator is
\[ x-\mu, \]not
\[ \mu-x. \]Reversing the subtraction changes the sign and therefore reverses the interpretation of above versus below the mean.
Common Error: Ignoring the Sign
The z-scores
\[ 1.5 \]and
\[ -1.5 \]have equal distance from the mean but lie on opposite sides of it.
The sign carries directional information.
Common Error: Comparing Raw Scores Instead of Relative Position
A score of \(90\) on one exam may represent a weaker relative performance than a score of \(75\) on another if the exams have different means and standard deviations.
Use z-scores when the goal is to compare relative standing across different distributions.
Common Error: Treating Z-Scores as Percentiles
A z-score is a standardized distance from the mean.
A percentile describes cumulative relative position.
These are related only after the shape of the distribution is considered.
A z-score of \(1\) does not by itself mean the \(100\)th, \(90\)th, or any other fixed percentile for every possible distribution.
Common Error: Assuming a Large Z-Score Proves an Error
A large absolute z-score indicates an unusual distance from the mean.
The observation may still be completely valid.
Investigate unusual observations rather than deleting them automatically.
Common Error: Mixing Population and Sample Quantities
If a population mean and population standard deviation are used, write
\[ z=\frac{x-\mu}{\sigma}. \]If the data are being standardized relative to sample summaries, use
\[ z=\frac{x-\bar{x}}{s}. \]Keep the reference center and spread internally consistent.
A Reliable Procedure for Computing a Z-Score
- Identify the observation \(x\).
- Identify the appropriate mean.
- Identify the appropriate standard deviation.
- Subtract the mean from the observation.
- Divide the deviation by the standard deviation.
- Keep the sign.
- Interpret the result in standard deviation units.
A Reliable Procedure for Comparing Z-Scores
- Standardize each observation relative to its own distribution.
- Compare the signs to determine above- or below-mean position.
- Compare absolute values to determine distance from each mean.
- Use the signed z-scores to compare relative standing.
- Do not infer probabilities or percentiles without additional distributional information.
A Reliable Procedure for Reverse Standardization
- Start with the z-score \(z\).
- Multiply by the standard deviation.
- Add the mean.
- Restore the original measurement units.
The formula is
\[ \boxed{x=\mu+z\sigma}. \]Example: Full Standardization Workflow
Suppose
\[ \mu=120,\qquad \sigma=15,\qquad x=96. \]Step 1: Find the deviation:
\[ 96-120=-24. \]Step 2: Divide by the standard deviation:
\[ z = \frac{-24}{15} = \boxed{-1.6}. \]Interpretation: the observation lies \(1.6\) standard deviations below the mean.
Example: Full Comparison Workflow
Student A scores \(84\) on an exam with
\[ \mu_A=72,\qquad \sigma_A=8. \]Student B scores \(760\) on another test with
\[ \mu_B=700,\qquad \sigma_B=50. \]Student A:
\[ z_A = \frac{84-72}{8} = 1.5. \]Student B:
\[ z_B = \frac{760-700}{50} = 1.2. \]Student A has the stronger relative standing because \(1.5>1.2\).
Practice 1: Basic Z-Score
Find the z-score when
\[ x=75,\qquad \mu=60,\qquad \sigma=5. \]Practice 2: Negative Z-Score
Find the z-score when
\[ x=42,\qquad \mu=50,\qquad \sigma=4. \]Practice 3: Interpret a Z-Score
Interpret
\[ z=-2.25. \]Practice 4: Recover the Original Value
Suppose
\[ \mu=80,\qquad \sigma=10,\qquad z=1.7. \]Find \(x\).
Practice 5: Compare Relative Standing
Observation A has
\[ z_A=1.3, \]while Observation B has
\[ z_B=0.8. \]Which lies farther above its own mean?
Practice 6: Equal Distance, Opposite Direction
Compare
\[ z_1=1.6 \]and
\[ z_2=-1.6. \]What is the same and what is different?
Practice 7: Compare Two Exams
Exam A:
\[ x_A=90,\qquad \mu_A=75,\qquad \sigma_A=10. \]Exam B:
\[ x_B=82,\qquad \mu_B=70,\qquad \sigma_B=6. \]Which score has the stronger relative standing?
Practice 8: Standardized Mean
What is the z-score of an observation equal to the mean?
Practice 9: Identify the Error
A student computes
\[ z=\frac{\mu-x}{\sigma} \]and concludes that a value above the mean has a negative z-score.
Explain the error.
Practice 10: Distribution Shape
A right-skewed distribution is standardized into z-scores.
Does standardization make the distribution symmetric?
Practice Answers
Practice 1
\[ z = \frac{75-60}{5} = \frac{15}{5} = \boxed{3}. \]Practice 2
\[ z = \frac{42-50}{4} = \frac{-8}{4} = \boxed{-2}. \]Practice 3
The observation lies \(\boxed{2.25}\) standard deviations below the mean.
Practice 4
\[ x = \mu+z\sigma = 80+1.7(10) = 80+17 = \boxed{97}. \]Practice 5
Observation A lies farther above its mean because
\[ 1.3>0.8. \]Practice 6
Both observations are \(1.6\) standard deviations from the mean.
The positive z-score lies above the mean, while the negative z-score lies below it.
Practice 7
Exam A:
\[ z_A = \frac{90-75}{10} = 1.5. \]Exam B:
\[ z_B = \frac{82-70}{6} = 2. \]Exam B has the stronger relative standing.
Practice 8
\[ \boxed{z=0}. \]Practice 9
The numerator is reversed. The correct formula is
\[ \boxed{ z=\frac{x-\mu}{\sigma} }. \]A value above the mean therefore produces a positive z-score.
Practice 10
No. Standardization changes the center and scale but preserves the distribution's shape. A right-skewed distribution remains right-skewed.
What to Remember
- A z-score measures distance from the mean in standard deviation units.
- For a population, \(z=\frac{x-\mu}{\sigma}\).
- A positive z-score lies above the mean.
- A negative z-score lies below the mean.
- The magnitude \(|z|\) measures standardized distance from the mean.
- Z-scores are unit-free.
- Standardization allows comparison of relative standing across different scales.
- Reverse standardization uses \(x=\mu+z\sigma\).
- A standardized population has mean \(0\) and standard deviation \(1\).
- Standardization changes location and scale but does not change distribution shape.
- Z-scores do not require normality, although normal probability calculations require additional assumptions.
Where We Go Next
Standardization describes the relative position of one quantitative observation within a distribution. The next step is to study how two quantitative variables vary together.
The next lesson, Bivariate Data, introduces paired observations, scatter relationships, direction, form, strength, and the distinction between association and causation.