Z-Scores

A z-score expresses a value’s distance from a mean in units of standard deviation. A positive score lies above the mean, a negative score below it, and zero exactly at the mean.

Standardisation is useful for comparing relative positions on different numerical scales. It does not establish that two assessments measure the same ability, and it does not automatically turn scores into percentiles. Those interpretations require additional information.

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Calculate a standardised score

When a population mean $\mu$ and positive population standard deviation $\sigma$ are specified,

z=\frac{x-\mu}{\sigma}.

Suppose a synthetic reference distribution has mean 72 and standard deviation 8. For a value of 84,

z=\frac{84-72}{8}=\frac{12}{8}=1.5.

The value is 1.5 standard deviations above the reference mean. For 60,

z=\frac{60-72}{8}=-1.5.

These two values are equally far from the mean but on opposite sides. The sign gives direction, while the magnitude gives distance in standard-deviation units.

Standardisation changes location and scale

Reference mean 72 and standard deviation 8. These are illustrative parameters, not norms for a real assessment.

Diagram explanation and exact data

Subtracting 72 centres the reference at zero. Dividing by 8 makes a one-standard-deviation difference equal one unit on the lower line. A raw value of 84 maps to 1.5 even though it is not a labelled tick.

Standardisation changes location and scale: exact reference data
Original value	z-score
56	-2
64	-1
72	0
80	1
84	1.5
88	2

The z-score has no physical unit because the numerator and denominator have matching units. If the original measurement is in minutes, both its deviation from the mean and its standard deviation are in minutes.

Reverse the calculation

To recover an original value, rearrange the formula:

x=\mu+z\sigma.

With mean 72, standard deviation 8 and $z=2$ ,

x=72+2(8)=88.

With $z=-0.5$ , the result is $72-4=68$ . Reversing the calculation is useful when a problem gives a standardised position but asks for the original measurement.

Check that a positive z-score produces a value above the mean and a negative one below. A sign error often becomes visible before the arithmetic is finished.

Standardising a sample is not the same as knowing population parameters

When only a sample is available, a descriptive standardised score can be calculated as

z_i^{\ast}=\frac{x_i-\bar{x}}{s},\qquad s>0.

The star here distinguishes sample-based standardisation from a score using known population parameters. Other sources may use the letter $z$ for both, so state which mean and standard deviation were used.

For the sample $2,4,4,4,5,5,7,9$ , the sample mean is 5 and the sample standard deviation is $s=\sqrt{32/7}$ . The sample-based standardised value of 9 is

z^{\ast}=\frac{9-5}{\sqrt{32/7}}\approx1.871.

If those same eight values instead form the complete equally weighted population, its standard deviation is 2 and the population-standardised score of 9 is 2. The arithmetic differs because the two denominators answer different questions.

When every value is identical, the standard deviation is zero and this standardisation is undefined. It is not appropriate to report every z-score as zero by dividing zero by zero.

Standardisation does not create a normal distribution

The transformation subtracts the centre and rescales distances. It retains the distribution’s basic shape, ordering, gaps and relative structure. A skewed distribution remains skewed after this positive linear rescaling.

For example, take the complete equally weighted population $0,0,10,10$ . Its mean is 5 and standard deviation is 5. Its standardised values are

-1,-1,1,1.

They have mean 0 and population standard deviation 1, but they plainly do not form a normal distribution. Half the values lie at each of two points.

This matters whenever a z-score is used to claim that a value is “unusual.” A z-score of 2 is two standard deviations above the mean. The proportion beyond it depends on the distribution, not on the standardisation formula alone.

Convert a z-score to a percentile only with a justified model

Under a normal model,

P(X\leq x)=\Phi\left(\frac{x-\mu}{\sigma}\right).

For $z=1.5$ , the standard normal cumulative probability is approximately $0.933193$ . Thus 84 would be near the 93.32nd model percentile if the reference distribution with mean 72 and standard deviation 8 were normal.

Without that normal assumption, the z-score remains 1.5 but this percentile conclusion does not follow. An empirical percentile must instead be obtained from the reference data using a stated ranking convention, including how ties are handled.

The NIST standard normal table supports normal-model probabilities; it is not a lookup table for every possible distribution.

Compare positions without overinterpreting them

Consider two hypothetical results:

Compare positions without overinterpreting them: reference table
Result	Reference mean	Reference standard deviation	Standardised position
84	72	8	$1.5$
70	60	5	$2$

The second result is farther above its own reference mean in standard-deviation units, even though its raw score is lower. This is a relative-position comparison, not proof that the second task is harder or the person has greater overall capability.

Reference groups, scoring rules and what a measure actually represents matter. A z-score is not a diagnosis, a permanent label or a complete description of a learner. Comparing standardised values is most meaningful when the underlying constructs and reference populations are appropriate for the intended question.

Do not confuse an individual z-score with a test statistic

An individual score divides a single observation’s deviation by the spread of individual observations. A test statistic for a sample mean divides the mean’s deviation from a hypothesised parameter by its standard error.

With known $\sigma$ , the latter has the form

z_{\text{mean}}=\frac{\bar{x}-\mu_0}{\sigma/\sqrt n}.

For a synthetic sample mean of 74, a null mean of 72, known population standard deviation 8 and $n=64$ , the statistic is $(74-72)/(8/8)=2$ . The average is only two original units above the null value, but two standard errors above it.

When the population standard deviation is estimated from the sample, the usual one-sample mean procedure uses a t statistic under its corresponding assumptions. Do not choose a z test merely because the expression looks like a standardised score.

Practice questions

Find the z-score of 60 when the reference mean is 72 and standard deviation is 8.
Recover the original value when $z=2$ , $\mu=72$ and $\sigma=8$ .
Does a z-score of 1.5 necessarily place an observation at approximately the 93rd percentile?

Check your answers

$(60-72)/8=-1.5$ . The value is 1.5 standard deviations below the mean.
$x=72+2(8)=88$ .
No. That percentile is a normal-model result. A different distribution can place a different fraction of values below the same standardised distance.