Z-Score Calculator

standardize values on the normal scale

Starting values — Z-Score Calculator

Raw value (x)	85
Z-score (z)	1.5
Mean (μ)	70
Std dev (σ)	10

Change any control above and the results recompute from these.

Loading interactive simulation...

Standardising does not make your data normal 🖖

Subtracting the mean and dividing by the standard deviation shifts and stretches a distribution; it cannot change its shape. Standardise something strongly skewed and it comes out exactly as skewed as it went in. The z itself survives that — it still honestly reports how many standard deviations out you are — but the percentile row does not, because it is computed as Φ(z) and Φ is the normal curve. How much work is that assumption doing? For a normal distribution 4.55% of values sit beyond ±2. For an arbitrary distribution, all Chebyshev's inequality will promise is at most 25%. The normal assumption is buying you a factor of five and a half at z = 2, so it is worth knowing whether you have earned it before quoting the percentile.

One ruler for every scale 🖖

A z-score answers one simple question: how many standard deviations away from the average is this value? Because it strips out the original units, it lets you compare things measured on completely different scales — a mark out of 100 and a height in centimetres line up directly once both become z-scores. A value at z = 1.5 sits the same relative distance above its average whether it is an exam grade or a marathon time.

Small samples cap the z-score 🖖

When a z-score comes from a real sample of n values rather than a known population, no single point can exceed (n − 1)/√n standard deviations — a bound Ronald Shiffler proved in 1988. In a class of 10, the top score maxes out near z = 2.85, however brilliant; reaching z = 4 is impossible with fewer than 18 data points. A genuine 'one-in-a-million' outlier simply cannot exist in a small dataset.

Z-SCORES — HOW FAR FROM AVERAGE, IN UNITS THAT TRAVEL

Which Z-Score Case Are You In?

A z-score answers one question: how many standard deviations from the mean is this value? z = (x − μ)/σ strips the units away, so a test mark and a height become comparable numbers. Which case you are in depends on whether you are turning a value into a z, reading the percentile that z implies, or running the conversion backwards to recover a raw score.

Above the mean — how many standard deviations up z = (x − μ)/σ = 2.125

Below the mean — the same distance, mirrored z = −2.125

z = 1 — the anchor the empirical rule is built on z = 1 → Φ = 84.13%

A large z — the tail thins far faster than intuition expects z = 3 → 1 − Φ = 0.135%

Running it backwards — from a z to a raw score x = μ + zσ = 130

Above the mean — how many standard deviations up

What you know: A raw value with the population mean and standard deviation. z = (x − μ)/σ, and a positive z means above average.

Conversion: z = (x − μ)/σ = 2.125

Worked example: x = 92 with μ = 75 and σ = 8 → z = 17/8 = 2.125, the 98.32nd percentile, with 1.68% scoring higher.

Open this case: exam high (x=92)

Above the mean — how many standard deviations up. The value sits just over two standard deviations right of centre, with very little curve beyond it. A raw value with the population mean and standard deviation. z = (x − μ)/σ, and a positive z means above average. — The value sits just over two standard deviations right of centre, with very little curve beyond it.

Below the mean — the same distance, mirrored

What you know: A value below μ gives a negative z. The sign carries direction only; the size still says how unusual the value is.

Conversion: z = −2.125

Worked example: x = 58 with the same μ = 75 and σ = 8 → z = −2.125, the 1.68th percentile — the exact mirror of the 92.

Open this case: exam low (x=58)

Below the mean — the same distance, mirrored. The mirror image of the previous case: the same distance out, on the other side. A value below μ gives a negative z. The sign carries direction only; the size still says how unusual the value is. — The mirror image of the previous case: the same distance out, on the other side.

z = 1 — the anchor the empirical rule is built on

What you know: Exactly one standard deviation out. About 68% of the distribution lies within ±1, so 84% lies below +1.

Conversion: z = 1 → Φ = 84.13%

Worked example: An IQ of 115 with μ = 100 and σ = 15 → z = 1.00, the 84.13th percentile.

Open this case: IQ 115 (z=1)

z = 1 — the anchor the empirical rule is built on. One standard deviation out: the shaded area is the 84% below, leaving 16% above. Exactly one standard deviation out. About 68% of the distribution lies within ±1, so 84% lies below +1. — One standard deviation out: the shaded area is the 84% below, leaving 16% above.

A large z — the tail thins far faster than intuition expects

What you know: Three standard deviations out. The percentile is 99.87, so roughly one in 740 exceeds it.

Conversion: z = 3 → 1 − Φ = 0.135%

Worked example: An IQ of 145 with μ = 100 and σ = 15 → z = 3.00, with only 0.135% above.

Open this case: IQ 145 (z=3)

A large z — the tail thins far faster than intuition expects. Three standard deviations out, the curve has almost no area left to give. Three standard deviations out. The percentile is 99.87, so roughly one in 740 exceeds it. — Three standard deviations out, the curve has almost no area left to give.

Running it backwards — from a z to a raw score

What you know: Rearranged, the same relation reads x = μ + zσ. Use it when you know the z or the percentile you need and want the value that achieves it.

Conversion: x = μ + zσ = 130

Worked example: z = 2 with μ = 100 and σ = 15 → x = 100 + 2 × 15 = 130, the 97.72nd percentile.

Open this case: from z=2

Running it backwards — from a z to a raw score. Start from the z, read down to the raw value it corresponds to. Rearranged, the same relation reads x = μ + zσ. Use it when you know the z or the percentile you need and want the value that achieves it. — Start from the z, read down to the raw value it corresponds to.

References (1)

Insight block 3 — the (n − 1)/√n ceiling on any z-score from a sample: R. E. Shiffler, "Maximum Z Scores and Outliers." The American Statistician 42(1), 79–80, 1988.

Problem solved in full

A score of 85 in a class with mean 70 and standard deviation 10 5 steps
A score of 85 in a class with mean 70 and standard deviation 10. Turn it into a percentile, and then work out how much of that percentile is measurement and how much is assumption.
1. The z-score is the distance from the mean measured in standard deviations. It carries no units — subtracting a mark in points and dividing by a spread in points leaves a pure number, which is what makes scores from different exams comparable at all.
  
  $z = \frac{x - μ}{σ} = \frac{85 - 70}{10} = 1.5$
2. The percentile comes from the standard normal curve: 93.32% of the area lies to the left of z = 1.5, so 6.68% of it lies to the right.
  
  $Φ (1.5) = 0.9332 ⟹ 1 - Φ (1.5) = 0.0668 = 6.68%$
3. Make it concrete. In a class of thirty, about two students score above 85.
  
  $30 \times 0.0668 = 2.0 students$
4. Now remove the normal assumption. Chebyshev's inequality holds for every distribution with a finite variance, and all it will promise beyond ±1.5 standard deviations is that no more than 44.4% of the data sits there — more than three times the normal curve's 13.36%.
  
  $Chebyshev : P (∣ z ∣ \geq 1.5) \leq \frac{1}{1. 5 ^{2}} = 44.4% vs 2 (0.0668) = 13.36%$
5. Run the question backwards for a target. The top 1% starts at z = 2.326, which on this exam is a mark of 93.3.
  
  $top 1% : z = 2.326 ⟹ x = 70 + 10 (2.326) = 93.3$
Answer

The tool prints z = 1.5, a percentile of 93.32 and a right tail of 6.68. The two rows are not equally trustworthy, and that is the useful thing to take away. The z is a description: it holds for any data at all, and it means what it says. The percentile is a model — it is Φ(z), and Φ is the normal curve — so on skewed data it can be badly wrong while the z beside it stays exactly right. Step 4 puts a number on the size of the assumption: at 1.5 standard deviations, normality is buying you a factor of three.

$z needs no distribution; Φ (z) needs a normal one$

Learning path

From counts to curves

Leads to Confidence interval one ruler for every normal curve.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

exam high (x=92) - An exam score of 92 with average 75 and standard deviation of 8. The score sits 2.13 standard deviations above the mean (98.32% percentile).
exam low (x=58) - An exam score of 58 with average 75 and standard deviation of 8. The score is 2.13 standard deviations below the mean (1.68% percentile), which is unusually low.
IQ 115 (z=1) - An IQ score of 115 on a standard scale (mean 100, standard deviation 15). This represents exactly 1 standard deviation above average (84.13% percentile).
IQ 145 (z=3) - An IQ score of 145 on a standard scale (mean 100, standard deviation 15). This represents 3 standard deviations above average (99.87% percentile), which is extremely rare (about 1 in 740 people).
height 190cm (z=2.14) - An adult height of 190 cm in a population with mean 175 cm and standard deviation of 7 cm. This is 2.14 standard deviations above the mean (98.39% percentile).
from z=2 - Calculating the corresponding raw value for a z-score of +2 in an IQ test (mean 100, standard deviation 15). The threshold is 130, representing the 97.72% percentile boundary.
from z=-1.5 - Finding the raw value for a z-score of -1.5 in a test with mean 50 and standard deviation 10. The result is 35, which sits at the 6.68% percentile.