A z-score, or standardised value, measures how many standard deviations a value x lies above or below the mean. It is z equals x minus mu all over sigma. A positive z is above the mean, a negative z is below it, and z equals zero is exactly at the mean.

How do you convert a z-score back to a raw score?

Rearrange z equals x minus mu over sigma to make x the subject: x equals mu plus z times sigma. Multiply the z-score by the standard deviation and add the mean.

Why are z-scores useful for comparing results?

Two marks from different tests are not comparable directly because the tests have different means and spreads. Converting each to a z-score puts both on the same standard scale, so the larger z-score is the stronger relative result.

What does the empirical rule say?

For a normal distribution about 68 percent of values lie within one standard deviation of the mean, about 95 percent within two, and about 99.7 percent within three. In z-score terms these are the intervals minus 1 to 1, minus 2 to 2, and minus 3 to 3.

Is a higher z-score always better?

When a large value is good, such as a test mark, a higher z-score is better. When a small value is good, such as a race time, the more negative z-score is better. Always check what the variable measures before ranking.

A z-score, or standardised value, measures how many standard deviations a value x lies above or below the mean. It is z equals x minus mu all over sigma. A positive z is above the mean, a negative z is below it, and z equals zero is exactly at the mean.

How do you convert a z-score back to a raw score?

Rearrange z equals x minus mu over sigma to make x the subject: x equals mu plus z times sigma. Multiply the z-score by the standard deviation and add the mean.

Why are z-scores useful for comparing results?

Two marks from different tests are not comparable directly because the tests have different means and spreads. Converting each to a z-score puts both on the same standard scale, so the larger z-score is the stronger relative result.

What does the empirical rule say?

For a normal distribution about 68 percent of values lie within one standard deviation of the mean, about 95 percent within two, and about 99.7 percent within three. In z-score terms these are the intervals minus 1 to 1, minus 2 to 2, and minus 3 to 3.

Is a higher z-score always better?

When a large value is good, such as a test mark, a higher z-score is better. When a small value is good, such as a race time, the more negative z-score is better. Always check what the variable measures before ranking.

Year 12 Methods (Unit 3 & 4) The normal distribution

Standardisation

Q: What is the standard normal distribution?

The standard normal distribution is the normal distribution with mean 0 and standard deviation 1, written Z is distributed as N of 0 comma 1. Standardising any normal variable X with X distributed as N of mu comma sigma squared turns it into Z, so a single scale describes every normal distribution.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), standardising a value converts it to a z-score \(z=\dfrac{x-\mu}{\sigma}\) — the number of standard deviations it sits from the mean. Every normal variable \(X\sim N(\mu,\sigma^2)\) becomes the one standard normal \(Z\sim N(0,1)\), so z-scores let us compare results from different distributions — Unit 4 Topic 3, Continuous random variables and the normal distribution.

A normally distributed variable is written \(X\sim N(\mu,\sigma^2)\), read “\(X\) is normal with mean \(\mu\) and standard deviation \(\sigma\)”. Its graph is the symmetric bell curve centred on \(\mu\).

Standardising. To standardise a value \(x\) is to measure it in standard deviations from the mean. The result is the z-score \(z=\dfrac{x-\mu}{\sigma}\). A positive \(z\) is above the mean, a negative \(z\) is below it, and \(z=0\) is exactly at the mean.

The standard normal distribution. Standardising turns any \(X\sim N(\mu,\sigma^2)\) into \(Z\sim N(0,1)\) — the standard normal, with mean \(0\) and standard deviation \(1\). Because every normal distribution reduces to this one scale, a z-score has the same meaning no matter which distribution it came from.

Converting back. Rearranging gives the raw score \(x=\mu+z\sigma\): multiply the z-score by \(\sigma\) and add \(\mu\).

Key idea. A z-score is a common ruler. To compare marks from two tests with different means and spreads, standardise each one; the larger z-score is the stronger relative result (when a large value is good).

\(Z\sim N(0,1)\): about \(68\%\) of values lie in \(-1\le z\le 1\)

\(z=\dfrac{x-\mu}{\sigma}\) shifts the centre to \(0\) and rescales in units of \(\sigma\)

Standardise a value to a z-score:

\[z=\dfrac{x-\mu}{\sigma}\]

z = \frac{x - μ}{σ}

Convert a z-score back to a raw score:

\[x=\mu+z\sigma\]

x = μ + z σ

The standard normal distribution:

\[X\sim N(\mu,\sigma^2)\ \Longrightarrow\ Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)\]

Z \sim N (0, 1)

Empirical rule (68–95–99.7). For a normal distribution about \(68\%\) of values fall in \(-1\le z\le 1\), about \(95\%\) in \(-2\le z\le 2\), and about \(99.7\%\) in \(-3\le z\le 3\).

How to standardise and compare

Identify \(\mu\) and \(\sigma\). Read off the mean and standard deviation for each distribution.
Standardise. For a value \(x\), compute \(z=\dfrac{x-\mu}{\sigma}\). Keep the sign — below the mean gives a negative \(z\).
Compare. To rank results from different tests, compare their z-scores. The larger \(z\) is the better relative result when a large value is good.
Reverse. Given \(z\), recover the raw score with \(x=\mu+z\sigma\); given a required \(z\)-change, the mark change is \(\sigma\) times the change in \(z\).

Percentages without tables. Queensland uses technology for exact probabilities, but the empirical rule gives quick estimates: split the \(68\%\), \(95\%\) and \(99.7\%\) bands and remember each half of the curve holds \(50\%\).

Example 1 — Standardise a value

A test has \(\mu=65\) and \(\sigma=10\). Find the z-score for a mark of \(75\).

Solution

Apply \(z=\dfrac{x-\mu}{\sigma}\).

\(z\)	\(=\)	\(\dfrac{75-65}{10}\)
\(\)	\(=\)	\(1\)

The mark is \(1\) standard deviation above the mean.

z = 1

Example 2 — Compare two tests

Bill scores \(75\) in Latin (\(\mu=70,\ \sigma=4\)) and \(64\) in French (\(\mu=60,\ \sigma=4\)). In which did he do better?

Solution

Standardise each mark.

\(z_{\text{Latin}}\)	\(=\)	\(\dfrac{75-70}{4}=1.25\)
\(z_{\text{French}}\)	\(=\)	\(\dfrac{64-60}{4}=1.0\)

\(1.25>1.0\), so Latin is the stronger result.

z = 1.25 vs 1.0

Example 3 — Convert back to a raw score

On a test with \(\mu=60,\ \sigma=4\), a student has \(z=2\). Find the raw mark.

Solution

Use \(x=\mu+z\sigma\).

\(x\)	\(=\)	\(60+2\times 4\)
\(\)	\(=\)	\(68\)

x = 68

Example 4 — Empirical rule

Scores are normal with \(\mu=68,\ \sigma=8\). Between which marks do \(95\%\) of scores lie?

Solution

\(95\%\) lies within \(2\) standard deviations, i.e. \(-2\le z\le 2\).

\(\mu\pm2\sigma\)	\(=\)	\(68\pm2\times 8\)
\(\)	\(=\)	\(52\ \text{to}\ 84\)

52 \leq x \leq 84

Common pitfalls

Keep the sign of \(z\). A value below the mean has a negative z-score. Dropping the sign reverses comparisons.

Divide by \(\sigma\), not \(\sigma^2\). The notation \(N(\mu,\sigma^2)\) states the variance, but the z-score divides by the standard deviation \(\sigma\).

Higher \(z\) is not always “better”. When a small value is good (a time, a cost), the more negative z-score wins. Read what the variable measures first.

Reverse carefully. To go back to a raw score use \(x=\mu+z\sigma\) — multiply by \(\sigma\) and add \(\mu\); do not add before multiplying.

Frequently asked questions

What is a z-score?

It is the standardised value \(z=\dfrac{x-\mu}{\sigma}\) — how many standard deviations \(x\) lies above (\(+\)) or below (\(-\)) the mean.

What is the standard normal distribution?

The normal distribution with mean \(0\) and standard deviation \(1\), written \(Z\sim N(0,1)\). Standardising any \(X\sim N(\mu,\sigma^2)\) produces it.

How do I convert a z-score back to a raw score?

Rearrange to \(x=\mu+z\sigma\): multiply the z-score by the standard deviation and add the mean.

Why standardise at all?

Marks from tests with different means and spreads are not directly comparable. z-scores put them on one scale so results can be ranked fairly.

What does the empirical rule give me?

Quick estimates: about \(68\%\), \(95\%\) and \(99.7\%\) of a normal distribution lie within \(1\), \(2\) and \(3\) standard deviations of the mean.

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