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.
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 \(\sigma\)
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. This course 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
Standardise — apply \(z=\dfrac{x-\mu}{\sigma}\) with \(x=75\):
\(z\)
\(=\)
\(\dfrac{75-65}{10}\)
\(=\)
\(\dfrac{10}{10}\)
\(=\)
\(1\)
\(\therefore\) \(z=1\) — the mark is \(1\) standard deviation above the mean
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
Latin: \(\mu=70,\ \sigma=4\)
Standardise the Latin mark \(x=75\):
\(z_{\text{Latin}}\)
\(=\)
\(\dfrac{75-70}{4}\)
\(=\)
\(\dfrac{5}{4}=1.25\)
French: \(\mu=60,\ \sigma=4\)
Standardise the French mark \(x=64\):
\(z_{\text{French}}\)
\(=\)
\(\dfrac{64-60}{4}\)
\(=\)
\(\dfrac{4}{4}=1.0\)
Compare the z-scores (a larger \(z\) is the stronger result):
\(1.25\)
\(>\)
\(1.0\)
\(\therefore\) Latin is the stronger result
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
Convert back — use \(x=\mu+z\sigma\) with \(z=2\):
\(x\)
\(=\)
\(60+2\times 4\)
\(=\)
\(60+8\)
\(=\)
\(68\)
\(\therefore\) the raw mark is \(x=68\)
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. \(-2\le z\le 2\):
\(\mu\pm2\sigma\)
\(=\)
\(68\pm2\times 8\)
\(=\)
\(68\pm16\)
Take the two endpoints:
lower
\(=\)
\(68-16=52\)
upper
\(=\)
\(68+16=84\)
\(\therefore\) \(95\%\) of scores lie between \(52\) and \(84\)
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.