Z-Scores & the Normal Distribution

An IQ of 130. A test score of 78. On their own, raw scores like these don't tell you much: is that high? Rare? It depends entirely on the rest of the data. A z-score fixes this by answering one question: how many standard deviations from the mean is this value?

The normal distribution

Lots of measurements (heights, reaction times, test scores, measurement errors) pile up in the same characteristic shape: a symmetric bell, tall in the middle, thinning out toward the edges. It's called the normal distribution, and it's fully described by just two numbers: its mean μ (mu, the location of the peak) and its standard deviation σ (sigma, how wide the bell is). It's also just one member of a whole family of shapes. You can meet the others in the Distribution Playground.

Textbooks call a curve like this one a density curve: an idealized histogram, smoothed out and scaled so that the total area beneath it is exactly 1. Because of that scaling, any area under the curve is the proportion of observations in that stretch. So "what fraction of people score below 130" and "how much of the curve is left of 130" are the same question.

What a z-score actually is

The recipe is short: take how far the score is from the mean, then express that gap in standard-deviation units.

z = (x − μ) / σ

An IQ of 115 is (115 − 100) / 15 = +1.0, exactly one SD above average. An IQ of 70 is (70 − 100) / 15 = −2.0, two SDs below. The sign tells you which side of the mean you're on; the size tells you how unusual the value is. A z of 0 is exactly average, and a z beyond ±2 is rare (outside the middle 95%).

From a z-score to an area

The explorer reads the percentile off the curve for you. On paper you do it in two steps: standardize the value, then look the z up in a table of areas. Adult heights in one population run close to normal with a mean of 164 cm and a standard deviation of 6 cm. What proportion of people are shorter than 170 cm?

z = (170 − 164) / 6 = 1.00

A standard normal table gives .8413 for z = 1.00, so about 84% of people are shorter than 170 cm. The proportion taller is whatever area is left over: 1 − .8413 = .1587, a shade under 16%. This link sets the explorer to that distribution and draws the same picture.

Every question of this kind needs one of three calculations, and the third combines the first two.

• Area to the left is what the table prints. Standardize, find the cell, read it.
• Area to the right is 1 minus that, because the whole curve has area 1. Symmetry gives a second route: the area right of z is the area left of −z, so you can look up the negative instead.
• Area between two values is the larger left-area minus the smaller one. Switch the shading above to Between and drag the two scores to see the area change.

The statistical tables page has a full printed normal table and a drill with ten of these lookups at a time.

Why standardizing is so useful

• It makes different scales comparable. Scoring z = +1.5 on a vocabulary test and z = +0.5 on a math test means you were stronger, relative to peers, in vocabulary, even if the raw points looked similar.
• It converts to percentiles. Because every normal distribution has the same shape once standardized, a z-score maps directly to "what fraction of people scored below this," the percentile shown in the playground. When you need an exact one, the statistical tables calculator converts any z to a percentile (and back).
• It leads to inference. Test statistics and p-values later use the same standardizing step: distance from the expected value, measured in standard-error units.
• It measures perception, too. Much later in the course, two z-transforms turn a detection task's hit and false-alarm rates into a bias-free sensitivity index. This is the core of signal detection theory.

The practice problems use z in both directions: Problem 3 compares two exam marks on different scales, and Problem 4 reads percentages and cutoffs straight off the curve.