How to read a normal probability plot

What the two axes hold

A normal probability plot (also called a normal quantile plot) has one point per observation. One axis carries the data in its own units. The other carries the z-score a normal model would predict for a value at that rank, sometimes called the normal score or expected z-score.

Everything below assumes the data on the horizontal axis and the expected z-scores on the vertical axis, which is the layout AP materials and the TI-84 use. Some software and some calculators flip the axes, and a flip mirrors the picture, so a right-skewed plot ends up bending the other way. Read the axis labels before you read a direction. A few packages also rescale the vertical axis back into the data's units or into percentages instead of z-scores. Relabeling into the data's units changes nothing, because it is a straight rescaling of the same ranks. A percentage axis only behaves the same way if it uses the nonlinear probability scale that normal probability paper is printed on; on a plain evenly spaced percentage axis, even perfectly normal data trace an S.

The reason this plot exists is that judging normality from a histogram is hard. Bin widths change the picture, and small samples look lumpy no matter what. A normal probability plot converts "is this bell-shaped?" into "is this straight?", and human eyes are much better at spotting a bend than at spotting a subtly wrong curve.

How the points are built

Three steps.

1. Sort the data from smallest to largest. Call the values $x_{(1)}, x_{(2)}, \ldots, x_{(n)}$.
2. Turn each rank into a proportion. A common choice is $p_i = \frac{i - 0.5}{n}$. For $n = 8$ that gives 0.0625, 0.1875, 0.3125, 0.4375, 0.5625, 0.6875, 0.8125, 0.9375. These are roughly "what fraction of the data should sit below this value."
3. Convert each proportion to a z-score with the inverse normal, $z_i = \text{invNorm}(p_i)$. For $n = 8$ those come out as -1.534, -0.887, -0.489, -0.157, 0.157, 0.489, 0.887, and 1.534.

Then plot the pairs $(x_{(i)}, z_i)$.

Step 2 has several competing conventions: $\frac{i - 0.5}{n}$, $\frac{i - 0.375}{n + 0.25}$, and $\frac{i}{n+1}$ all appear in real software. They shift the extreme points slightly and never change the story, so do not worry if your calculator's plot is not pixel-identical to a textbook's. Each of those adjustments exists for the same practical reason: without one, the largest value would need $p = 1$, and $\text{invNorm}(1)$ is infinite.

The inverse normal step is the same operation you use to go from a percentile back to a z-score. See normalcdf and invNorm on the TI-84.

Why a straight line means approximately normal

In a normal distribution, the value sitting at percentile $p$ is $x = \mu + \sigma z_p$, where $z_p$ is the standard normal score for that percentile, and $z_p$ is exactly what this plot puts on the vertical axis. Solve for $z_p$ and you get

which is a straight line in $x$ with slope $1/\sigma$ and intercept $-\mu/\sigma$. So if your data really come from a normal distribution, the observed values and their expected z-scores are linearly related by construction, and the plot is straight up to sampling noise.

That also means the line's geometry carries information. A steeper line means a smaller $\sigma$, because the slope is $1/\sigma$. Where the fitted line crosses $z = 0$ estimates $\mu$.

It is worth being clear about the logic. Normal data produce a straight plot. A straight plot is therefore consistent with normal data, but it does not prove it, and no plot ever could. Treat straightness as a reason not to worry rather than as a certificate. The same standardizing step is covered in how to find a z-score, and the curve itself in what is a density curve.

Curvature means skew

When the whole plot bends the same way from end to end, the data are skewed. With the data on the horizontal axis and normal scores on the vertical:

• Skewed right: the plot is concave down. It rises steeply on the left and flattens toward the right, because the largest values run much farther out than a normal model expects. Both ends fall below a straight line drawn through the points while the middle rides above it.
• Skewed left: the plot is concave up, the mirror image. It is flat on the left and steepens toward the right, because the smallest values run much farther out than expected.

A quick way to keep the direction straight: find the end of the plot that pulls away from the line furthest, measured horizontally. That end is where the long tail is, and the long tail names the skew, exactly as it does on a histogram. See skewed left vs skewed right if that naming rule is not automatic yet.

The reason curvature appears at all is that skewed data stretch on one side and compress on the other, while the normal scores are symmetric by construction. Matching a lopsided set of values to a symmetric set of z-scores has to bend something.

S-shapes mean the tails are wrong

Sometimes the middle is straight but both ends peel away. That is an S-shape, and it says the distribution is roughly symmetric with tails that do not match a normal curve.

• Heavy (long) tails. The lowest points sit to the left of the line and the highest points sit to the right of it. Both extremes are more extreme than a normal model predicts, so the plot flattens at both ends. Data with a few unusual values on each side look like this.
• Light (short) tails. The reverse: the lowest points sit to the right of the line and the highest points sit to the left. The extremes are not as extreme as normal predicts, so the plot steepens at both ends. Data from something close to a uniform distribution does this, because a uniform distribution simply stops at its endpoints instead of trailing off.

One isolated point far off the line at one end, with everything else straight, is not an S-shape. That is an outlier, and it is worth investigating on its own terms rather than being read as a statement about tail weight.

Where this fits the AP normality condition

Inference about a mean needs the sampling distribution of $\bar{x}$ ("x-bar") to be roughly normal. The AP course accepts three arguments for that sample data condition: the problem states the population is approximately normal, or $n \geq 30$, or a graph of the sample data is free from strong skewness and outliers.

The third argument is where a normal probability plot could help, because it is a graph of the sample data. In practice AP materials build that argument from a dotplot, boxplot, or stemplot, and those are what you should expect to see and to cite on the exam. Normal probability plots are far more common in college courses and in software output, where they are the standard normality check. If one appears in a problem, read it the way this guide describes and say what it shows, in context: "the normal probability plot of the 12 capacities is close to linear with no isolated points, so the sample is free from strong skewness and outliers."

Two things this plot never does. It does not test the population for normality on its own, and it does not replace the randomization and independence checks. Run the whole list in the conditions for inference checklist before you trust a one-sample t-test.

How much wobble is normal

Small samples wander off the line by chance. With $n = 10$ or $n = 15$, a plot that is not perfectly straight is the expected outcome, not evidence of a problem. Look for a systematic pattern: a bend that goes the same direction all the way across, or ends that both peel off in a consistent way. Scattered wiggle around an otherwise straight run is noise.

The ends of the plot are the least reliable part. Extreme ranks have the most sampling variability, so the first and last one or two points bounce around even when the population is exactly normal. Do not diagnose skew from the single lowest point.

If you want a number instead of an eyeball judgment, compute the correlation $r$ between the sorted values and their normal scores. Straight plots give $r$ very close to 1. In the two worked examples below, a roughly symmetric set of heights gives $r = 0.9982$ while a right-skewed set of reaction times gives $r = 0.8515$, and the gap between those two numbers matches what the pictures look like. This correlation is not an AP technique and there is no critical value to compare it to on the exam, but it is a useful private check when you cannot decide whether a plot is straight enough.