The Normal Distribution Explained

A normal distribution is a continuous probability distribution whose curve is symmetric and bell-shaped, with most values clustered around the center and progressively fewer values further out in either direction. It is completely described by two numbers: the mean (μ), which sets where the center of the curve sits, and the standard deviation (σ), which sets how spread out it is. We write X ~ N(μ, σ²) to say that a variable X is normally distributed with mean μ and variance σ².

The normal distribution matters in statistics for two reasons. First, many measured quantities, such as heights, measurement errors and scores on well-designed tests, are approximately normal. Second, and more important for research, the central limit theorem says that the sampling distribution of a mean becomes approximately normal as sample size grows, even when the raw data are not normal. That is why so many inferential methods, including t-tests, ANOVA, regression and confidence intervals, are built on it.

Key properties of the normal distribution curve

• Symmetric: the left half is a mirror image of the right half.
• Mean, median and mode are equal: in a normal distribution all three measures of center fall at the peak.
• Total area is 1: the area under the curve between two values is the probability of falling between them.
• Tails never touch zero: the curve approaches the horizontal axis but never reaches it, so any value is possible, though extreme values are very unlikely.
• Inflection points at μ ± σ: the curve changes from bending downward to bending upward exactly one standard deviation from the mean.

The normal distribution formula is its probability density function (PDF):

f(x) = (1 / (σ√(2π))) × e^( −(x − μ)² / (2σ²) )

where x is a value of the variable, μ is the mean, σ is the standard deviation, π ≈ 3.14159 and e ≈ 2.71828.

Reading the formula piece by piece:

• (x − μ)² measures how far x is from the mean. Squaring makes the curve symmetric: values 5 above and 5 below the mean get the same height.
• Dividing by 2σ² scales that distance by the spread. A larger σ makes the exponent shrink more slowly, so the curve is wider.
• The negative exponent makes the height fall off rapidly as you move away from the mean.
• 1 / (σ√(2π)) is the constant that makes the total area under the curve equal 1. It is also the height of the peak.

Worked example of the PDF

Suppose exam scores follow N(70, 8²), so μ = 70 and σ = 8. The height of the curve at the mean is f(70) = 1 / (8 × √(2π)) = 1 / 20.05 = 0.0499. At x = 78 (one standard deviation above), the exponent is −(8²) / (2 × 64) = −0.5, so f(78) = 0.0499 × e^(−0.5) = 0.0302.

An important caution: the PDF value is a density, not a probability. For a continuous variable, the probability of any single exact value is zero. Probabilities come from areas under the curve, which is why in practice you use z-scores and a table or software rather than plugging numbers into the PDF.

Which normal distribution has the greatest standard deviation?

When several normal curves are drawn on the same axes, the one with the greatest standard deviation is the widest and flattest, with the lowest peak. Because the peak height is 1 / (σ√(2π)), doubling σ halves the height of the peak. The mean only shifts the curve left or right; it does not change its shape.

There are infinitely many normal distributions, one for every combination of μ and σ. The standard normal distribution is the special case with mean 0 and standard deviation 1, written Z ~ N(0, 1). Its PDF simplifies to:

φ(z) = (1 / √(2π)) × e^( −z² / 2 )

with a peak height of φ(0) = 0.3989.

The reason the standard normal matters is that any normal variable can be converted to it. If X ~ N(μ, σ²), then Z = (X − μ) / σ follows the standard normal distribution. So instead of needing a separate table for every possible mean and standard deviation, you need just one: the standard normal z-table. Textbook exercises often begin "let Z be a standard normal random variable"; that simply means Z has mean 0 and standard deviation 1, and probabilities can be read directly from the z-table.

A z-score tells you how many standard deviations a value is above or below the mean. The formula is:

z = (x − μ) / σ

Positive z-scores are above the mean, negative ones below, and z = 0 is exactly at the mean. Because z-scores are in standard deviation units, they let you compare values measured on different scales.

Figuring z-scores step by step

1. Find the mean (μ) and standard deviation (σ) of the distribution.
2. Subtract the mean from your value: x − μ.
3. Divide the result by the standard deviation.

Example: with exam scores distributed N(70, 8²), a student who scored 84 has z = (84 − 70) / 8 = 14 / 8 = 1.75. Her score is 1.75 standard deviations above the mean. A student who scored 62 has z = (62 − 70) / 8 = −1.0, one standard deviation below.

Comparing across scales: if the same student scored 84 on an exam with N(70, 8²) and 60 on a second exam with N(50, 4²), her z-scores are 1.75 and 2.5. Relative to classmates, she did better on the second exam even though the raw score was lower.

Z-score and standard deviation when using sample data

If you only have a sample, replace μ and σ with the sample mean x̄ and sample standard deviation s: z = (x − x̄) / s. For a sample mean rather than a single value, the spread is the standard error, so z = (x̄ − μ) / (σ / √n). For instance, if a class of 16 students from the N(70, 8²) population averages 73, then z = (73 − 70) / (8 / √16) = 3 / 2 = 1.5.

The empirical rule summarizes the standard deviation normal distribution percentages that every student should know:

• About 68% of values lie within 1 standard deviation of the mean (exactly 68.27%).
• About 95% lie within 2 standard deviations (95.45%).
• About 99.7% lie within 3 standard deviations (99.73%).

Because the curve is symmetric, the same percentages apply below the mean. A related number you will meet constantly: exactly 95% of values fall within ±1.96 standard deviations, which is where the 1.96 in 95% confidence intervals comes from.

Use the rule for quick mental estimates and sanity checks. If a variable has mean 70 and standard deviation 8 but values of 20 and 120 appear regularly, either the distribution is not normal or something is wrong with the data.

To find a probability from a normal distribution, convert the value to a z-score, then look up the area to the left of that z in a standard normal table or with software. A standard z-table gives P(Z ≤ z), the cumulative area to the left.

How to use the standard normal distribution table

In a full z-table, the row gives z to one decimal place and the column gives the second decimal. For z = 1.75, go to row 1.7 and column 0.05, where you will find .9599. For negative z, either use the negative half of the table or use symmetry: P(Z ≤ −z) = 1 − P(Z ≤ z).

Three rules cover almost every question:

• Area to the left: P(X ≤ x) = table value for z.
• Area to the right: P(X > x) = 1 − table value.
• Area between two values: P(a < X < b) = table value for z_b − table value for z_a.

Worked examples with exam scores N(70, 8²)

1. What proportion score below 82? z = (82 − 70) / 8 = 1.5. P(Z ≤ 1.5) = .9332, so about 93.3%.
2. What proportion score above 90? z = (90 − 70) / 8 = 2.5. P(Z > 2.5) = 1 − .9938 = .0062, about 0.6%.
3. What proportion score between 66 and 80? z = (66 − 70) / 8 = −0.5 and z = (80 − 70) / 8 = 1.25. P(Z ≤ −0.5) = 1 − .6915 = .3085 and P(Z ≤ 1.25) = .8944, so the answer is .8944 − .3085 = .5859, about 58.6%.
4. What score marks the top 10%? This runs the process backward. Find the z with .9000 to its left: z ≈ 1.2816. Then x = μ + zσ = 70 + 1.2816 × 8 = 80.25. A score of about 80.3 or higher is in the top 10%.
5. What is the chance a class of 16 averages 73 or more? From the previous section, z = 1.5, so P = 1 − .9332 = .0668, about 6.7%.

In software, you can skip the table. In Excel, =NORM.DIST(82, 70, 8, TRUE) returns .9332 and =NORM.INV(0.9, 70, 8) returns 80.25. In R, pnorm(82, 70, 8) and qnorm(0.9, 70, 8) do the same.

Many analyses assume normality, but it is important to know what must be normal. For a t-test, it is the outcome within each group (or the difference scores, for a paired test). For linear regression and ANOVA, it is the residuals, not the raw variables. Check the right thing.

Visual checks (start here)

• Histogram: look for a roughly symmetric, single-peaked shape. Long tails on one side indicate skew.
• Q-Q plot: plots your data's quantiles against those expected from a normal distribution. Points close to the diagonal line suggest normality; a curve indicates skew, and an S-shape indicates tails that are too heavy or too light.
• Boxplot: a median far from the center of the box, or many outliers on one side, signals asymmetry.

Numerical checks

• Skewness and kurtosis: a normal distribution has skewness 0 and excess kurtosis 0. Values close to zero (many texts use a working range of roughly −1 to +1 for skewness) suggest only mild departures.
• Shapiro–Wilk test: tests the null hypothesis that the data come from a normal distribution; p < .05 suggests non-normality. It is the most widely recommended formal test for small and moderate samples.
• Kolmogorov–Smirnov test (with Lilliefors correction): an alternative, generally less powerful than Shapiro–Wilk.

Interpret tests with care

Formal normality tests depend on sample size. With small samples they have little power and may miss real problems; with large samples they flag trivial departures that do not matter, because the central limit theorem makes mean-based methods robust. Base your judgment on plots plus the purpose of the analysis, not on a single p-value.

If the data are not normal

• Check for data entry errors and genuine outliers first.
• Consider a transformation (log or square root for right-skewed data such as income or reaction times).
• Use a nonparametric test, such as Mann–Whitney U or Wilcoxon signed-rank, instead of a t-test.
• Use bootstrapped confidence intervals or a model that fits the data's actual distribution (for example, Poisson regression for counts).

Our guide to the t-test explains how robust that test is to departures from normality.

• Standardized scores: many psychological and educational tests are scaled so scores are approximately normal with a chosen mean and standard deviation, which makes percentiles easy to compute with z-scores.
• Confidence intervals: a large-sample 95% confidence interval for a mean is x̄ ± 1.96 × SE, which comes straight from the standard normal distribution.
• Hypothesis tests: z-tests and the large-sample behavior of many test statistics rely on the normal distribution, and the t distribution converges to it as degrees of freedom grow.
• Measurement error: random errors in measurement are often modeled as normal with mean zero.
• Quality control: process limits are often set at μ ± 3σ, beyond which only about 0.27% of values would fall if the process were stable and normal.

Not everything is normal. Incomes, waiting times and counts are usually right-skewed; proportions are bounded between 0 and 1; and Likert items have only a few categories. Assuming normality where it clearly does not hold leads to misleading results. If you are deciding how to describe a variable before testing it, see descriptive vs inferential statistics.

• Using the variance instead of the standard deviation. In N(70, 64), the 64 is the variance; σ = 8. Divide by σ, not σ², when computing z.
• Forgetting which side the table measures. Most tables give the area to the left. For "greater than" questions, subtract from 1.
• Treating the PDF value as a probability. f(x) is a height, not an area, and can even exceed 1 for very narrow distributions.
• Using σ instead of σ / √n for a sample mean. Questions about averages of samples need the standard error.
• Assuming normality without checking. Always look at the data before relying on z-based probabilities.
• Rounding z too early. Keep at least two decimals for z and four for probabilities.