The uniform distribution, explained

What a uniform distribution is

A uniform distribution spreads probability evenly, so no outcome is favored over any other. Graphed, it is flat: in the Fall 2026 AP framework, a distribution whose frequencies are all about the same, with no prominent peak, is called approximately uniform.

There are two versions, and mixing them up causes most of the trouble students have here. The discrete uniform applies to a countable list of outcomes, such as the six faces of a fair die, and gives each outcome the same probability. The continuous uniform applies to every value in an interval, such as any real number between 0 and 10, and gives each equal-width slice of that interval the same probability.

The distinction matters because probability is computed differently in each case. For a discrete uniform you add up probabilities of individual outcomes. For a continuous uniform you measure area under a flat curve, and no individual value carries any probability at all. If the discrete and continuous split is new, see discrete vs continuous variables.

The discrete uniform: a fair die

Let $X$ be the number showing on one roll of a fair six-sided die. A probability distribution lists every possible value with its probability, and here all six probabilities are equal.

Two checks confirm this is a legitimate probability distribution. Every probability is between 0 and 1, and the probabilities add to $6 \times \frac{1}{6} = 1$, which the AP framework requires of any discrete probability distribution.

Probabilities of events come from counting outcomes. $P(X > 4) = P(X = 5) + P(X = 6) = \frac{2}{6} \approx 0.333$, and $P(X \text{ is even}) = \frac{3}{6} = 0.5$. In general, for $k$ of $n$ equally likely outcomes the probability is $\frac{k}{n}$, which is the reason the fair-die model shows up constantly in probability questions.

The mean of a discrete uniform

The mean of a random variable is its expected value, written $E(X)$ or $\mu_X$ (mu-sub-X), and the AP formula is a weighted sum:

Read that as: multiply each possible value by its probability and add the results. For the fair die, every probability is $\frac{1}{6}$, so

The expected value is 3.5 even though no face shows 3.5, because it is a long-run average rather than a prediction of one roll. There is also a shortcut worth knowing: for a discrete uniform on consecutive whole numbers from $a$ to $b$, symmetry puts the mean at the midpoint, $\frac{a + b}{2}$. For the die that gives $\frac{1 + 6}{2} = 3.5$, matching the formula. You can check work like this in the expected value calculator.

The continuous uniform: probability is area

A continuous random variable can take any value in an interval, and every interval inside its domain has a probability attached. You cannot list the outcomes, so probability is represented by area under a density curve instead of by a table.

For a continuous uniform on the interval from $a$ to $b$, the density curve is a horizontal line, a rectangle sitting on the interval. Total probability must equal 1, so the rectangle's area must equal 1. Its width is $b - a$, which forces its height to be

Every probability is then just a smaller rectangle. For any two values $c$ and $d$ inside the interval, the chance that $X$ lands between them is that piece's width times the height:

So probability under a continuous uniform is pure proportion. Ask what fraction of the total width your target region covers, and that fraction is the probability.

One consequence catches people out. A single point has width 0, so the rectangle above it has area $0 \times \text{height} = 0$, and the probability of hitting one exact value is 0. That is not saying the value is impossible; it is saying that a point carries no area, which is true under every continuous model rather than just this one.

The practical consequence saves you time. Because the endpoints contribute nothing, $P(3 < X < 7)$ and $P(3 \le X \le 7)$ are the same number. You never have to worry about whether an inequality is strict when the variable is continuous.

Discrete variables are the opposite. $P(X > 4)$ and $P(X \ge 4)$ differ for a die, because the single outcome 4 carries a real probability of $\frac{1}{6}$. Checking whether the endpoint counts is one of the first things to settle in any discrete probability question.

The mean and spread of a continuous uniform

A flat curve is symmetric about its middle, so the mean sits at the midpoint of the interval:

The median is the same value, since half the area lies on each side. For a uniform on 0 to 10 the mean and median are both $\frac{0 + 10}{2} = 5$.

The spread has a formula too. The standard deviation, written $\sigma$ (sigma), of a continuous uniform on $a$ to $b$ is $\sigma = \frac{b - a}{\sqrt{12}}$, which for the interval 0 to 10 gives $\frac{10}{\sqrt{12}} \approx 2.887$. Treat that one as background: the uniform is not a named probability model in the Fall 2026 AP framework, which builds its continuous work around the normal distribution instead.

Uniform vs normal: random does not mean bell-shaped

Students often assume that anything random is normally distributed, and the uniform distribution is the cleanest counterexample. Both shapes are symmetric, and both put the mean and median at the center, so on those two features they agree completely. Everything else differs.

The row that matters most is the third one. Under a normal model, values near the mean are more likely than values far out in the tails, which is what makes z-scores and the empirical rule useful. Under a uniform model on 0 to 10, the slice from 0 to 1 and the slice from 4.5 to 5.5 both have probability 0.1, so no region is unusual and a z-score tells you much less.

There is a bridge between the two, and it is one of the better surprises in the course. Take repeated samples from a uniform population and the distribution of the sample means becomes approximately normal as the sample size grows. The population shape does not have to be normal for the central limit theorem to deliver a normal sampling distribution.

Where uniform distributions actually turn up

Random number generators produce continuous uniform values between 0 and 1, which is why simulations start there. Assigning digits 0 through 9 to outcomes in a simulation uses a discrete uniform, and so does any fair draw: a shuffled deck, a lottery ball, a randomly chosen row of a random digit table.

Random assignment in an experiment leans on the same idea. Every subject is equally likely to land in each treatment group, which is what balances other variables across the groups on average.

One warning about real data. A histogram that looks roughly flat is described as approximately uniform, but that is a description of a sample, not a claim that the underlying model is uniform. Small samples produce lumpy graphs, so treat a flat-looking histogram as weak evidence and check what the data-generating process actually was.

Common mistakes

• Using the density height as a probability. The height on 0 to 10 is 0.1, but that is not the chance of any outcome; probability comes from multiplying height by width.
• Forgetting to divide by the total width. $P(3 < X < 7)$ is $\frac{4}{10}$, not 4.
• Counting endpoints in the continuous case. There is no difference between $<$ and $\le$ when the variable is continuous.
• Assuming a value outside the interval has some small chance. Outside $a$ to $b$ the density is 0, so the probability is exactly 0.
• Reaching for the empirical rule. The 68-95-99.7 percentages belong to the normal curve and are wrong for a flat one.
• Confusing uniform with random. Random means governed by chance; uniform means every equal-width slice is equally likely, which is one particular way for chance to behave.