Ask a computer to "pick a random number between 0 and 1" and it reaches for exactly one distribution: the continuous uniform. It is the simplest continuous distribution there is — and the most democratic. Every value in a chosen interval [a, b] is equally likely; no stretch of the interval is favoured over any other of the same length. There is no peak, no tail, no lopsidedness — just a perfectly flat, level playing field from a to b.

Because it is so even-handed, it turns up wherever "we know the range but nothing more" is the honest description: the fractional part of a rounding error, the position of a bus somewhere along a timetable window, the raw output of a random-number generator. It is also the launch pad for every other random distribution a computer produces — feed uniform numbers through the right transformation and out come normal, exponential, or any shape you like.

Unlike the normal distribution, whose bell concentrates values near a centre, the uniform spreads its chances out perfectly flatly. We write

to say X is uniform on the interval from a to b.

A flat density: how "equally likely" looks

For a continuous distribution we describe chance with a probability density f(x) — a curve whose area gives probability. "Every value equally likely" forces that curve to be constant across [a, b] and zero everywhere else:

Why the exact height \tfrac{1}{b-a}? Because the total area must be 1 — some value certainly occurs. The graph is just a rectangle of width (b - a) sitting on the interval, and a rectangle's area is width times height. To make that area equal 1, the height has to be exactly \tfrac{1}{b-a}. Widen the interval and the rectangle must get shorter; narrow it and the rectangle must grow taller — always keeping its area pinned at one.

Now the magic simplification. Since the height is constant, the probability of landing in some sub-interval [c, d] is just the area of a smaller rectangle — and that is simply what fraction of the whole interval it covers:

Probability becomes a matter of measuring length. Half the interval carries half the probability; a tenth of the interval carries a tenth. Nothing to integrate, nothing to look up — just "how big a slice is it?"

See it — and shade a probability

The top panel is the flat density; the shaded rectangle is P(c \le X \le d), and its area (printed above it) is exactly the fraction of the interval it covers. The bottom panel is the running total — the cumulative distribution F(x) = P(X \le x). Because chance accumulates at a constant rate, that graph is a perfectly straight line climbing from 0 at a to 1 at b. Slide a and b to reshape the interval, and c, d to move the shaded region.

The straight-line CDF has a tidy formula on the interval — it is just how far along you are, as a fraction:

(with F(x) = 0 below a and F(x) = 1 above b). Its constant slope is the density \tfrac{1}{b-a} — a flat density and a straight-line CDF are two views of the same evenness.

Centre and spread

Where is the "average" of a flat distribution? By symmetry it can only be the midpoint. Balance the rectangle on a fingertip and it tips exactly halfway along:

The spread is measured by the variance, and for the uniform it has a famous, clean form:

Notice it depends only on the width (b - a), not on where the interval sits: sliding the whole rectangle left or right changes the mean but never the spread. The mysterious 12 is not a fudge — it drops straight out of integrating x^2 across the interval, and it makes the standard deviation about 0.29 of the interval's width.

Worked example 1 — a random number on [0, 10]

Let X \sim U(0, 10). The interval has width b - a = 10, so the density is a flat f(x) = \tfrac{1}{10} = 0.1 across the whole stretch.

A probability. What is P(3 \le X \le 7)? The sub-interval [3, 7] has length 4, so we take the fraction of the whole:

No integral needed — the middle 40% of the interval carries 40% of the probability.

The mean. The midpoint of [0, 10] is \mu = \tfrac{0 + 10}{2} = 5.

The variance. Using \sigma^2 = \tfrac{(b-a)^2}{12},

so the standard deviation is \sigma = \sqrt{8.33} \approx 2.89 — right at that "0.29 \times the width" we predicted.

Worked example 2 — waiting for a bus

Buses on your route are so frequent that, from the moment you arrive at the stop, the next one is equally likely to come at any instant in the next 15 minutes. Let W be your waiting time, so W \sim U(0, 15). What is the chance you wait less than 5 minutes?

The lucky outcomes are the sub-interval [0, 5], of length 5, out of a total window of length 15. So it is simply the proportion:

A one-in-three chance of a short wait — and by the same reasoning your expected wait is the midpoint, \tfrac{0 + 15}{2} = 7.5 minutes. Notice we wrote W < 5 with a strict "<", but W \le 5 would give the identical answer — see the Watch out! below for why.