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Generated on: July 20, 2026 at 09:33 PM

Uniform Distribution Calculator

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Minimum0
Maximum10
Interval lower bound2
Interval upper bound5

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F(s)
s F(s)

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Mathematics

Uniform Distribution Calculator

Calculate the probability density, mean, variance, and interval probability P(c ≤ X ≤ d) of a continuous uniform distribution on [a, b]. Enter the bounds and the interval.

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F(s)
s F(s)
\begin{aligned} P(c \le X \le d) &= \dfrac{d - c}{b - a} \\ &= \dfrac{(5) - (2)}{10 - (0)} \\ &= ? \end{aligned}
\begin{aligned} F(c) = P(X \le c) &= \dfrac{c - a}{b - a} \\ &= \dfrac{(2) - (0)}{10 - (0)} \\ &= ? \end{aligned}
\begin{aligned} \mu &= \dfrac{a + b}{2} \\ &= \dfrac{(0) + (10)}{2} \\ &= ? \end{aligned}
\begin{aligned} \sigma^2 &= \dfrac{(b - a)^2}{12} \\ &= \dfrac{(10 - (0))^2}{12} \\ &= ? \end{aligned}
\begin{aligned} f(x) &= \dfrac{1}{b - a} \\ &= \dfrac{1}{10 - (0)} \\ &= ? \end{aligned}

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Uniform distribution calculator

The continuous uniform distribution is the simplest continuous probability model: every value in an interval [a, b] is equally likely, and nothing outside it can occur. It is the natural description of "a value somewhere between these limits, with no preference" — rounding error, a random arrival time within a fixed window, or a random point on a line segment. From the two bounds and an interval [c, d], this calculator returns the probability density, the mean, the variance, the cumulative probability P(X ≤ c), and the interval probability P(c ≤ X ≤ d).

The uniform distribution

The distribution is fully described by two parameters:

  • a — the minimum, the smallest value X can take
  • b — the maximum, the largest value X can take

Because probability is spread evenly, the probability density function is constant on [a, b] and zero elsewhere:

f(x)={1b−aa≤x≤b0otherwisef(x) = \begin{cases} \dfrac{1}{b - a} & a \le x \le b \\[6pt] 0 & \text{otherwise} \end{cases}

The density has to integrate to 1 over the interval of width (b − a), which forces the constant height to be exactly 1 / (b − a).

Cumulative distribution

The cumulative distribution function (CDF) gives the probability that X is at or below a value x. For the uniform distribution it is a straight line rising from 0 at a to 1 at b:

F(x)=P(X≤x)=x−ab−a,a≤x≤bF(x) = P(X \le x) = \frac{x - a}{b - a}, \quad a \le x \le b

Below a it is 0; above b it is 1. This linear ramp is what the chart plots.

Interval probability

The probability that X lands between c and d is the fraction of the total width that the interval [c, d] covers:

P(c≤X≤d)=F(d)−F(c)=d−cb−aP(c \le X \le d) = F(d) - F(c) = \frac{d - c}{b - a}

When c or d falls outside [a, b], it is clamped to the bound before measuring, so the probability never exceeds 1.

Mean and variance

Because the distribution is symmetric, the expected value sits at the midpoint:

μ=a+b2\mu = \frac{a + b}{2}

The variance measures the spread. For the continuous uniform distribution it is:

σ2=(b−a)212\sigma^2 = \frac{(b - a)^2}{12}

The standard deviation is therefore (b − a) / √12 ≈ 0.2887 × (b − a). Doubling the interval width quadruples the variance.

QuantityFormula
Density f(x)1 / (b − a)
Mean μ(a + b) / 2
Variance σ²(b − a)² / 12
CDF F(x)(x − a) / (b − a)

Worked example

A bus arrives at a stop at a time that is uniformly distributed between a = 0 and b = 15 minutes after you arrive. What is the probability you wait between c = 2 and d = 5 minutes?

First the density:

f(x)=115−0=115≈0.0667f(x) = \frac{1}{15 - 0} = \frac{1}{15} \approx 0.0667

The interval probability:

P(2≤X≤5)=5−215−0=315=0.2P(2 \le X \le 5) = \frac{5 - 2}{15 - 0} = \frac{3}{15} = 0.2

So there is a 20% chance the wait falls in that window. The expected wait is:

μ=0+152=7.5 minutes\mu = \frac{0 + 15}{2} = 7.5 \text{ minutes}

with variance:

σ2=(15−0)212=22512=18.75\sigma^2 = \frac{(15 - 0)^2}{12} = \frac{225}{12} = 18.75

giving a standard deviation of about 4.33 minutes.

Density is not probability

The value f(x) = 1 / (b − a) is a density, not a probability, and it can be greater than 1 when the interval is narrow — for a = 0, b = 0.5 the density is 2. This is fine: a density becomes a probability only after being multiplied by a width. The probability of any single exact value is zero, so P(c ≤ X ≤ d) equals P(c < X < d); the endpoints contribute nothing.

Frequently Asked Questions (FAQ)

What is a continuous uniform distribution?

A continuous uniform distribution assigns equal probability to every value in an interval [a, b] and zero probability outside it. Because the density is flat, any two sub-intervals of the same width are equally likely. It models situations where an outcome is equally likely to land anywhere within known limits — such as the rounding error of a measurement or a random arrival time within a fixed window.

What is the difference between the PDF and the CDF here?

The probability density function (PDF) is constant, f(x) = 1 / (b − a), and represents density rather than probability. The cumulative distribution function (CDF), F(x) = (x − a) / (b − a), is a straight line rising from 0 at a to 1 at b, and gives the probability that X is at or below x. The interval probability P(c ≤ X ≤ d) equals F(d) − F(c).

How are the mean and variance of a uniform distribution calculated?

The mean is the midpoint of the interval, μ = (a + b) / 2, because the distribution is symmetric. The variance is σ² = (b − a)² / 12; the standard deviation is therefore (b − a) / √12 ≈ 0.2887 × (b − a). Widening the interval increases the variance quadratically.

What is the probability of a single exact value?

For any continuous distribution, including the uniform, the probability of any single exact value is zero. Only intervals carry non-zero probability. That is why P(c ≤ X ≤ d) uses the interval width — the endpoints c and d themselves contribute nothing, so P(c ≤ X ≤ d) equals P(c < X < d).

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