Compute mean, standard deviation, and probability — or solve the inverse problem
This page provides two calculators for the continuous uniform distribution \( X \sim \mathrm{Uniform}(a, b) \):
Requirement: the lower bound must satisfy \( a < b \). If \( a = b \) the distribution is degenerate; if \( a > b \) the PDF would be negative.
The continuous uniform distribution models a random variable that is equally likely to take any value in the interval \( [a, b] \). Its probability density function is:
\[ f(x) = \begin{cases} \dfrac{1}{b-a} & a \le x \le b \\[1em] 0 & x < a \text{ or } x > b \end{cases} \]![Uniform density curve: constant height 1/(b − a) on the interval [a, b] and zero outside](https://www.analyzemath.com/probabilities/calculators/continous-uniform-probability-distribution.png)
| Quantity | Formula |
|---|---|
| Cumulative probability | \( P(X < x_1) = \displaystyle\int_a^{x_1} \frac{1}{b-a}\,dx = \frac{x_1 - a}{b - a} \;\; (a \le x_1 \le b) \) |
| Mean \( \mu \) | \( \mu = \dfrac{a + b}{2} \) |
| Variance \( \sigma^2 \) | \( \sigma^2 = \dfrac{(b - a)^2}{12} \) |
| Standard deviation \( \sigma \) | \( \sigma = \dfrac{b - a}{\sqrt{12}} \) |
Edge cases for \(P(X < x_1)\). If \(x_1 < a\), the probability is 0. If \(x_1 > b\), it is 1. For a continuous distribution, \(P(X < x_1) = P(X \le x_1)\) — the strict/non-strict distinction doesn't matter.
Given \( a \), \( b \), and \( x_1 \), compute the mean, standard deviation, and \( P(X < x_1) \).
Given \( a \), \( b \), and a target probability \( p \in [0, 1] \), find \( x_1 \) such that \( P(X < x_1) = p \). This computes the \( p \)-th quantile of the distribution.
Mean:
\[ \mu = \frac{1 + 5}{2} = 3 \]Standard deviation:
\[ \sigma = \frac{5 - 1}{\sqrt{12}} = \frac{4}{\sqrt{12}} \approx 1.1547 \]Probability:
\[ P(X < 2) = \frac{2 - 1}{5 - 1} = \frac{1}{4} = 0.25 \]Answer: \( \mu = 3 \), \( \sigma \approx 1.1547 \), \( P(X < 2) = 0.25 \).
Solve \( \dfrac{x_1 - 4}{10 - 4} = 0.25 \):
\[ x_1 - 4 = 1.5 \;\Rightarrow\; x_1 = 5.5 \]Check: \( P(X < 5.5) = \dfrac{5.5 - 4}{6} = \dfrac{1.5}{6} = 0.25 \; \checkmark \)
Let \( X \sim \mathrm{Uniform}(2, 8) \). Compute the mean, standard deviation, and \( P(X < 5) \).
\( \mu = (2+8)/2 = 5 \).
\( \sigma = (8-2)/\sqrt{12} = 6/\sqrt{12} \approx 1.7321 \).
\( P(X < 5) = (5-2)/(8-2) = 3/6 = 0.5 \).
For the same \( X \sim \mathrm{Uniform}(2, 8) \), find the value \( x_1 \) such that \( P(X < x_1) = 0.75 \).
Solve \( (x_1 - 2)/6 = 0.75 \):
\[ x_1 - 2 = 4.5 \;\Rightarrow\; x_1 = 6.5 \]For \( X \sim \mathrm{Uniform}(0, 10) \), find \( P(2 \le X \le 7) \).
Use \( P(2 \le X \le 7) = P(X < 7) - P(X < 2) \):
\[ = \frac{7}{10} - \frac{2}{10} = \frac{5}{10} = 0.5 \]Or directly: \( \text{width} / (b - a) = (7 - 2)/(10 - 0) = 5/10 = 0.5 \).
For \(X \sim \mathrm{Uniform}(a, b)\), \(f(x) = 1/(b-a)\) for \(a \le x \le b\), and 0 elsewhere. The constant height reflects equal likelihood across the interval.
\( \mu = (a+b)/2 \) (the midpoint). \( \sigma^2 = (b-a)^2/12 \), so \( \sigma = (b-a)/\sqrt{12} \).
\( P(X < x_1) = \dfrac{x_1 - a}{b-a} \) for \( a \le x_1 \le b \). If \(x_1 < a\), it's 0; if \(x_1 > b\), it's 1.
Given \( p \in [0,1] \), find \( x_1 \) such that \( P(X < x_1) = p \). The solution is \( x_1 = a + p(b-a) \) — the \(p\)-th quantile of the uniform distribution.
Last updated: October 20, 2026