Mean and Variance of the Continuous Uniform Distribution

Derives and applies the formulas $E[X]=\dfrac{a+b}{2}$ and $\mathrm{Var}(X)=\dfrac{(b-a)^2}{12}$ for $X\sim U(a,b)$ , including inverse problems where the endpoints must be recovered from the mean and variance.

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Tutorial

Mean of the Continuous Uniform Distribution

If $X \sim U(a,b)$ , then $X$ has constant density $f(x) = \dfrac{1}{b-a}$ on the interval $[a,b]$ . The mean of $X$ is

E[X] = \int_a^b x \cdot \dfrac{1}{b-a} \, dx = \dfrac{1}{b-a} \cdot \left[ \dfrac{x^2}{2} \right]_a^b = \dfrac{b^2 - a^2}{2(b-a)} = \dfrac{(b-a)(b+a)}{2(b-a)} = \dfrac{a+b}{2}.

So the mean of any continuous uniform random variable on $[a,b]$ is just the midpoint of the interval:

E[X] = \dfrac{a+b}{2}.

For example, if $X \sim U(1, 9)$ , then

E[X] = \dfrac{1+9}{2} = 5.