The continuous uniform distributionEdexcel A-Level Further Maths: Revision notes

A continuous random variable $X$ has a continuous uniform distribution on $[a,b]$, written $X\sim U[a,b]$, when every value in the interval is equally likely. Its probability density function is constant: $f(x)=\begin{cases}\dfrac{1}{b-a}, & a\le x\le b,\\[4pt] 0, & \text{otherwise.}\end{cases}$ The graph of $f$ is a rectangle of width $b-a$ and height $\frac{1}{b-a}$, which is why it is also called the rectangular distribution; the total area is 1. Typical models are waiting times for a regular service and rounding errors.

The mean is $\mathrm{E}(X)=\int_a^b\frac{x}{b-a}\,dx=\left[\frac{x^2}{2(b-a)}\right]_a^b=\frac{b^2-a^2}{2(b-a)}=\frac{a+b}{2},$ the midpoint of the interval. For the variance, first $\mathrm{E}(X^2)=\int_a^b\frac{x^2}{b-a}\,dx=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3},$ so $\mathrm{Var}(X)=\frac{a^2+ab+b^2}{3}-\frac{(a+b)^2}{4}=\frac{(b-a)^2}{12}.$ The median equals the mean, because the distribution is symmetrical.

For $x<a$ no probability lies below $x$; for $x>b$ all of it does. For $a\le x\le b$, $F(x)=\int_a^x\frac{1}{b-a}\,dt=\frac{x-a}{b-a}.$ So $F(x)=\begin{cases}0,&x<a,\\[2pt]\dfrac{x-a}{b-a},&a\le x\le b,\\[6pt]1,&x>b.\end{cases}$ This is a straight line from $(a,0)$ to $(b,1)$. Percentiles follow directly: the $p$th percentile is $a+\frac{p}{100}(b-a)$.

Probabilities are rectangle areas, so for $a\le c\le d\le b$: $\mathrm{P}(c<X<d)=\frac{d-c}{b-a}.$ Example: $X\sim U[2,10]$. Then $\mathrm{P}(X>7)=\frac{10-7}{8}=\frac38$. Because the distribution is continuous, $\mathrm{P}(X=c)=0$ and $<$ and $\le$ make no difference. If the interval you want runs outside $[a,b]$, cut it back to the part inside $[a,b]$ first.

If the mean and variance (or a probability) are given, form equations in $a$ and $b$. Example: $\mathrm{E}(Y)=9$ and $\mathrm{Var}(Y)=12$ give $a+b=18$ and $(b-a)^2=144$. Since $b>a$, $b-a=12$, so $a=3$ and $b=15$. In context, state the assumption behind the model: every value in the interval is equally likely, and no value outside it can occur. Uniform models suit rounding errors (a measurement recorded to the nearest $0.1$ has error uniform on $[-0.05,0.05]$) and waiting times at regular intervals.