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

A continuous uniform (rectangular) distribution on $[a,b]$ gives every sub-interval of the same length the same probability. We write $X\sim U[a,b]$. The pdf is a constant: $f(x)=\begin{cases}\dfrac{1}{b-a}&a\le x\le b,\\[4pt]0&\text{otherwise.}\end{cases}$ The graph is a rectangle of width $b-a$ and height $\frac{1}{b-a}$, so its area is 1. Use it to model a quantity that is equally likely to lie anywhere in a range, such as a rounding error or the waiting time for a service with random arrivals.

Integrating the constant density from $a$ to $x$: $F(x)=\int_a^x\frac{1}{b-a}\,dt=\frac{x-a}{b-a},\quad a\le x\le b,$ with $F(x)=0$ for $x<a$ and $F(x)=1$ for $x>b$. This is a straight line from $0$ at $x=a$ to $1$ at $x=b$. For any interval, $P(c<X<d)=F(d)-F(c)=\frac{d-c}{b-a}$, the fraction of the range covered. Example: $X\sim U[2,8]$, $P(X\le3.5)=\frac{1.5}{6}=\frac14$.

By symmetry the mean is the midpoint of the interval. The derivation is a standard integral: $E(X)=\int_a^b\frac{x}{b-a}\,dx=\frac{b^2-a^2}{2(b-a)}=\frac{a+b}{2}.$ For the variance, find $E(X^2)$ and subtract the square of the mean: $E(X^2)=\int_a^b\frac{x^2}{b-a}\,dx=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3},$ $\text{Var}(X)=\frac{a^2+ab+b^2}{3}-\frac{(a+b)^2}{4}=\frac{(b-a)^2}{12}.$ Check: $U[2,8]$ has mean 5 and variance $\frac{36}{12}=3$.

To find $a$ and $b$ from given moments, form two equations: $a+b=2E(X)$, and $(b-a)^2=12\,\text{Var}(X)$. Take the positive root for $b-a$ because $b>a$. Worked example: $E(X)=12$, $\text{Var}(X)=12$. Then $a+b=24$ and $(b-a)^2=144$, so $b-a=12$ and $a=6$, $b=18$. Then $P(X>15)=\frac{18-15}{12}=\frac14$. State the model assumption when you use it: 'assuming the rod length is equally likely to be anywhere in the interval'.