3.2 Uniform Distribution: $\mathrm{𝖴𝗇𝗂𝖿}(a,b)$

A continuous random variable for which all outcomes in a given range have equal chance of occurring is said to be uniformly distributed. Specifically, a random variable $X$ has a Uniform distribution over the interval $(a,b)$ if the pdf and cdf are given by

where $𝜽=(a,b)$, which we write $X\sim \mathrm{𝖴𝗇𝗂𝖿}(a,b)$ or sometimes $X\sim U(a,b)$. This pdf for two different sets of parameter values is illustrated on Figure 3.1 (First Link, Second Link).

The reason that this pdf is appropriate for such random outcomes is that for all $x$ and $x+c$ such that

so the probability of $X$ falling in any interval of length $c$ in the range $(a,b)$ is the same for all $x$, i.e. independent of the position $x$ and proportional to the interval length $c$.

The $r$-th moment of the $\mathrm{𝖴𝗇𝗂𝖿}(a,b)$ distribution is

$𝖤\left[X^r\right]={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{t}^r{f}_X(t)dt={\displaystyle {\int }_a^b}{\displaystyle \frac{t^r}{b-a}}dt={\displaystyle \frac{b^{r+1}-a^{r+1}}{(r+1)(b-a)}}.$

Hence the expectation (taking $r=1$ in the result above) is

$𝖤\left[X\right]={\displaystyle \frac{b^2-a^2}{2(b-a)}}={\displaystyle \frac{b+a}{2}},$

and the variance is

$\mathrm{𝖵𝖺𝗋}\left[X\right]={\displaystyle \frac{b^3-a^3}{3(b-a)}}-{\left[{\displaystyle \frac{b+a}{2}}\right]}^2={\displaystyle \frac{{(b-a)}^2}{12}}.$

These results seem logical as if all values in the interval $(a,b)$ are equally likely then the expected value should be the average of the endpoints. Similarly the wider the interval the more variable the outcomes, hence the larger variance.

Solution.  Let $X$ model the time to stepping on the gum. If $X\sim \mathrm{𝖴𝗇𝗂𝖿}(0,20)$ then $F(2)=0.1$ and $1-F(18)=0.1$.