The discrete uniform distributionEdexcel International A Level Maths: Revision notes

A discrete random variable $X$ has a discrete uniform distribution on $\{1,2,\ldots,n\}$ when each of the $n$ values is equally likely: $\mathrm{P}(X=x)=\frac1n,\quad x=1,2,\ldots,n.$ Typical models are a fair die ($n=6$), a fair spinner or a number drawn at random from a list of equally likely numbers. The model is only valid if every outcome really is equally likely.

Because every value has probability $\frac1n$, probabilities are found by counting the favourable values: $\mathrm{P}(X\le k)=\frac kn,\qquad \mathrm{P}(a\le X\le b)=\frac{b-a+1}{n}.$ Example: for a spinner numbered 1 to 8, $\mathrm{P}(X\ge6)$ counts the values 6, 7, 8, so it is $\frac38$. Check whether the inequality is strict ($>$) or not ($\ge$) before counting.

The mean of the discrete uniform distribution on $\{1,\ldots,n\}$ is $\mathrm{E}(X)=\frac{n+1}{2}.$ It is the midpoint of the values, because the distribution is symmetrical about it. It comes from $\mathrm{E}(X)=\frac1n(1+2+\cdots+n)=\frac1n\cdot\frac{n(n+1)}{2}$. For a fair six-sided die, $\mathrm{E}(X)=3.5$. The mean need not be a possible value of $X$.

The variance of the discrete uniform distribution on $\{1,\ldots,n\}$ is $\mathrm{Var}(X)=\frac{n^2-1}{12}.$ It comes from $\mathrm{Var}(X)=\mathrm{E}(X^2)-[\mathrm{E}(X)]^2$ with $\mathrm{E}(X^2)=\frac{(n+1)(2n+1)}{6}$. For a fair die, $\mathrm{Var}(X)=\frac{35}{12}=2.92$ (3 s.f.), and the standard deviation is $\sqrt{\frac{35}{12}}=1.71$. Worked example: a spinner numbered 1 to 8 has $\mathrm{E}(X)=4.5$ and $\mathrm{Var}(X)=\frac{64-1}{12}=5.25$.

If $\mathrm{E}(X)=\mu$ then $\frac{n+1}{2}=\mu$, so $n=2\mu-1$. If $\mathrm{Var}(X)=\sigma^2$ then $n^2=12\sigma^2+1$, and $n$ is the positive root. Example: $\mathrm{E}(Y)=7.5$ gives $n=14$, and then $\mathrm{Var}(Y)=\frac{196-1}{12}=16.25$. Example: $\mathrm{Var}(X)=14$ gives $n^2=169$, so $n=13$ (reject $-13$). Inequalities such as $\frac{m^2-1}{12}>14$ give bounds on $m$; $m$ must be an integer.

If $X$ takes the $n$ equally likely values $0,1,\ldots,n-1$, the shape is the same, shifted by 1. The mean becomes $\frac{n-1}{2}$ and the variance is unchanged at $\frac{n^2-1}{12}$. For a spinner numbered 0 to 9, $\mathrm{E}(X)=4.5$ and $\mathrm{Var}(X)=\frac{100-1}{12}=8.25$. When in doubt, work from first principles with $\mathrm{E}(X)$ and $\mathrm{E}(X^2)$. To judge whether $X$ lies within one standard deviation of the mean, find $\mu\pm\sigma$ and count the values inside.