15.3 Uniform

Discrete

Discrete uniform distribution The discrete uniform distribution is defined to be a constant value for all possibilities. Mathematically this is written

$p(x_{i})=\frac{1}{N}$

where $N$ is the total number of possibilities, labeled $x_{1}$ to $x_{N}$ . The picture of the distribution is shown in Figure C.1

Continuous

Continuous uniform distribution The continuous uniform distribution is defined to be a constant between a minimum and maximum value, and zero everywhere else. Mathematically this is written

$p(x)=\frac{1}{\rm max-min} \text{ for }{\rm min}<x<{\rm max}$

. The picture of the distribution is shown in Figure C.2.

Note

Continuous uniform distribution

The continuous uniform distribution is defined to be a constant between a minimum and maximum value, and zero everywhere else. Mathematically this is written

$p(x)=\frac{1}{\rm max-min} \text{ for }{\rm min}<x<{\rm max}$

Example 1 You call a plumber, and they say that they can come anytime in the next 4 hours. The probability of them arriving at any particular time can be represented with a uniform distribution. What is the probability that they arrive in the first 20 minutes of the second hour?

Show solution

Figure C.3. Continuous uniform distribution for the plumber example (Example C.1).

In order to ask questions about total probability from a continuous distribution you take the area under the curve between the relevant values. In this case it'd be the area under the curve from the time $t=2 {\rm hr}$ and $t=2 {\rm hr} + 20 {\rm minutes} = 2.333 {\rm hr}$ , as shown in Figure C.3. The area under the curve is just the area of the shaded region between times $t=2{\rm hr}$ and $t=2.333 {\rm hr}$ , or just the area of a rectangle - $A={\rm base} \times {\rm height}$ . The base of the rectangle is the length of time, or

$\begin{eqnarray*} \text{\emph{base}}=0.333 {\rm hr} \end{eqnarray*}$

The height of the rectangle is given by the constant value of the uniform distribution, or

$\begin{eqnarray*} \text{\emph{height}} = \frac{1}{{\rm max}-{\rm min}} = \frac{1}{4 {\rm hr} - 0 {\rm hr}} = 0.25 \frac{1}{\rm hr} \end{eqnarray*}$

So the total probability of the plumber coming in the first 20 minutes of the second hour is

$\begin{eqnarray*} P(2<t<2.25) = \left(0.333 {\rm hr}\right) \times \left(0.25 \frac{1}{\rm hr}\right) = 0.0833 \end{eqnarray*}$

Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.