6.2 Uniform Distribution

The continuous uniform distribution models the probability that is the same on an interval from a to b. We use the following probability density function (PDF) to graph a straight line.

The probability is found by taking the area between two points within the rectangle formed from the x-axis, between the endpoints a and b, the length, and f(x) = 1/(b-a), the height. When working with continuous distributions it is helpful to draw a picture of the distribution, then shade in the area of the probability that you are trying to find. See Figure 6-5.

Uniform density function drawn on axes labeled f(x) and x: a rectangle of height 1/(b-a) sits above the x-axis between the points a and b.

Figure 6-5

If a continuous random variable X has a uniform distribution with starting point a and ending point b then the distribution is denoted as X~U(a,b).

Definition

To find the probability (area) under the uniform distribution, use the following formulas.

$P (X \geq x) = P (X > x) = (\frac{1}{b - a}) \cdot (b - x)$
$P (X \leq x) = P (X < x) = (\frac{1}{b - a}) \cdot (x - a)$
$P (x_{1} \leq X \leq x_{2}) = P (x_{1} < X < x_{2}) = (\frac{1}{b - a}) \cdot (x_{2} - x_{1})$

Example 1

The arrival time between trains at a train stop is uniformly distributed between 0 and 15 minutes. A student does not check the schedule and has arrived at the train stop.

# Train arrival ~ Uniform(0, 15) minutes
punif(10, 0, 15, lower.tail = FALSE)   # P(X > 10)  -> 0.3333
punif(8, 0, 15) - punif(2, 0, 15)      # P(2 <= X <= 8) -> 0.4

Compute the probability they wait more than 10 minutes.
Compute the probability of waiting between 2 and 8 minutes.

Show solution

a) First plug in the endpoints a = 0 and b = 15 into the PDF to get the height of the rectangle. The height is f(x)= $\frac{1}{15 - 0} = \frac{1}{15}$ . Draw and label the distribution with the a, b and the height as in Figure 6- 6. The probability is the area of the shaded rectangle P(X > 10). Draw a vertical line at x = 10. We want x values that are greater than 10, so shade the area to the right of 10, stopping at b = 15. To find the area of the shaded rectangle in Figure 6-6, we can take the length times the height. The length would be b – a = 15 – 10 = 5 and the height is f(x) = 1/15.

Figure 6-6

The area of the shaded rectangle is 5 ( $\frac{1}{15}$ ) = $\frac{1}{3}$ = 0.3333 or P(X > 10) = 0.3333, which is the probability of waiting more than 10 minutes. Note that this would be the same if we asked P(X ≥ 10) = 0.3333 since there is no area at the line X = 10.

b) The area will be length times height. Draw the picture and shade the rectangle between 2 and 8, see Figure 6-7. The length is b – a = 8 – 2 = 6 and the height is still f(x) = 1/15. P(2 ≤ X ≤ 8) = 6( $\frac{1}{15}$ ) = 0.4

Figure 6-7

Adapted from Mostly Harmless Statistics by Rachel Webb (Portland State University), hosted on LibreTexts (stats.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.