←
The Uniform Distribution

5.2 The Uniform Distribution

The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints.

The mathematical statement of the uniform distribution is

f(x) = 1 b−a 1 b−a for a ≤ x ≤ b

where a = the lowest value of x and b = the highest value of x.

Formulas for the theoretical mean and standard deviation are

μ= a+b 2 μ= a+b 2 and σ= (b−a) 2 12 σ= (b−a) 2 12

Example 5.2

The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.

Problem

a. What is the probability that a person waits fewer than 12.5 minutes?

Solution

a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U(0, 15). Write the probability density function. f (x) = 1 15 − 0 1 15 − 0 = 1 15 1 15 for 0 ≤ x ≤ 15.

Find P (x < 12.5). Draw a graph.

P(x<k)=(base)(height)=(12.5-0)( 1 15 )=0.8333 P(x<k)=(base)(height)=(12.5-0)( 1 15 )=0.8333

The probability a person waits less than 12.5 minutes is 0.8333.

This shows the graph of the function f(x) = 1/15. A horiztonal line ranges from the point (0, 1/15) to the point (15, 1/15). A vertical line extends from the x-axis to the end of the line at point (15, 1/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = 12.5.
Figure 5.11

Problem

b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ.

Solution

b. μ = a + b 2 a + b 2 = 15 + 0 2 15 + 0 2 = 7.5. On the average, a person must wait 7.5 minutes.

σ = (b-a)2 12 = (15-0)2 12 (b-a)2 12= (15-0)2 12 = 4.3. The Standard deviation is 4.3 minutes.

Problem

c. Ninety percent of the time, the time a person must wait falls below what value?

NOTE

This asks for the 90th percentile.

Solution

c. Find the 90th percentile. Draw a graph. Let k = the 90th percentile.

P(x<k)=(base)(height)=(k−0)( 1 15 ) P(x<k)=(base)(height)=(k−0)( 1 15 )

0.90=( k )( 1 15 ) 0.90=( k )( 1 15 )

k=(0.90)(15)=13.5 k=(0.90)(15)=13.5

The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.

f(X)=1/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/15 on the y-axis, a vertical upward line from an arbitrary point on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The area of this probability region is equal to 0.90.
Figure 5.12

Try It 5.2

The total duration of baseball games in the major league in a typical season is uniformly distributed between 447 hours and 521 hours inclusive.

  1. Find a and b and describe what they represent.
  2. Write the distribution.
  3. Find the mean and the standard deviation.
  4. What is the probability that the duration of games for a team in a single season is between 480 and 500 hours?