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The Uniform Distribution

September 9, 2024 | by Bloom Code Studio

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) = 1b−a1b−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+b2μ=a+b2 and σ=(b−a)212−−−−−√σ=(b−a)212

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) = 115 − 0115 − 0 = 115115 for 0 ≤ x ≤ 15.

Find P (x < 12.5). Draw a graph.

P(x<k)=(base)(height)=(12.5−0)(115)=0.8333P(x<k)=(base)(height)=(12.5-0)(115)=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 + b2a + b2 = 15 + 0215 + 02 = 7.5. On the average, a person must wait 7.5 minutes.

σ = (b−a)212−−−−−√=(15−0)212−−−−−−√(b-a)212=(15-0)212 = 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)(115)P(x<k)=(base)(height)=(k−0)(115)

0.90=(k)(115)0.90=(k)(115)

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

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