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The Uniform Distribution

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

Summary: Continuous Random Variable: Uniform Distribution is part of the collection col10555 written by Barbara Illowsky and Susan Dean. It describes the properties of the Uniform Distribution with contributions from Roberta Bloom.

Example 1

The previous problem is an example of the uniform probability distribution.

Illustrate the uniform distribution. The data that follows are 55 smiling times, in seconds, of an eight-week old baby.

Table 1
10.4 19.6 18.8 13.9 17.8 16.8 21.6 17.9 12.5 11.1 4.9
12.8 14.8 22.8 20.0 15.9 16.3 13.4 17.1 14.5 19.0 22.8
1.3 0.7 8.9 11.9 10.9 7.3 5.9 3.7 17.9 19.2 9.8
5.8 6.9 2.6 5.8 21.7 11.8 3.4 2.1 4.5 6.3 10.7
8.9 9.4 9.4 7.6 10.0 3.3 6.7 7.8 11.6 13.8 18.6

sample mean = 11.49 and sample standard deviation = 6.23

We will assume that the smiling times, in seconds, follow a uniform distribution between 0 and 23 seconds, inclusive. This means that any smiling time from 0 to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.

Let XX = length, in seconds, of an eight-week old baby's smile.

The notation for the uniform distribution is

XX ~ U(a, b)U(a,b) where aa = the lowest value of xx and bb = the highest value of xx.

The probability density function is f⁢x = 1b-a fx=1b-a for a≤x≤b a x b.

For this example, xx ~ U(0, 23)U(0,23) and f⁢x = 123-0 fx=123-0 for 0≤x≤23 0 x 23.

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

For this problem, the theoretical mean and standard deviation are

μ=0+23 2=11.50 μ 0+23 2 11.50 seconds and σ= (23-0)2 12 =6.64 σ (23-0)2 12 6.64 seconds

Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation.

Example 2

Problem 1

What is the probability that a randomly chosen eight-week old baby smiles between 2 and 18 seconds?

[ Show Solution ]

Problem 2

Find the 90th percentile for an eight week old baby's smiling time.

[ Show Solution ]

Problem 3

Find the probability that a random eight week old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN 8 SECONDS.

[ Show Solution ]

Example 3

Uniform: The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between 0 and 15 minutes, inclusive.

Problem 1

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

[ Show Solution ]

Problem 2

On the average, how long must a person wait?

Find the mean, μμ, and the standard deviation, σσ.

[ Show Solution ]

Problem 3

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

Note:

This asks for the 90th percentile.

[ Show Solution ]

Example 4

Uniform: Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let XX = the time, in minutes, it takes a nine-year old child to eat a donut. Then XX ~ U(0.5, 4)U(0.5,4).

Problem 1

The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______.

[ Show Solution ]

Problem 2

Find the probability that a different nine-year old child eats a donut in more than 2 minutes given that the child has already been eating the donut for more than 1.5 minutes.

The second probability question has a conditional (refer to "Probability Topics"). You are asked to find the probability that a nine-year old child eats a donut in more than 2 minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see the first example). You must reduce the sample space. First way: Since you already know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5a = 0.5 minutes. Your starting point is 1.5 minutes.

Write a new f(x):

f⁢x =14-1.5 =25fx=14-1.5=25 for 1.5≤x≤4 1.5 x 4.

Find P(x>2|x>1.5) P(x 2|x 1.5). Draw a graph.

f(X)=2/5 graph displaying a boxed region consisting of a horizontal line extending to the right from point 2/5 on the y-axis, a vertical upward line from points 1.5 and 4 on the x-axis, and the x-axis. A shaded region from points 2-4 occurs within this area.

P(x>2|x>1.5) =(base)(new height) = (4-2) (2/5) =? P(x 2|x 1.5)=(base)(new height)=(4-2)(2/5)=?

[ Show Solution ]

The probability that a nine-year old child eats a donut in more than 2 minutes given that the child has already been eating the donut for more than 1.5 minutes is 4545.

Second way: Draw the original graph for xx ~ U(0.5, 4)U(0.5,4). Use the conditional formula

P(x>2|x>1.5)= P(x> 2AND x>1.5) P ( x > 1.5 ) = P ( x > 2 ) P ( x > 1.5 ) = 2 3.5 2.5 3.5 = 0.8 =45 P(x 2|x 1.5) P(x> 2AND x>1.5) P ( x 1.5 ) P ( x 2 ) P ( x 1.5 ) 2 3.5 2.5 3.5 0.8 =45

Example 5

Uniform: Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and 4 hours. Let xx = the time needed to fix a furnace. Then xx ~ U(1.5, 4)U(1.5,4).

  1. Find the problem that a randomly selected furnace repair requires more than 2 hours.
  2. Find the probability that a randomly selected furnace repair requires less than 3 hours.
  3. Find the 30th percentile of furnace repair times.
  4. The longest 25% of repair furnace repairs take at least how long? (In other words: Find the minimum time for the longest 25% of repair times.) What percentile does this represent?
  5. Find the mean and standard deviation

Problem 1

Find the probability that a randomly selected furnace repair requires longer than 2 hours.

[ Show Solution ]

Problem 2

Find the probability that a randomly selected furnace repair requires less than 3 hours. Describe how the graph differs from the graph in the first part of this example.

[ Show Solution ]

Problem 3

Find the 30th percentile of furnace repair times.

[ Show Solution ]

Problem 4

The longest 25% of furnace repair times take at least how long? (Find the minimum time for the longest 25% of repairs.)

[ Show Solution ]

Problem 5

Find the mean and standard deviation

[ Show Solution ]

**Example 5 contributed by Roberta Bloom

Glossary

Conditional Probability:
The likelihood that an event will occur given that another event has already occurred.
Uniform Distribution:
A continuous random variable (RV) that has equally likely outcomes over the domain, a<x<ba<x<b size 12{a<x<b} {}. Often referred as the Rectangular distribution because the graph of the pdf has the form of a rectangle. Notation: X~U(a,b)X~U(a,b) size 12{X "~" U \( a,b \) } {}. The mean is μ=a+b2μ=a+b2 size 12{μ= { {a+b} over {2} } } {} and the standard deviation is σ= (b-a)2 12 σ (b-a)2 12 The probability density function is f⁢x = 1b-a fx=1b-a for a<x<b a x b or a≤x≤b a x b. The cumulative distribution is P(X≤x)=x−ab−aP(X≤x)=x−ab−a size 12{P \( X <= x \) = { {x-a} over {b-a} } } {}.

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