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Continuous Random Variables: The Uniform Distribution

Module by: Susan Dean, Barbara Illowsky, Ph.D.

Summary: This module describes the properties of the Uniform Distribution which describes a set of data for which all values have an equal probability.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

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 fX = 1b-a fX=1b-a for a≤X≤b a X b.

For this example, XX ~ U(0, 23)U(0,23) and fX = 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?

Solution

Find P(2<X<18) P(2 X 18).

P(2<X<18)=(base)(height)=(18-2)⋅123=1623 P(2 X 18) (base)(height) (18-2)⋅123 1623.

f(X) graph displaying a boxed region consisting of a horizontal line extending to the right from midway on the y-axis, a vertical upward line from point 23 on the x-axis, and the x and y-axes. A shaded region from points 2-18 occurs within this area.

Problem 2

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

Solution

Ninety percent of the smiling times fall below the 90th percentile, kk, so P(X<k)=0.90 P(X k) 0.90

P(X<k)=0.90 P(X k) 0.90

(base)(height)=0.90(base)(height)=0.90

(k-0)⋅123=0.90(k-0)⋅123=0.90

k=23⋅0.90=20.7 k 23⋅0.90 20.7

f(X)=1/23 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/23 on the y-axis, a vertical upward line from point 23 on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The shaded region probability area is equal to 0.90.

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.

Solution

Find P(X>12|X>8) P(X 12|X 8) There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than 8 seconds.

Write a new fXfX: fX = 1 23-8 =115 fX 1 23-8 115

for 8<X<23 8 X 23

P(X>12|X>8)=(23-12)⋅115=1115 P(X 12|X 8) (23-12)⋅115 1115

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 points 8 and 23 on the x-axis, and the x-axis. A shaded region from points 12-23 occurs within this area.

For the second way, use the conditional formula from Probability Topics with the original distribution XX ~ U(0,23)U(0,23):

P ( A | B ) = P ( A AND B ) P ( B ) P(A|B)= P ( A AND B ) P ( B ) For this problem, AA is ( X > 12 ) ( X 12 ) and B B is ( X > 8 ) ( X 8 ) .

So, P(X>12|X>8)= (X>12 AND X >8) P ( X > 8 ) = P ( X > 12 ) P ( X > 8 ) = 11 23 15 23 = 0.733 P(X 12|X 8) (X>12 AND X >8) P ( X 8 ) P ( X 12 ) P ( X 8 ) 11 23 15 23 0.733

f(X)=1/23 graph displaying a conditional boxed region consisting of a horizontal red line extending to the right from point 1/23 on the y-axis, a vertical red upward line from point 23 on the x-axis, and the x and y-axes. Two vertical upward lines from points 8 and 12 on the x-axis occur within this area.

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?

Solution

Let XX = the number of minutes a person must wait for a bus. aa = 0 and bb = 15. X~U(0,15)X~U(0,15). Write the probability density function. fX = 115-0 =115 fX=115-0=115 for 0≤X≤15 0 X 15.

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

P(X<k)=(base)(height) =(12.5-0) ⋅115 =0.8333 P(X k) (base)(height)=(12.5-0)⋅115=0.8333

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

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 point 15 on the x-axis, and the x and y-axes. A shaded region from points 0-12.5 occurs within this area.

Problem 2

On the average, how long must a person wait?

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

Solution

μ= a+b 2 = 15+0 2 = 7.5μ= a+b 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 = 4.3σ= (b-a)2 12= (15-0)2 12=4.3. The Standard deviation is 4.3 minutes.

Problem 3

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

Note:

This asks for the 90th percentile.

Solution

Find the 90th percentile. Draw a graph. Let kk = 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

kk is sometimes called a critical value.

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.

Example 4

Uniform: The average number of donuts a nine-year old child eats per month is uniformly distributed from 0.5 to 4 donuts, inclusive. Let XX = the average number of donuts a nine-year old child eats per month. Then XX ~ U(0.5, 4)U(0.5,4).

Problem 1

The probability that a randomly selected nine-year old child eats an average of more than two donuts is _______.

Solution

0.5714

Problem 2

Find the probability that a different nine-year old child eats an average of more than two donuts given that his or her amount is more than 1.5 donuts.

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

Write a new f(X):

fX =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)=?

Solution

4545

The probability that a nine-year old child eats an average of more than 2 donuts when he/she has already eaten more than 1.5 donuts 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

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 fX = 1b-a fX=1b-a for 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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