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# Continuous Random Variables: The Uniform Distribution

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

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## Example 1

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

 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 15.9 16.3 13.4 17.1 14.5 19 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 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. This means that any smiling time between 0 and 23 seconds is equally likely. You can see this for yourself by constructing the histogram from the data.

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

The notation for the uniform distribution is

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

For this example, XX ~ U(0, 23)U(0,23). 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 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.

#### Problem 2

Find the 90th percentile for an eight week old'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=230.90=20.7 k 230.90 20.7

#### Problem 3

Find the probability that a random eight week old 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

For the second way, use the conditional formula from Chapter 3 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

## 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.

### 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.

### 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= k115 0.90=k115

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.

## Example 4

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

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.

### Problem 2

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 __________.

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:
Continuous random variable (RV) that appears to have equally likely outcomes over the domain, a<x<ba<x<b size 12{a<x<b} {}. Often referred as Rectangular distribution because graph of its pdf has form of 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 variance is σ2=(ba)212σ2=(ba)212 size 12{s rSup { size 8{2} } = { { $$b-a$$ rSup { size 8{2} } } over {"12"} } } {}, the probability density function is f(x)=1ba,aXbf(x)=1ba,aXb size 12{f $$x$$ = { {1} over {b-a} } ," "a <= X <= b} {}, and cumulative distribution is P(Xx)=xabaP(Xx)=xaba size 12{P $$X <= x$$ = { {x-a} over {b-a} } } {}.

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