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Inside Collection (Textbook):

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

The Uniform Distribution

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.

 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, 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 axb a x b.

For this example, xx ~ U(0, 23)U(0,23) and fx = 123-0 fx=123-0 for 0x23 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.

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=230.90=20.7 k 230.90 20.7

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

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

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 0x15 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: 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 _______.

0.5714

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):

fx =14-1.5 =25fx=14-1.5=25 for 1.5x4 1.5 x 4.

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

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 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.

Solution

To find fxfx: fx = 1 4-1.5 =12.5 fx 1 4-1.5 12.5 so fxfx = 0.4 = 0.4

P(x>2) = (base)(height) = (4 − 2)(0.4) = 0.8

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.

Solution

P(x<3) P(x 3) = (base)(height) = (3 − 1.5)(0.4) = 0.6

The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x=1.5 and x=3. Note that the shaded area starts at x=1.5 rather than at x=0; since X~U(1.5,4), x can not be less than 1.5.

Problem 3

Find the 30th percentile of furnace repair times.

Solution

P(x<k)=0.30 P(x k) 0.30

P(x<k)=(base)(height) =(k-1.5) (0.4) P(x k) (base)(height)=(k-1.5)(0.4)

• 0.3 = (k − 1.5) (0.4) ; Solve to find k:
• 0.75 = k − 1.5 , obtained by dividing both sides by 0.4
• k = 2.25 , obtained by adding 1.5 to both sides

The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.

Problem 4

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

Solution

P(x>k)=0.25 P(x k) 0.25

P(x>k)=(base)(height) =(4-k) (0.4) P(x k) (base)(height)=(4-k)(0.4)

• 0.25 = (4 − k)(0.4) ; Solve for k:
• 0.625 = 4 − k , obtained by dividing both sides by 0.4
• −3.375 = −k , obtained by subtracting 4 from both sides
• k=3.375

The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).

Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times.

Problem 5

Find the mean and standard deviation

Solution

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

μ=1.5+4 2=2.75 μ 1.5+4 2 2.75 hours and σ= (4-1.5)2 12 =0.7217 σ (4-1.5)2 12 0.7217 hours

**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 fx = 1b-a fx=1b-a for a<x<b a x b or axb a x b. The cumulative distribution is P(Xx)=xabaP(Xx)=xaba size 12{P $$X <= x$$ = { {x-a} over {b-a} } } {}.

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