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Introductory Statistics

5.2 The Uniform Distribution

Introductory Statistics5.2 The Uniform Distribution
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  1. Preface
  2. 1 Sampling and Data
    1. Introduction
    2. 1.1 Definitions of Statistics, Probability, and Key Terms
    3. 1.2 Data, Sampling, and Variation in Data and Sampling
    4. 1.3 Frequency, Frequency Tables, and Levels of Measurement
    5. 1.4 Experimental Design and Ethics
    6. 1.5 Data Collection Experiment
    7. 1.6 Sampling Experiment
    8. Key Terms
    9. Chapter Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  3. 2 Descriptive Statistics
    1. Introduction
    2. 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
    3. 2.2 Histograms, Frequency Polygons, and Time Series Graphs
    4. 2.3 Measures of the Location of the Data
    5. 2.4 Box Plots
    6. 2.5 Measures of the Center of the Data
    7. 2.6 Skewness and the Mean, Median, and Mode
    8. 2.7 Measures of the Spread of the Data
    9. 2.8 Descriptive Statistics
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. Solutions
  4. 3 Probability Topics
    1. Introduction
    2. 3.1 Terminology
    3. 3.2 Independent and Mutually Exclusive Events
    4. 3.3 Two Basic Rules of Probability
    5. 3.4 Contingency Tables
    6. 3.5 Tree and Venn Diagrams
    7. 3.6 Probability Topics
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Bringing It Together: Practice
    13. Homework
    14. Bringing It Together: Homework
    15. References
    16. Solutions
  5. 4 Discrete Random Variables
    1. Introduction
    2. 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
    3. 4.2 Mean or Expected Value and Standard Deviation
    4. 4.3 Binomial Distribution
    5. 4.4 Geometric Distribution
    6. 4.5 Hypergeometric Distribution
    7. 4.6 Poisson Distribution
    8. 4.7 Discrete Distribution (Playing Card Experiment)
    9. 4.8 Discrete Distribution (Lucky Dice Experiment)
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. References
    16. Solutions
  6. 5 Continuous Random Variables
    1. Introduction
    2. 5.1 Continuous Probability Functions
    3. 5.2 The Uniform Distribution
    4. 5.3 The Exponential Distribution
    5. 5.4 Continuous Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  7. 6 The Normal Distribution
    1. Introduction
    2. 6.1 The Standard Normal Distribution
    3. 6.2 Using the Normal Distribution
    4. 6.3 Normal Distribution (Lap Times)
    5. 6.4 Normal Distribution (Pinkie Length)
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  8. 7 The Central Limit Theorem
    1. Introduction
    2. 7.1 The Central Limit Theorem for Sample Means (Averages)
    3. 7.2 The Central Limit Theorem for Sums
    4. 7.3 Using the Central Limit Theorem
    5. 7.4 Central Limit Theorem (Pocket Change)
    6. 7.5 Central Limit Theorem (Cookie Recipes)
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  9. 8 Confidence Intervals
    1. Introduction
    2. 8.1 A Single Population Mean using the Normal Distribution
    3. 8.2 A Single Population Mean using the Student t Distribution
    4. 8.3 A Population Proportion
    5. 8.4 Confidence Interval (Home Costs)
    6. 8.5 Confidence Interval (Place of Birth)
    7. 8.6 Confidence Interval (Women's Heights)
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. Solutions
  10. 9 Hypothesis Testing with One Sample
    1. Introduction
    2. 9.1 Null and Alternative Hypotheses
    3. 9.2 Outcomes and the Type I and Type II Errors
    4. 9.3 Distribution Needed for Hypothesis Testing
    5. 9.4 Rare Events, the Sample, Decision and Conclusion
    6. 9.5 Additional Information and Full Hypothesis Test Examples
    7. 9.6 Hypothesis Testing of a Single Mean and Single Proportion
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. Solutions
  11. 10 Hypothesis Testing with Two Samples
    1. Introduction
    2. 10.1 Two Population Means with Unknown Standard Deviations
    3. 10.2 Two Population Means with Known Standard Deviations
    4. 10.3 Comparing Two Independent Population Proportions
    5. 10.4 Matched or Paired Samples
    6. 10.5 Hypothesis Testing for Two Means and Two Proportions
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  12. 11 The Chi-Square Distribution
    1. Introduction
    2. 11.1 Facts About the Chi-Square Distribution
    3. 11.2 Goodness-of-Fit Test
    4. 11.3 Test of Independence
    5. 11.4 Test for Homogeneity
    6. 11.5 Comparison of the Chi-Square Tests
    7. 11.6 Test of a Single Variance
    8. 11.7 Lab 1: Chi-Square Goodness-of-Fit
    9. 11.8 Lab 2: Chi-Square Test of Independence
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. Solutions
  13. 12 Linear Regression and Correlation
    1. Introduction
    2. 12.1 Linear Equations
    3. 12.2 Scatter Plots
    4. 12.3 The Regression Equation
    5. 12.4 Testing the Significance of the Correlation Coefficient
    6. 12.5 Prediction
    7. 12.6 Outliers
    8. 12.7 Regression (Distance from School)
    9. 12.8 Regression (Textbook Cost)
    10. 12.9 Regression (Fuel Efficiency)
    11. Key Terms
    12. Chapter Review
    13. Formula Review
    14. Practice
    15. Homework
    16. Bringing It Together: Homework
    17. References
    18. Solutions
  14. 13 F Distribution and One-Way ANOVA
    1. Introduction
    2. 13.1 One-Way ANOVA
    3. 13.2 The F Distribution and the F-Ratio
    4. 13.3 Facts About the F Distribution
    5. 13.4 Test of Two Variances
    6. 13.5 Lab: One-Way ANOVA
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  15. A | Review Exercises (Ch 3-13)
  16. B | Practice Tests (1-4) and Final Exams
  17. C | Data Sets
  18. D | Group and Partner Projects
  19. E | Solution Sheets
  20. F | Mathematical Phrases, Symbols, and Formulas
  21. G | Notes for the TI-83, 83+, 84, 84+ Calculators
  22. H | Tables
  23. Index

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.

Example 5.2

The data in Table 5.1 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 20.0 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
Table 5.1

The sample mean = 11.49 and the sample standard deviation = 6.23.

We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero 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 X = length, in seconds, of an eight-week-old baby's smile.

The notation for the uniform distribution is

X ~ U(a, b) where a = the lowest value of x and b = the highest value of x.

The probability density function is f(x) = 1 ba 1 ba for axb.

For this example, X ~ U(0, 23) and f(x) = 1 230 1 230 for 0 ≤ X ≤ 23.

Formulas for the theoretical mean and standard deviation are

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

For this problem, the theoretical mean and standard deviation are

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

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

Try It 5.2

The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and b. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.

1 12 4 10 4 14 11
7 11 4 13 2 4 6
3 10 0 12 6 9 10
5 13 4 10 14 12 11
6 10 11 0 11 13 2
Table 5.2

Example 5.3

a. Refer to Example 5.2. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?

Solution 5.3

P(2 < x < 18) = (base)(height) = (18 – 2) ( 1 23 ) ( 1 23 ) = 16 23 16 23 .

This graph shows a uniform distribution. The horizontal axis ranges from 0 to 15. The distribution is modeled by a rectangle extending from x = 0 to x = 15. A region from x = 2 to x = 18 is shaded inside the rectangle.
Figure 5.11

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

Solution 5.3

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

P(x<k)=0.90 P(x<k)=0.90

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

(k0)( 1 23 )=0.90 (k0)( 1 23 )=0.90

k=( 23 )( 0.90 )=20.7 k=( 23 )( 0.90 )=20.7

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 = k. The shaded area represents P(x < k) = 0.90.
Figure 5.12

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

Solution 5.3

c. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds.

Find 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 eight seconds.

Write a new f(x): f(x) = 1 23  8 1 23  8 = 1 15 1 15 for 8 < x < 23

P(x > 12|x > 8) = (23 − 12) ( 1 15 ) ( 1 15 ) = 11 15 11 15

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.
Figure 5.13

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

P(A|B) = P(A AND B) P(B) P(A AND B) P(B)

For this problem, A is (x > 12) and B is (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 = 11 15 (x>12 AND x>8) P(x>8) = P(x>12) P(x>8) = 11 23 15 23 = 11 15

This diagram shows a horizontal X axis that intersects a vertical F of x axis at the origin. The X axis runs from 0 to 24 while the Y axis only has the fraction one twenty third located about two thirds of the way to the top. A rectangular box extends horizontally from 0 to about 23.7 on the X axis. The box extends vertically up to the fraction one twenty third on the F of x axis. The area of the box between 8 and 12 on the X axis is shaded.
Figure 5.14
Try It 5.3

A distribution is given as X ~ U (0, 20). What is P(2 < x < 18)? Find the 90th percentile.

Example 5.4

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

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

Solution 5.4

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) = 1 15  0 1 15  0 = 1 15 1 15 for 0 ≤ x ≤ 15.

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

P(x<k)=(base)(height)=(12.5-0)( 1 15 )=0.8333 P(x<k)=(base)(height)=(12.5-0)( 1 15 )=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 5.15

b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ.

Solution 5.4

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

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

This asks for the 90th percentile.
Solution 5.4

c. Find the 90th percentile. Draw a graph. Let k = the 90th percentile.

P(x<k)=(base)(height)=(k0)( 1 15 ) P(x<k)=(base)(height)=(k0)( 1 15 )

0.90=( k )( 1 15 ) 0.90=( k )( 1 15 )

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

k 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.
Figure 5.16
Try It 5.4

The total duration of baseball games in the major league in the 2011 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 for the 2011 season is between 480 and 500 hours?
  5. What is the 65th percentile for the duration of games for a team for the 2011 season?

Example 5.5

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

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

Solution 5.5

a. 0.5714

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

The second question has a conditional probability. You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example 5.3). You must reduce the sample space. First way: Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes.

Write a new f(x):

f(x) = 1 41.5 1 41.5 = 2 5 2 5 for 1.5 ≤ x ≤ 4.

Find 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.
Figure 5.17

P(x > 2|x > 1.5) = (base)(new height) = (4 − 2) ( 2 5 ) =45 ( 2 5 ) =45

Solution 5.5

b. 4545

The probability that a nine-year old child eats a donut in more than two 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 X ~ U (0.5, 4). Use the conditional formula

P(x > 2|x > 1.5) =   P(x>2 AND x>1.5) P(x>1.5) = P(x>2) P(x>1.5) = 2 3.5 2.5 3.5 =0.8= 4 5   P(x>2 AND x>1.5) P(x>1.5) = P(x>2) P(x>1.5) = 2 3.5 2.5 3.5 =0.8= 4 5

Try It 5.5

Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let X = the time, in minutes, it takes a student to finish a quiz. Then X ~ U (6, 15).

Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes.

Example 5.6

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 four hours. Let x = the time needed to fix a furnace. Then x ~ U (1.5, 4).

  1. Find the probability that a randomly selected furnace repair requires more than two hours.
  2. Find the probability that a randomly selected furnace repair requires less than three hours.
  3. Find the 30th percentile of furnace repair times.
  4. The longest 25% of furnace repair times 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
Solution 5.6

a. To find f(x): f (x) = 1 4  1.5 1 4  1.5 = 1 2.5 1 2.5 so f(x) = 0.4

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

This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 2 to x = 4.
Figure 5.18 Uniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time x is greater than two
Solution 5.6

b. 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.

This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = 3.
Figure 5.19 Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time x is less than three
Solution 5.6

c.

This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = k. The shaded area represents P(x < k) = 0.3.
Figure 5.20 Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.


P (x < k) = 0.30
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.

Solution 5.6

d.

Figure 5.21 Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times.


P(x > k) = 0.25
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 four 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.

Solution 5.6

e. μ= a+b 2 μ= a+b 2 and σ= (ba) 2 12 σ= (ba) 2 12
μ=1.5+4 2=2.75 μ 1.5+4 2 2.75 hours and σ= (41.5) 2 12 =0.7217 σ= (41.5) 2 12 =0.7217 hours

Try It 5.6

The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let X = the time needed to change the oil on a car.

  1. Write the random variable X in words. X = __________________.
  2. Write the distribution.
  3. Graph the distribution.
  4. Find P (x > 19).
  5. Find the 50th percentile.
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