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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
1.

Uniform Distribution

3.

Normal Distribution

5.

P(6 < x < 7)

7.

one

9.

zero

11.

one

13.

0.625

15.

The probability is equal to the area from x = 3 2 3 2 to x = 4 above the x-axis and up to f(x) = 1 3 1 3 .

17.

It means that the value of x is just as likely to be any number between 1.5 and 4.5.

19.

1.5 ≤ x ≤ 4.5

21.

0.3333

23.

zero

25.

0.6

27.

b is 12, and it represents the highest value of x.

29.

six

31.
This graph shows a uniform distribution. The horizontal axis ranges from 0 to 12. The distribution is modeled by a rectangle extending from x = 0 to x = 12. A region from x = 9 to x = 12 is shaded inside the rectangle.
Figure 5.52
33.

4.8

35.

X = The age (in years) of cars in the staff parking lot

37.

0.5 to 9.5

39.

f(x) = 1 9 1 9 where x is between 0.5 and 9.5, inclusive.

41.

μ = 5

43.
  1. Check student’s solution.
  2. 3.5 7 3.5 7
45.
  1. Check student's solution.
  2. k = 7.25
  3. 7.25
47.

No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.

49.

five

51.

f(x) = 0.2e-0.2x

53.

0.5350

55.

6.02

57.

f(x) = 0.75e-0.75x

59.
This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.75) on the y-axis and approaches the x-axis at the right edge of the graph. The decay parameter, m, equals 0.75.
Figure 5.53
61.

0.4756

63.

The mean is larger. The mean is 1 m = 1 0.75 1.33 1 m = 1 0.75 1.33 , which is greater than 0.9242.

65.

continuous

67.

m = 0.000121

69.
  1. Check student's solution
  2. P(x < 5,730) = 0.5001
71.
  1. Check student's solution.
  2. k = 2947.73
73.

Age is a measurement, regardless of the accuracy used.

75.
  1. X ~ U(1, 9)
  2. Check student’s solution.
  3. f(x)= 1 8 f(x)= 1 8 where 1x9 1x9
  4. five
  5. 2.3
  6. 15 32 15 32
  7. 333 800 333 800
  8. 2 3 2 3
  9. 8.2
77.
  1. X represents the length of time a commuter must wait for a train to arrive on the Red Line.
  2. X ~ U(0, 8)
  3. Graph the probability distribution.
  4. f ( x ) = 1 8 f ( x ) = 1 8 where 0 x 8 0 x 8
  5. four
  6. 2.31
  7. 1 8 1 8
  8. 1 8 1 8
  9. 3.2
79.

d

81.

b

83.
  1. The probability density function of X is 1 2516 = 1 9 1 2516 = 1 9 .
    P(X > 19) = (25 – 19) ( 1 9 ) ( 1 9 ) = 6 9 6 9 = 2 3 2 3 .
    Figure 5.54
  2. P(19 < X < 22) = (22 – 19) ( 1 9 ) ( 1 9 ) = 3 9 3 9 = 1 3 1 3 .
    Figure 5.55
  3. The area must be 0.25, and 0.25 = (width) ( 1 9 ) ( 1 9 ) , so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75.
  4. This is a conditional probability question. P(x > 21| x > 18). You can do this two ways:
    • Draw the graph where a is now 18 and b is still 25. The height is 1 (2518) 1 (2518) = 1 7 1 7
      So, P(x > 21|x > 18) = (25 – 21) ( 1 7 ) ( 1 7 ) = 4/7.
    • Use the formula: P(x > 21|x > 18) = P(x>21 AND x>18) P(x>18) P(x>21 AND x>18) P(x>18)
      = P(x>21) P(x>18) P(x>21) P(x>18) = (2521) (2518) (2521) (2518) = 4 7 4 7 .
85.
  1. P(X > 650) = 700650 700300 = 50 400 = 1 8 700650 700300 = 50 400 = 1 8 = 0.125.
  2. P(400 < X < 650) = 650400 700300 = 250 400 650400 700300 = 250 400 = 0.625
  3. 0.10 = width 700300 width 700300 , so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days.
87.
  1. X = the useful life of a particular car battery, measured in months.
  2. X is continuous.
  3. X ~ Exp(0.025)
  4. 40 months
  5. 360 months
  6. 0.4066
  7. 14.27
89.
  1. X = the time (in years) after reaching age 60 that it takes an individual to retire
  2. X is continuous.
  3. X ~ Exp ( 1 5 ) ( 1 5 )
  4. five
  5. five
  6. Check student’s solution.
  7. 0.1353
  8. before
  9. 18.3
91.

a

93.

c

95.

Let T = the life time of a light bulb.

The decay parameter is m = 1/8, and T ∼ Exp(1/8). The cumulative distribution function is P(T<t)=1 e t 8 P(T<t)=1 e t 8

  1. Therefore, P(T < 1) = 1 – e 1 8 1 8 ≈ 0.1175.
  2. We want to find P(6 < t < 10).
    To do this, P(6 < t < 10) – P(t < 6)
    = =( 1 e 1 8 *10 )( 1 e 1 8 *6 ) =( 1 e 1 8 *10 )( 1 e 1 8 *6 ) ≈ 0.7135 – 0.5276 = 0.1859
    Figure 5.56
  3. We want to find 0.70 =P(T>t)=1( 1 e t 8 )= e t 8 . =P(T>t)=1( 1 e t 8 )= e t 8 .
    Solving for t, e t 8 t 8 = 0.70, so t 8 t 8 = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years.
    Or use t = ln(area_to_the_right) (m) = ln(0.70) 1 8 2.85 years ln(area_to_the_right) (m) = ln(0.70) 1 8 2.85 years .
    Figure 5.57
  4. We want to find 0.02 = P(T < t) = 1 – e t 8 t 8 .
    Solving for t, e t 8 t 8 = 0.98, so t 8 t 8 = ln(0.98), and t = –8ln(0.98) ≈ 0.1616 years, or roughly two months.
    The warranty should cover light bulbs that last less than 2 months.
    Or use ln(area_to_the_right) (m) = ln(10.2) 1 8 ln(area_to_the_right) (m) = ln(10.2) 1 8 = 0.1616.
  5. We must find P(T < 8|T > 7).
    Notice that by the rule of complement events, P(T < 8|T > 7) = 1 – P(T > 8|T > 7).
    By the memoryless property (P(X > r + t|X > r) = P(X > t)).
    So P(T > 8|T > 7) = P(T > 1) = 1( 1 e 1 8 )= e 1 8 0.8825 1( 1 e 1 8 )= e 1 8 0.8825
    Therefore, P(T < 8|T > 7) = 1 – 0.8825 = 0.1175.
97.

Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3.
Therefore, (X = 0) = 3 0 e 3 0! 3 0 e 3 0! = e–3 ≈ 0.0498

NOTE

You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is 1 3 1 3 season. For the exponential, µ = 1 3 1 3 .
Therefore, m = 1 μ 1 μ = 3 and TExp(3).

  1. The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – e–3) = e–3 ≈ 0.0498.
  2. Let T = duration of time between no-hitters. We find P(T > 2|T > 1), and by the memoryless property this is simply P(T > 1), which we found to be 0.0498 in part a.
  3. Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P(X > 3) = 1 – P(X ≤ 3) = 0.3528.
99.
  1. 100 9 100 9 = 11.11
  2. P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532.
  3. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = 1 9 1 9 . The cumulative distribution function of X is P( X<x )=1 e x 9 P( X<x )=1 e x 9 . Thus hus, P(X > 20) = 1 - P(X ≤ 20) = 1( 1 e 20 9 )0.1084. 1( 1 e 20 9 )0.1084.

Note

We could also deduce that each person arriving has a 8/9 chance of not having Type B blood. So the probability that none of the first 20 people arrive have Type B blood is ( 8 9 ) 20 0.0948 ( 8 9 ) 20 0.0948 . (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)

101.

Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7 1 7 . The cdf is P(T < t) = 1 e t 7 1 e t 7

  1. P(T < 2) = 1 - 1 e 2 7 1 e 2 7 ≈ 0.2485.
  2. P(T > 15) = 1P( T<15 )=1( 1 e 15 7 ) e 15 7 0.1173 1P( T<15 )=1( 1 e 15 7 ) e 15 7 0.1173 .
  3. P(T > 15|T > 10) = P(T > 5) = 1( 1 e 5 7 )= e 5 7 0.4895 1( 1 e 5 7 )= e 5 7 0.4895 .
  4. Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7 30 7 , X ∼ Poisson ( 30 7 ) ( 30 7 ) . Find P(X > 8) = 1 – P(X ≤ 8) ≈ 0.0311.
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