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Table of contents
  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 Levels of Measurement
    5. 1.4 Experimental Design and Ethics
    6. Key Terms
    7. Chapter Review
    8. Homework
    9. References
    10. Solutions
  3. 2 Descriptive Statistics
    1. Introduction
    2. 2.1 Display Data
    3. 2.2 Measures of the Location of the Data
    4. 2.3 Measures of the Center of the Data
    5. 2.4 Sigma Notation and Calculating the Arithmetic Mean
    6. 2.5 Geometric Mean
    7. 2.6 Skewness and the Mean, Median, and Mode
    8. 2.7 Measures of the Spread of the Data
    9. Key Terms
    10. Chapter Review
    11. Formula Review
    12. Practice
    13. Homework
    14. Bringing It Together: Homework
    15. References
    16. 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 and Probability Trees
    6. 3.5 Venn Diagrams
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Bringing It Together: Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  5. 4 Discrete Random Variables
    1. Introduction
    2. 4.1 Hypergeometric Distribution
    3. 4.2 Binomial Distribution
    4. 4.3 Geometric Distribution
    5. 4.4 Poisson Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  6. 5 Continuous Random Variables
    1. Introduction
    2. 5.1 Properties of Continuous Probability Density Functions
    3. 5.2 The Uniform Distribution
    4. 5.3 The Exponential Distribution
    5. Key Terms
    6. Chapter Review
    7. Formula Review
    8. Practice
    9. Homework
    10. References
    11. 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 Estimating the Binomial with the Normal Distribution
    5. Key Terms
    6. Chapter Review
    7. Formula Review
    8. Practice
    9. Homework
    10. References
    11. Solutions
  8. 7 The Central Limit Theorem
    1. Introduction
    2. 7.1 The Central Limit Theorem for Sample Means
    3. 7.2 Using the Central Limit Theorem
    4. 7.3 The Central Limit Theorem for Proportions
    5. 7.4 Finite Population Correction Factor
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  9. 8 Confidence Intervals
    1. Introduction
    2. 8.1 A Confidence Interval for a Population Standard Deviation, Known or Large Sample Size
    3. 8.2 A Confidence Interval for a Population Standard Deviation Unknown, Small Sample Case
    4. 8.3 A Confidence Interval for A Population Proportion
    5. 8.4 Calculating the Sample Size n: Continuous and Binary Random Variables
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. 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 Full Hypothesis Test Examples
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  11. 10 Hypothesis Testing with Two Samples
    1. Introduction
    2. 10.1 Comparing Two Independent Population Means
    3. 10.2 Cohen's Standards for Small, Medium, and Large Effect Sizes
    4. 10.3 Test for Differences in Means: Assuming Equal Population Variances
    5. 10.4 Comparing Two Independent Population Proportions
    6. 10.5 Two Population Means with Known Standard Deviations
    7. 10.6 Matched or Paired Samples
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  12. 11 The Chi-Square Distribution
    1. Introduction
    2. 11.1 Facts About the Chi-Square Distribution
    3. 11.2 Test of a Single Variance
    4. 11.3 Goodness-of-Fit Test
    5. 11.4 Test of Independence
    6. 11.5 Test for Homogeneity
    7. 11.6 Comparison of the Chi-Square Tests
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  13. 12 F Distribution and One-Way ANOVA
    1. Introduction
    2. 12.1 Test of Two Variances
    3. 12.2 One-Way ANOVA
    4. 12.3 The F Distribution and the F-Ratio
    5. 12.4 Facts About the F Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  14. 13 Linear Regression and Correlation
    1. Introduction
    2. 13.1 The Correlation Coefficient r
    3. 13.2 Testing the Significance of the Correlation Coefficient
    4. 13.3 Linear Equations
    5. 13.4 The Regression Equation
    6. 13.5 Interpretation of Regression Coefficients: Elasticity and Logarithmic Transformation
    7. 13.6 Predicting with a Regression Equation
    8. 13.7 How to Use Microsoft Excel® for Regression Analysis
    9. Key Terms
    10. Chapter Review
    11. Practice
    12. Solutions
  15. A | Statistical Tables
  16. B | Mathematical Phrases, Symbols, and Formulas
  17. 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

24.

0.6

26.

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

28.

six

30.
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 5.38
33.

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

35.

0.5 to 9.5

37.

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

39.

μ = 5

41.
  1. Check student’s solution.
  2. 3.5 7 3.5 7
43.
  1. Check student's solution.
  2. k = 7.25
  3. 7.25
45.

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.

47.

five

49.

f(x) = 0.2e-0.2x

51.

0.5350

53.

6.02

55.

f(x) = 0.75e-0.75x

57.
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 5.39
59.

0.4756

61.

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.

63.

continuous

65.

m = 0.000121

67.
  1. Check student's solution
  2. P(x < 5,730) = 0.5001
69.
  1. Check student's solution.
  2. k = 2947.73
71.

Age is a measurement, regardless of the accuracy used.

73.
  1. Check student’s solution.
  2. f(x)= 1 8 f(x)= 1 8 where 1≤x≤9 1≤x≤9
  3. five
  4. 2.3
  5. 15 32 15 32
  6. 333 800 333 800
  7. 2 3 2 3
75.
  1. X represents the length of time a commuter must wait for a train to arrive on the Red Line.
  2. Graph the probability distribution.
  3. f(x)= 1 8 f(x)= 1 8 where 0 ≤ x ≤ 8
  4. four
  5. 2.31
  6. 1 8 1 8
  7. 1 8 1 8
77.

d

78.

b

80.
  1. The probability density function of X is 1 25−16 = 1 9 1 25−16 = 1 9 .
    P(X > 19) = (25 – 19) ( 1 9 ) ( 1 9 ) = 6 9 6 9 = 2 3 2 3 .
    Figure 5.40
  2. P(19 < X < 22) = (22 – 19) ( 1 9 ) ( 1 9 ) = 3 9 3 9 = 1 3 1 3 .
    Figure 5.41
  3. 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 (25−18) 1 (25−18) = 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∩x>18) P(x>18) P(x>21∩x>18) P(x>18)
      = P(x>21) P(x>18) P(x>21) P(x>18) = (25−21) (25−18) (25−21) (25−18) = 4 7 4 7 .
82.
  1. P(X > 650) = 700−650 700−300 = 50 400 = 1 8 700−650 700−300 = 50 400 = 1 8 = 0.125
  2. P(400 < X < 650) = 650−400 700−300 = 250 400 650−400 700−300 = 250 400 = 0.625
84.
  1. X = the useful life of a particular car battery, measured in months.
  2. X is continuous.
  3. 40 months
  4. 360 months
  5. 0.4066
  6. 14.27
86.
  1. X = the time (in years) after reaching age 60 that it takes an individual to retire
  2. X is continuous.
  3. five
  4. five
  5. Check student’s solution.
  6. 0.1353
  7. before
  8. 18.3
88.

a

90.

c

92.

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 T ∼ Exp(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.
94.
  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, 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.)

96.

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) = 1−P( T<15 )=1−( 1− e − 15 7 )≈ e − 15 7 ≈0.1173 1−P( 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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