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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 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.
  1. P(L′) = P(S)
  2. P(M S)
  3. P(F L)
  4. P(M||L)
  5. P(L||M)
  6. P(S||F)
  7. P(F||L)
  8. P(F L)
  9. P(M S)
  10. P(F)
3.

P(N) = 15 42 15 42 = 5 14 5 14 = 0.36

5.

P(C) = 5 42 5 42 = 0.12

7.

P(G) = 20 150 20 150 = 2 15 2 15 = 0.13

9.

P(R) = 22 150 22 150 = 11 75 11 75 = 0.15

11.

P(O) = 150-22-38-20-28-26 150 150-22-38-20-28-26 150 = 16 150 16 150 = 8 75 8 75 = 0.11

13.

P(E) = 47 194 47 194 = 0.24

15.

P(N) = 23 194 23 194 = 0.12

17.

P(S) = 12 194 12 194 = 6 97 6 97 = 0.06

19.

13 52 13 52 = 1 4 1 4 = 0.25

21.

3 6 3 6 = 1 2 1 2 = 0.5

23.

P(R)= 4 8 =0.5 P(R)= 4 8 =0.5

25.

P(O H)

27.

P(H||I)

29.

P(N||O)

31.

P(I N)

33.

P(I)

35.

The likelihood that an event will occur given that another event has already occurred.

37.

1

39.

the probability of landing on an even number or a multiple of three

41.

P(J) = 0.3

43.

P ( Q R ) = P ( Q ) P ( R ) P(QR)=P(Q)P(R)

0.1 = (0.4)P(R)

P(R) = 0.25

45.

0.376

47.

C||L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.

49.

L C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.

51.

0.6492

53.

No, because P(L C) does not equal 0.

55.

P(musician is a male had private instruction) = 15 130 15 130 = 3 26 3 26 = 0.12

57.

P(being a female musician  learning music in school) = 38 130 38 130 = 19 65 19 65 = 0.29

P(being a female musician)P(learning music in school) = ( 72 130 )( 62 130 ) ( 72 130 )( 62 130 ) = 4,464 16,900 4,464 16,900 = 1,116 4,225 1,116 4,225 = 0.26

No, they are not independent because P(being a female musician learning music in school) is not equal to P(being a female musician)P(learning music in school).

58.
This is a tree diagram with two branches. The first branch, labeled Cancer, shows two lines: 0.4567 C and 0.5433 C'. The second branch is labeled False Positive. From C, there are two lines: 0 P and 1 P'. From C', there are two lines: 0.51 P and 0.49 P'.
Figure 3.21
60.

35,065100,45035,065100,450

62.

To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is 4,715100,4504,715100,450.

64.

To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is 471515,273471515,273.

66.
  1. This is a tree diagram with branches showing probabilities of each draw. The first branch shows two lines: 5/8 Green and 3/8 Yellow. The second branch has a set of two lines (5/8 Green and 3/8 Yellow) for each line of the first branch.
    Figure 3.22
  2. P(GG) = ( 5 8 )( 5 8 ) ( 5 8 )( 5 8 ) = 25 64 25 64
  3. P(at least one green) = P(GG) + P(GY) + P(YG) = 25 64 25 64 + 15 64 15 64 + 15 64 15 64 = 55 64 55 64
  4. P(G||G) = 5 8 5 8
  5. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.
68.
  1. <2020–64>64Totals
    Female 0.0244 0.3954 0.0661 0.486
    Male 0.0259 0.4186 0.0695 0.514
    Totals 0.0503 0.8140 0.1356 1
    Table 3.22
  2. P(F) = 0.486
  3. P(>64 || F) = 0.1361
  4. P(>64 and F) = P(F) P(>64|F) = (0.486)(0.1361) = 0.0661
  5. P(>64 || F) is the percentage of female drivers who are 65 or older and P(>64 F) is the percentage of drivers who are female and 65 or older.
  6. P(>64) = P(>64 F) + P(>64 M) = 0.1356
  7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 F) = 0.0661.
70.
  1. Car, truck or van Walk Public transportation Other Totals
    Alone 0.7318
    Not alone 0.1332
    Totals 0.8650 0.0390 0.0530 0.0430 1
    Table 3.23
  2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = 0.7318 + 0.0390 = 0.7708.
  3. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771
  4. (0.1332)(1,000) = 133
73.
  1. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%
  2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.
75.

0

77.

0.3571

79.

0.2142

81.

Physician (83.7)

83.

83.7 − 79.6 = 4.1

85.

P(Occupation < 81.3) = 0.5

87.
  1. The Forum Research surveyed 1,046 Torontonians.
  2. 58%
  3. 42% of 1,046 = 439 (rounding to the nearest integer)
  4. 0.57
  5. 0.60.
89.
  1. P(Betting on two line that touch each other on the table) = 6 38 6 38
  2. P(Betting on three numbers in a line) = 3 38 3 38
  3. P(Bettting on one number) = 1 38 1 38
  4. P(Betting on four number that touch each other to form a square) = 4 38 4 38
  5. P(Betting on two number that touch each other on the table ) = 2 38 2 38
  6. P(Betting on 0-00-1-2-3) = 5 38 5 38
  7. P(Betting on 0-1-2; or 0-00-2; or 00-2-3) = 3 38 3 38
91.
  1. {G1, G2, G3, G4, G5, Y1, Y2, Y3}
  2. 5 8 5 8
  3. 2 3 2 3
  4. 2 8 2 8
  5. 6 8 6 8
  6. No, because P(G E) does not equal 0.
93.

NOTE

The coin toss is independent of the card picked first.

  1. {(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)}
  2. P(A) = P(blue)P(head) = ( 3 10 ) ( 3 10 ) ( 1 2 ) ( 1 2 ) = 3 20 3 20
  3. Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P(A B) = 0
  4. No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A C) = P(A) = 3 20 3 20
95.
  1. S = {(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
  2. 4 8 4 8
  3. Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P(A B) = 0.
97.
  1. If Y and Z are independent, then P(Y Z) = P(Y)P(Z), so P(Y Z) = P(Y) + P(Z) - P(Y)P(Z).
  2. 0.5
99.

iii i iv ii

101.
  1. P(R) = 0.44
  2. P(R||E) = 0.56
  3. P(R||O) = 0.31
  4. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P(R||E) ≠ P(R).
  5. No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P(R||E) > P(R).
103.
  1. P(type O Rh-) = P(type O) + P(Rh-) - P(type O Rh-)

    0.52 = 0.43 + 0.15 - P(type O Rh-); solve to find P(type O Rh-) = 0.06

    6% of people have type O, Rh- blood

  2. P(NOT(type O Rh-)) = 1 - P(type O Rh-) = 1 - 0.06 = 0.94

    94% of people do not have type O, Rh- blood

105.
  1. Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts.
  2. P(C N) = P(C) + P(N) - P(C N) = 0.36 + 0.12 - 0.08 = 0.40
  3. P(NEITHER chocolate NOR nuts) = 1 - P(C N) = 1 - 0.40 = 0.60
107.

0

109.

10 67 10 67

111.

10 34 10 34

113.

d

115.
  1. Race and sex 1–14 15–24 25–64 Over 64 TOTALS
    White, male 210 3,360 13,610 4,870 22,050
    White, female 80 580 3,380 890 4,930
    Black, male 10 460 1,060 140 1,670
    Black, female 0 40 270 20 330
    All others 100
    TOTALS 310 4,650 18,780 6,020 29,760
    Table 3.24
  2. Race and sex 1–14 15–24 25–64 Over 64 TOTALS
    White, male 210 3,360 13,610 4,870 22,050
    White, female 80 580 3,380 890 4,930
    Black, male 10 460 1,060 140 1,670
    Black, female 0 40 270 20 330
    All others 10 210 460 100 780
    TOTALS 310 4,650 18,780 6,020 29,760
    Table 3.25
  3. 22,050 29,760 22,050 29,760
  4. 330 29,760 330 29,760
  5. 2,000 29,760 2,000 29,760
  6. 23,720 29,760 23,720 29,760
  7. 5,010 6,020 5,010 6,020
117.

b

119.
  1. 26 106 26 106
  2. 33 106 33 106
  3. 21 106 21 106
  4. ( 26 106 ) ( 26 106 ) + ( 33 106 ) ( 33 106 ) - ( 21 106 ) ( 21 106 ) = ( 38 106 ) ( 38 106 )
  5. 21 33 21 33
121.

a

124.
  1. P(C) = 0.4567
  2. not enough information
  3. not enough information
  4. No, because over half (0.51) of men have at least one false positive text
126.
  1. ( J K ) = P ( J ) + P ( K ) P ( J K ) ; 0.45 = 0.18 + 0.37 P ( J K ) ; solve to find P ( J K ) = 0.10 (JK)=P(J)+P(K)P(JK);0.45=0.18+0.37P(JK);solve to findP(JK)=0.10
  2. P ( NOT ( J K ) ) = 1 P ( J K ) = 1 0.10 = 0.90 P(NOT(JK))=1P(JK)=10.10=0.90
  3. P ( NOT ( J K ) ) = 1 P ( J K ) = 1 0.45 = 0.55 P(NOT(JK))=1P(JK)=10.45=0.55
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