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Introductory Statistics

6.1 The Standard Normal Distribution

Introductory Statistics6.1 The Standard Normal 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 standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:

x = μ + (z)(σ) = 5 + (3)(2) = 11

The z-score is three.

The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation z = xμ σ xμ σ produces the distribution Z ~ N(0, 1). The value x in the given equation comes from a normal distribution with mean μ and standard deviation σ.

Z-Scores

If X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is:

z= x  μ σ z= x  μ σ

The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero.

Example 6.1

Suppose X ~ N(5, 6). This says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then:

z= xμ σ = 175 6 =2 z= xμ σ = 175 6 =2

This means that x = 17 is two standard deviations (2σ) above or to the right of the mean μ = 5.

Notice that: 5 + (2)(6) = 17 (The pattern is μ + = x)

Now suppose x = 1. Then: z = xμ σ xμ σ = 15 6 15 6 = –0.67 (rounded to two decimal places)

This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1)

Summarizing, when z is positive, x is above or to the right of μ and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative x is less than μ.

Try It 6.1

What is the z-score of x, when x = 1 and X ~ N(12,3)?

Example 6.2

Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks.

a. Suppose a person lost ten pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

Solution 6.2

a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five.

b. Suppose a person gained three pounds (a negative weight loss). Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean.

Solution 6.2

b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean.

c. Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z?

Solution 6.2

c. z = yμ σ yμ σ = 42 1 42 1 = 2 where µ = 2 and σ = 1.

The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two (of their own) standard deviations to the right of their respective means.

The z-score allows us to compare data that are scaled differently. To understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.

Try It 6.2

Fill in the blanks.

Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16,4). Suppose Jerome scores ten points in a game. The z–score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).

The Empirical RuleIf X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule states the following:

  • About 68% of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean).
  • About 95% of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean).
  • About 99.7% of the x values lie between –3σ and +3σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean.
  • The z-scores for +1σ and –1σ are +1 and –1, respectively.
  • The z-scores for +2σ and –2σ are +2 and –2, respectively.
  • The z-scores for +3σ and –3σ are +3 and –3 respectively.

The empirical rule is also known as the 68-95-99.7 rule.

This frequency curve illustrates the empirical rule. The normal curve is shown over a horizontal axis. The axis is labeled with points -3s, -2s, -1s, m, 1s, 2s, 3s.  Vertical lines connect the axis to the curve at each labeled point. The peak of the curve aligns with the point m.
Figure 6.3

Example 6.3

The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).

a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

Solution 6.3

a. –0.32, 0.32, left, 170

b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

Solution 6.3

b. 177.98 cm, 1.27, right

Try It 6.3

Use the information in Example 6.3 to answer the following questions.

  1. Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
  2. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = –2. What is the male’s height? The z-score (z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

Example 6.4

From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males from 1984 to 1985. Then Y ~ N(172.36, 6.34).

The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).

Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations?

Solution 6.4

The z-score for x = -160.58 is z = –1.5.
The z-score for y = 162.85 is z = –1.5.
Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.

Try It 6.4

In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then X ~ N(496, 114).

Find the z-scores for x1 = 325 and x2 = 366.21. Interpret each z-score. What can you say about x1 = 325 and x2 = 366.21 as they compare to their respective means and standard deviations?

Example 6.5

Suppose x has a normal distribution with mean 50 and standard deviation 6.

  • About 68% of the x values lie within one standard deviation of the mean. Therefore, about 68% of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively.
  • About 95% of the x values lie within two standard deviations of the mean. Therefore, about 95% of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively.
  • About 99.7% of the x values lie within three standard deviations of the mean. Therefore, about 95% of the x values lie between –3σ = (–3)(6) = –18 and 3σ = (3)(6) = 18 from the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively.
Try It 6.5

Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68% of the values lie?

Example 6.6

From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males in 1984 to 1985. Then Y ~ N(172.36, 6.34).

  1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
  2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively.
  3. About 99.7% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
Solution 6.6
  1. About 68% of the values lie between 166.02 cm and 178.7 cm. The z-scores are –1 and 1.
  2. About 95% of the values lie between 159.68 cm and 185.04 cm. The z-scores are –2 and 2.
  3. About 99.7% of the values lie between 153.34 cm and 191.38 cm. The z-scores are –3 and 3.
Try It 6.6

The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points.

  1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
  2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
  3. About 99.7% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
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