1. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 1 Chapter Descriptive Statistics 2

2. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 2 Chapter Outline 2.1 Frequency Distributions and Their Graphs 2.2 More Graphs and Displays 2.3 Measures of Central Tendency 2.4 Measures of Variation 2.5 Measures of Position

3. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 3 Section 2.4 Measures of Variation.

4. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 4 Section 2.4 Objectives How to find the range of a data set How to find the variance and standard deviation of a population and of a sample How to use the Empirical Rule and Chebychev’s Theorem to interpret standard deviation How to approximate the sample standard deviation for grouped data How to use the coefficient of variation to compare variation in different data sets.

5. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 5 Range The difference between the maximum and minimum data entries in the set. The data must be quantitative. Range = (Max. data entry) – (Min. data entry).

6. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 6 Example: Finding the Range A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the range of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42.

7. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 7 Solution: Finding the Range Ordering the data helps to find the least and greatest salaries. 37 38 39 41 41 41 42 44 45 47 Range = (Max. salary) – (Min. salary) = 47 – 37 = 10 The range of starting salaries is 10 or $10,000. minimum maximum.

8. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 8 Deviation, Variance, and Standard Deviation Deviation The difference between the data entry, x, and the mean of the data set. Population data set:  Deviation of x = x – μ Sample data set:  Deviation of x = x – x.

9. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 9 Example: Finding the Deviation A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the deviation of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42 Solution: First determine the mean starting salary..

10. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 10 Solution: Finding the Deviation Determine the deviation for each data entry. Salary ($1000s), xDeviation: x – μ 4141 – 41.5 = –0.5 3838 – 41.5 = –3.5 3939 – 41.5 = –2.5 4545 – 41.5 = 3.5 4747 – 41.5 = 5.5 4141 – 41.5 = –0.5 4444 – 41.5 = 2.5 4141 – 41.5 = –0.5 3737 – 41.5 = –4.5 4242 – 41.5 = 0.5 Σx = 415 Σ(x – μ) = 0.

11. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 11 Deviation, Variance, and Standard Deviation Population Variance Population Standard Deviation Sum of squares, SS x.

12. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 12 Finding the Population Variance & Standard Deviation In Words In Symbols 1.Find the mean of the population data set. 2.Find deviation of each entry. 3.Square each deviation. 4.Add to get the sum of squares. x – μ (x – μ) 2 SS x = Σ(x – μ) 2.

13. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 13 Finding the Population Variance & Standard Deviation 5.Divide by N to get the population variance. 6.Find the square root to get the population standard deviation. In Words In Symbols.

14. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 14 Example: Finding the Population Standard Deviation A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the population variance and standard deviation of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42 Recall μ = 41.5..

15. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 15 Solution: Finding the Population Standard Deviation Determine SS x N = 10 Salary, xDeviation: x – μSquares: (x – μ) 2 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3838 – 41.5 = –3.5(–3.5) 2 = 12.25 3939 – 41.5 = –2.5(–2.5) 2 = 6.25 4545 – 41.5 = 3.5(3.5) 2 = 12.25 4747 – 41.5 = 5.5(5.5) 2 = 30.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 4444 – 41.5 = 2.5(2.5) 2 = 6.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3737 – 41.5 = –4.5(–4.5) 2 = 20.25 4242 – 41.5 = 0.5(0.5) 2 = 0.25 Σ(x – μ) = 0 SS x = 88.5.

16. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 16 Solution: Finding the Population Standard Deviation Population Variance Population Standard Deviation The population standard deviation is about 3.0, or $3000..

17. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 17 Deviation, Variance, and Standard Deviation Sample Variance Sample Standard Deviation.

18. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 18 Finding the Sample Variance & Standard Deviation In Words In Symbols 1.Find the mean of the sample data set. 2.Find deviation of each entry. 3.Square each deviation. 4.Add to get the sum of squares..

19. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 19 Finding the Sample Variance & Standard Deviation 5.Divide by n – 1 to get the sample variance. 6.Find the square root to get the sample standard deviation. In Words In Symbols.

20. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 20 Example: Finding the Sample Standard Deviation The starting salaries are for the Chicago branches of a corporation. The corporation has several other branches, and you plan to use the starting salaries of the Chicago branches to estimate the starting salaries for the larger population. Find the sample standard deviation of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42.

21. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 21 Solution: Finding the Sample Standard Deviation Determine SS x n = 10 Salary, xDeviation: x – μSquares: (x – μ) 2 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3838 – 41.5 = –3.5(–3.5) 2 = 12.25 3939 – 41.5 = –2.5(–2.5) 2 = 6.25 4545 – 41.5 = 3.5(3.5) 2 = 12.25 4747 – 41.5 = 5.5(5.5) 2 = 30.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 4444 – 41.5 = 2.5(2.5) 2 = 6.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3737 – 41.5 = –4.5(–4.5) 2 = 20.25 4242 – 41.5 = 0.5(0.5) 2 = 0.25 Σ(x – μ) = 0 SS x = 88.5.

22. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 22 Solution: Finding the Sample Standard Deviation Sample Variance Sample Standard Deviation The sample standard deviation is about 3.1, or $3100..

23. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 23 Example: Using Technology to Find the Standard Deviation Sample office rental rates (in dollars per square foot per year) for Miami’s central business district are shown in the table. Use a calculator or a computer to find the mean rental rate and the sample standard deviation. (Adapted from: Cushman & Wakefield Inc.) Office Rental Rates 35.0033.5037.00 23.7526.5031.25 36.5040.0032.00 39.2537.5034.75 37.7537.2536.75 27.0035.7526.00 37.0029.0040.50 24.5033.0038.00.

24. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 24 Solution: Using Technology to Find the Standard Deviation Sample Mean Sample Standard Deviation.

25. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 25 Interpreting Standard Deviation Standard deviation is a measure of the typical amount an entry deviates from the mean. The more the entries are spread out, the greater the standard deviation..

26. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 26 Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics: About 68% of the data lie within one standard deviation of the mean. About 95% of the data lie within two standard deviations of the mean. About 99.7% of the data lie within three standard deviations of the mean..

27. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 27 Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) 68% within 1 standard deviation 34% 99.7% within 3 standard deviations 2.35% 95% within 2 standard deviations 13.5%.

28. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 28 Example: Using the Empirical Rule In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64.3 inches, with a sample standard deviation of 2.62 inches. Estimate the percent of the women whose heights are between 59.06 inches and 64.3 inches..

29. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 29 Solution: Using the Empirical Rule Because the distribution is bell-shaped, you can use the Empirical Rule. 34% + 13.5% = 47.5% of women are between 59.06 and 64.3 inches tall..

30. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 30 Chebychev’s Theorem The portion of any data set lying within k standard deviations (k > 1) of the mean is at least: k = 2: In any data set, at least of the data lie within 2 standard deviations of the mean. k = 3: In any data set, at least of the data lie within 3 standard deviations of the mean..

31. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 31 Example: Using Chebychev’s Theorem The age distribution for Florida is shown in the histogram. Apply Chebychev’s Theorem to the data using k = 2. What can you conclude?.

32. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 32 Solution: Using Chebychev’s Theorem k = 2: μ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age can’t be negative) μ + 2σ = 39.2 + 2(24.8) = 88.8 At least 75% of the population of Florida is between 0 and 88.8 years old..

33. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 33 Standard Deviation for Grouped Data Sample standard deviation for a frequency distribution When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class. where n= Σf (the number of entries in the data set).

34. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 34 Example: Finding the Standard Deviation for Grouped Data You collect a random sample of the number of children per household in a region. Find the sample mean and the sample standard deviation of the data set. Number of Children in 50 Households 13111 12210 11000 15036 30311 11601 36612 23011 41122 03024.

35. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 35 xfxf 0100(10) = 0 1191(19) = 19 272(7) = 14 373(7) =21 424(2) = 8 515(1) = 5 646(4) = 24 Solution: Finding the Standard Deviation for Grouped Data First construct a frequency distribution. Find the mean of the frequency distribution. Σf = 50 Σ(xf )= 91 The sample mean is about 1.8 children..

36. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 36 Solution: Finding the Standard Deviation for Grouped Data Determine the sum of squares. xf 0100 – 1.8 = –1.8(–1.8) 2 = 3.243.24(10) = 32.40 1191 – 1.8 = –0.8(–0.8) 2 = 0.640.64(19) = 12.16 272 – 1.8 = 0.2(0.2) 2 = 0.040.04(7) = 0.28 373 – 1.8 = 1.2(1.2) 2 = 1.441.44(7) = 10.08 424 – 1.8 = 2.2(2.2) 2 = 4.844.84(2) = 9.68 515 – 1.8 = 3.2(3.2) 2 = 10.2410.24(1) = 10.24 646 – 1.8 = 4.2(4.2) 2 = 17.6417.64(4) = 70.56.

37. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 37 Solution: Finding the Standard Deviation for Grouped Data Find the sample standard deviation. The standard deviation is about 1.7 children..

38. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 38 Coefficient of Variation Coefficient of Variation (CV) Describes the standard deviation of a data set as a percent of the mean. Population data set:  Sample data set: .

39. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 39 Example: Comparing Variation in Different Data Sets The table shows the population heights (in inches) and weights (in pounds) of the members of a basketball team. Find the coefficient of variation for the heights and the weighs. Then compare the results..

40. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 40 Solution: Comparing Variation in Different Data Sets The mean height is   72.8 inches with a standard deviation of   3.3 inches. The coefficient of variation for the heights is.

41. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 41 Solution: Comparing Variation in Different Data Sets The mean weight is   187.8 pounds with a standard deviation of   17.7 pounds. The coefficient of variation for the weights is. The weights (9.4%) are more variable than the heights (4.5%).

42. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 42 Section 2.4 Summary Found the range of a data set Found the variance and standard deviation of a population and of a sample Used the Empirical Rule and Chebychev’s Theorem to interpret standard deviation Approximated the sample standard deviation for grouped data Used the coefficient of variation to compare variation in different data sets.