1. Chapter 2
Descriptive Statistics

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. Section 2.4
Measures of Variation .

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. Range
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. 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) .

7. Solution: Finding the Range
Ordering the data helps to find the least and greatest salaries.
Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
The range of starting salaries is 10 or $10,000.
minimum
maximum .

8. Deviation, Variance, and Standard 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. 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)
Solution:
First determine the mean starting salary. .

10. Solution: Finding the Deviation
Determine the deviation for each data entry.
Salary ($1000s), x
Deviation: x – μ 41
41 – 41.5 = –0.5 38
38 – 41.5 = –3.5 39
39 – 41.5 = –2.5 45
45 – 41.5 = 3.5 47
47 – 41.5 = 5.5 44
44 – 41.5 = 2.5 37
37 – 41.5 = –4.5 42
42 – 41.5 = 0.5
Σx = 415
Σ(x – μ) = 0 .

11. Deviation, Variance, and Standard Deviation
Population Variance
Population Standard Deviation
Sum of squares, SSx .

12. Finding the Population Variance & Standard Deviation
In Words In Symbols
Find the mean of the population data set.
Find deviation of each entry.
Square each deviation.
Add to get the sum of squares.
x – μ
(x – μ)2
SSx = Σ(x – μ)2 .

13. Finding the Population Variance & Standard Deviation
In Words In Symbols
Divide by N to get the population variance.
Find the square root to get the population standard deviation. .

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) Recall μ = 41.5. .

15. Solution: Finding the Population Standard Deviation
Determine SSx
N = 10
Salary, x
Deviation: x – μ
Squares: (x – μ)2 41
41 – 41.5 = –0.5
(–0.5)2 = 0.25 38
38 – 41.5 = –3.5
(–3.5)2 = 12.25 39
39 – 41.5 = –2.5
(–2.5)2 = 6.25 45
45 – 41.5 = 3.5
(3.5)2 = 12.25 47
47 – 41.5 = 5.5
(5.5)2 = 30.25 44
44 – 41.5 = 2.5
(2.5)2 = 6.25 37
37 – 41.5 = –4.5
(–4.5)2 = 20.25 42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5 .

16. Solution: Finding the Population Standard Deviation
Population Variance
Population Standard Deviation
The population standard deviation is about 3.0, or $3000. .

17. Deviation, Variance, and Standard Deviation
Sample Variance
Sample Standard Deviation .

18. Finding the Sample Variance & Standard Deviation
In Words In Symbols
Find the mean of the sample data set.
Find deviation of each entry.
Square each deviation.
Add to get the sum of squares. .

19. Finding the Sample Variance & Standard Deviation
In Words In Symbols
Divide by n – 1 to get the sample variance.
Find the square root to get the sample standard deviation. .

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) .

21. Solution: Finding the Sample Standard Deviation
Determine SSx
n = 10
Salary, x
Deviation: x – μ
Squares: (x – μ)2 41
41 – 41.5 = –0.5
(–0.5)2 = 0.25 38
38 – 41.5 = –3.5
(–3.5)2 = 12.25 39
39 – 41.5 = –2.5
(–2.5)2 = 6.25 45
45 – 41.5 = 3.5
(3.5)2 = 12.25 47
47 – 41.5 = 5.5
(5.5)2 = 30.25 44
44 – 41.5 = 2.5
(2.5)2 = 6.25 37
37 – 41.5 = –4.5
(–4.5)2 = 20.25 42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5 .

22. Solution: Finding the Sample Standard Deviation
Sample Variance
Sample Standard Deviation
The sample standard deviation is about 3.1, or $3100. .

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
35.00
33.50
37.00
23.75
26.50
31.25
36.50
40.00
32.00
39.25
37.50
34.75
37.75
37.25
36.75
27.00
35.75
26.00
29.00
40.50
24.50
33.00
38.00 .

24. Solution: Using Technology to Find the Standard Deviation
Sample Mean
Sample Standard Deviation .

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. 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. Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)
99.7% within 3 standard deviations
2.35%
95% within 2 standard deviations
13.5%
68% within 1 standard deviation
34% .

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 inches and 64.3 inches. .

29. Solution: Using the Empirical Rule
Because the distribution is bell-shaped, you can use the Empirical Rule.
34% % = 47.5% of women are between and 64.3 inches tall. .

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. 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. Solution: Using Chebychev’s Theorem
k = 2: μ – 2σ = 39.2 – 2(24.8) = (use 0 since age can’t be negative) μ + 2σ = (24.8) = 88.8
At least 75% of the population of Florida is between 0 and 88.8 years old. .

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. 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 1 3 2 5 6 4 .

35. Solution: Finding the Standard Deviation for Grouped Data
First construct a frequency distribution.
Find the mean of the frequency distribution. x f xf 10
0(10) = 0 1 19
1(19) = 19 2 7
2(7) = 14 3
3(7) =21 4
4(2) = 8 5
5(1) = 5 6
6(4) = 24
The sample mean is about 1.8 children.
Σf = 50
Σ(xf )= 91 .

36. Solution: Finding the Standard Deviation for Grouped Data
Determine the sum of squares. x f 10
0 – 1.8 = –1.8
(–1.8)2 = 3.24
3.24(10) = 32.40 1 19
1 – 1.8 = –0.8
(–0.8)2 = 0.64
0.64(19) = 12.16 2 7
2 – 1.8 = 0.2
(0.2)2 = 0.04
0.04(7) = 0.28 3
3 – 1.8 = 1.2
(1.2)2 = 1.44
1.44(7) = 10.08 4
4 – 1.8 = 2.2
(2.2)2 = 4.84
4.84(2) = 9.68 5
5 – 1.8 = 3.2
(3.2)2 = 10.24
10.24(1) = 10.24 6
6 – 1.8 = 4.2
(4.2)2 = 17.64
17.64(4) = 70.56 .

37. Solution: Finding the Standard Deviation for Grouped Data
Find the sample standard deviation.
The standard deviation is about 1.7 children. .

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. 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. 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. Solution: Comparing Variation in Different Data Sets
The mean weight is   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. 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 .