1. Section 2.4
Measures of Variation
Larson/Farber 4th ed.

2. 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
Larson/Farber 4th ed.

3. 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
Larson/Farber 4th ed.

4. 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.
Larson/Farber 4th ed.

5. Deviation, Variance, and Standard Deviation
Sample Variance
Sample Standard Deviation
Larson/Farber 4th ed.

6. 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.
Larson/Farber 4th ed.

7. 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.
Larson/Farber 4th ed.

8. 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)
Larson/Farber 4th ed.

9. 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
Larson/Farber 4th ed.

10. Solution: Finding the Sample Standard Deviation
Sample Variance
Sample Standard Deviation
The sample standard deviation is about 3.1, or $3100.
Larson/Farber 4th ed.

11. 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
Larson/Farber 4th ed.

12. Solution: Using Technology to Find the Standard Deviation
Sample Mean
Sample Standard Deviation
Larson/Farber 4th ed.

13. 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.
Larson/Farber 4th ed.

14. 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.
Larson/Farber 4th ed.

15. 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.
Larson/Farber 4th ed.

16. 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?
Larson/Farber 4th ed.

17. 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.
Larson/Farber 4th ed.

18. 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)
Larson/Farber 4th ed.

19. 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
Larson/Farber 4th ed.

20. 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
Larson/Farber 4th ed.

21. 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
Larson/Farber 4th ed.

22. Solution: Finding the Standard Deviation for Grouped Data
Find the sample standard deviation.
The standard deviation is about 1.7 children.
Larson/Farber 4th ed.

23. Mean of Grouped Data Mean of a Frequency Distribution Approximated by
where x and f are the midpoints and frequencies of a class, respectively
Larson/Farber 4th ed.

24. Finding the Mean of a Frequency Distribution
In Words In Symbols
Find the midpoint of each class.
Find the sum of the products of the midpoints and the frequencies.
Find the sum of the frequencies.
Find the mean of the frequency distribution.
Larson/Farber 4th ed.

25. Example: Find the Mean of a Frequency Distribution
Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session.
Class
Midpoint
Frequency, f
7 – 18
12.5 6
19 – 30
24.5 10
31 – 42
36.5 13
43 – 54
48.5 8
55 – 66
60.5 5
67 – 78
72.5
79 – 90
84.5 2
Larson/Farber 4th ed.

26. Solution: Find the Mean of a Frequency Distribution
Class
Midpoint, x
Frequency, f
(x∙f)
7 – 18
12.5 6
12.5∙6 = 75.0
19 – 30
24.5 10
24.5∙10 = 245.0
31 – 42
36.5 13
36.5∙13 = 474.5
43 – 54
48.5 8
48.5∙8 = 388.0
55 – 66
60.5 5
60.5∙5 = 302.5
67 – 78
72.5
72.5∙6 = 435.0
79 – 90
84.5 2
84.5∙2 = 169.0
n = 50
Σ(x∙f) =
Larson/Farber 4th ed.

27. The Shape of Distributions
Symmetric Distribution
A vertical line can be drawn through the middle of a graph of the distribution and the resulting halves are approximately mirror images.
Larson/Farber 4th ed.

28. The Shape of Distributions
Uniform Distribution (rectangular)
All entries or classes in the distribution have equal or approximately equal frequencies.
Symmetric.
Larson/Farber 4th ed.

29. The Shape of Distributions
Skewed Left Distribution (negatively skewed)
The “tail” of the graph elongates more to the left.
The mean is to the left of the median.
Larson/Farber 4th ed.

30. The Shape of Distributions
Skewed Right Distribution (positively skewed)
The “tail” of the graph elongates more to the right.
The mean is to the right of the median.
Larson/Farber 4th ed.

31. © 2010 Pearson Prentice Hall. All rights reserved
The interest rate boxplot indicates that the distribution is skewed left.
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32. © 2010 Pearson Prentice Hall. All rights reserved
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33. 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.
Larson/Farber 4th ed.

34. 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%
Larson/Farber 4th ed.

35. 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 inches, with a sample standard deviation of 2.71 inches. Estimate the percent of the women whose heights are between 64 inches and inches.
Larson/Farber 4th ed.

36. Solution: Using the Empirical Rule
Because the distribution is bell-shaped, you can use the Empirical Rule.
34%
13.5%
55.87
58.58
61.29 64
66.71
69.42
72.13
34% % = 47.5% of women are between 64 and inches tall.
Larson/Farber 4th ed.

37. © 2010 Pearson Prentice Hall. All rights reserved
EXAMPLE Using the Empirical Rule
The following data represent the serum HDL cholesterol of the 54 female patients of a family doctor.
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38. © 2010 Pearson Prentice Hall. All rights reserved
(a) Using a TI-83 plus graphing calculator, we find
(b)
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39. © 2010 Pearson Prentice Hall. All rights reserved
(c) According to the Empirical Rule, 99.7% of the patients that have serum HDL within 3 standard deviations of the mean.
(e) 45 out of the 54 or 83.3% of the patients have a serum HDL between 34.0 and 69.1.
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