Measures of Variation 1 Section 2.4

Section 2.4 Objectives 2 Determine the range of a data set Determine the variance and standard deviation of a population and of a sample Use the Empirical Rule and Chebychev’s Theorem to interpret standard deviation Approximate the sample standard deviation for grouped data

Range 3 The difference between the maximum and minimum data entries in the set. The data must be quantitative. Range = (Max. data entry) – (Min. data entry)

Example: Finding the Range 4 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)

Solution: Finding the Range 5 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

Deviation, Variance, and Standard Deviation 6 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

Example: Finding the Deviation 7 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.

Solution: Finding the Deviation 8 Determine the deviation for each data entry. Salary ($1000s), xDeviation: x – μ 4141 – 41.5 = – – 41.5 = – – 41.5 = – – 41.5 = – 41.5 = – 41.5 = – – 41.5 = – 41.5 = – – 41.5 = – – 41.5 = 0.5 Σx = 415 Σ(x – μ) = 0

Deviation, Variance, and Standard Deviation 9 Population Variance Population Standard Deviation Sum of squares, SS x

Finding the Population Variance & Standard Deviation 10 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

Finding the Population Variance & Standard Deviation 11 5.Divide by N to get the population variance. 6.Find the square root to get the population standard deviation. In Words In Symbols

Example: Finding the Population Standard Deviation 12 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.

Solution: Finding the Population Standard Deviation 13 Determine SS x N = 10 Note that SS x = Σ (x – μ ) 2 Salary, x Deviation: x – μ Squares: (x – μ ) – 41.5 = –0.5(–0.5) 2 = – 41.5 = –3.5(–3.5) 2 = – 41.5 = –2.5(–2.5) 2 = – 41.5 = 3.5(3.5) 2 = – 41.5 = 5.5(5.5) 2 = – 41.5 = –0.5(–0.5) 2 = – 41.5 = 2.5(2.5) 2 = – 41.5 = –0.5(–0.5) 2 = – 41.5 = –4.5(–4.5) 2 = – 41.5 = 0.5(0.5) 2 = 0.25 Σ(x – μ) = 0 SS x = 88.5

Solution: Finding the Population Standard Deviation 14 Population Variance Population Standard Deviation The population standard deviation is about 3.0, or $3000.

Deviation, Variance, and Standard Deviation 15 Sample Variance Sample Standard Deviation

Finding the Sample Variance & Standard Deviation 16 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.

Finding the Sample Variance & Standard Deviation 17 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

Sample Standard Deviation Shortcut Formula n ( n - 1) s = n (  x 2 ) - (  x ) 2 18

Symbols for Standard Deviation Sample Population   x x  n s S x x  n-1 Book Some graphics calculators Some non-graphics calculators Textbook Some graphics calculators Some non-graphics calculators 19

Example: Finding the Sample Standard Deviation 20 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)

Solution: Finding the Sample Standard Deviation 21 Determine SS x n = 10 Note that Salary, xDeviation:Squares: 4141 – 41.5 = –0.5(–0.5) 2 = – 41.5 = –3.5(–3.5) 2 = – 41.5 = –2.5(–2.5) 2 = – 41.5 = 3.5(3.5) 2 = – 41.5 = 5.5(5.5) 2 = – 41.5 = –0.5(–0.5) 2 = – 41.5 = 2.5(2.5) 2 = – 41.5 = –0.5(–0.5) 2 = – 41.5 = –4.5(–4.5) 2 = – 41.5 = 0.5(0.5) 2 = 0.25 Σ( ) = 0 SS x = 88.5

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

Example: Using Technology to Find the Standard Deviation 23 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

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

Interpreting Standard Deviation 25 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.

minimum ‘usual’ value  (mean) - 2 (standard deviation) minimum  x - 2(s) maximum ‘usual’ value  (mean) + 2 (standard deviation) maximum  x + 2(s) Usual Sample Values

Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) 27 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.

Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) 28 68% within 1 standard deviation 34% 99.7% within 3 standard deviations 2.35% 95% within 2 standard deviations 13.5%

Example: Using the Empirical Rule 29 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.

Solution: Using the Empirical Rule % 13.5% Because the distribution is bell-shaped, you can use the Empirical Rule. 34% % = 47.5% of women are between 64 and inches tall.

Chebychev’s Theorem 31 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.

Example: Using Chebychev’s Theorem 32 The age distribution for Florida is shown in the histogram. Apply Chebychev’s Theorem to the data using k = 2. What can you conclude?

Solution: Using Chebychev’s Theorem 33 k = 2: μ – 2 σ = 39.2 – 2(24.8) = (use 0 since age can’t be negative) μ + 2 σ = (24.8) = 88.8 At least 75% of the population of Florida is between 0 and 88.8 years old.

Estimation of Standard Deviation Range Rule of Thumb x - 2 s x x + 2 s Range  4 s or (minimum usual value) (maximum usual value) Range 4 s  = highest value - lowest value 4 34

Standard Deviation for Grouped Data 35 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)

Example: Finding the Standard Deviation for Grouped Data 36 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

Solution: Finding the Standard Deviation for Grouped Data 37 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. xfxf 0100(10) = (19) = (7) = (7) =21 424(2) = 8 515(1) = 5 646(4) = 24

Solution: Finding the Standard Deviation for Grouped Data 38 Determine the sum of squares. xf 0100 – 1.8 = –1.8(–1.8) 2 = (10) = – 1.8 = –0.8(–0.8) 2 = (19) = – 1.8 = 0.2(0.2) 2 = (7) = – 1.8 = 1.2(1.2) 2 = (7) = – 1.8 = 2.2(2.2) 2 = (2) = – 1.8 = 3.2(3.2) 2 = (1) = – 1.8 = 4.2(4.2) 2 = (4) = 70.56

Solution: Finding the Standard Deviation for Grouped Data 39 Find the sample standard deviation. The standard deviation is about 1.7 children.

Standard Deviation from a Frequency Table Shortcut Formula n ( n - 1) S =S = n [  ( f x 2 )] -[  ( f x )] 2 40

Practice Questions Q(2.11) Compute the sample variance and sample Standard deviation.

Practice Questions Q(2.12) Compute the variance and standard deviation of the given grouped data f Number

Practice Questions Q(2.13) The mean of a distribution is 20 and the standard deviation is 2. Use Chebyshev’s Theorem to answer the following questions. (1) At least what percentage of the values will fall between 10 and 30? (2) At least what percentage of the values will fall between 12 and 28? 43

Practice Questions Q(2.14) The average U.S yearly per capita consumption of citrus fruits is 26.8 pounds. Suppose that the distribution of fruits amount consumed is bell-shaped with standard deviation of 4.2 pounds. What percentage of Americans would you expect to consume more than 31 pounds of citrus fruit per year? 44

Section 2.4 Summary 45 Determined the range of a data set Determined 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