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##### 13.6 Use Measures of Central Tendency and Dispersion

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**13.6 Use Measures of Central Tendency and Dispersion**Essential Question: How do you compare measures of central tendency and dispersion?**1.**Order from least to greatest: 10.14, 11.2, 10.1, 10.08, 11.21 2. Jenna scored the following points in her last five basketball games: 28, 18, 24, 22, and 18. What is Jenna’s average score for the 5 games? 22 points 10.08, 10.1, 10.14, 11.2, 11.21 ANSWER ANSWER Lesson 13.6, For use with pages 875-878 Warm-up Exercises**Descriptions the Middle**• Mean = average = x • x = x1 + x2 + … … + xn n • Median = the middle when the data set is ordered • Odd pieces of data single number • Even pieces of data mean of middle two numbers • Mode = most common data • Can be one mode, no mode, or multiple modes Measures of Central Tendency**Descriptions of the Spread**• Range = the difference between the greatest value and the least • Mean Absolute Deviation = the average variation of the data from the mean • M.A.D. = |x1 – x| + |x2 – x| + … … + |xn – x| n Measures of Dispersion**1000 + 1000 + 1181 + 1191 + 1200 + 1268 + 1328 + 2584**8 = 10,752 = x 8 EXAMPLE 1 Compare measures of central tendency The heights (in feet) of 8 waterfalls in the state of Washington are listed below. Which measure of central tendency best represents the data? 1000, 1000, 1181, 1191, 1200, 1268, 1328, 2584 SOLUTION = 1344**ANSWER**The median best represents the data. The mode is significantly less than most of the data, and the mean is significantly greater than most of the data. EXAMPLE 1 Compare measures of central tendency The median is the mean of the two middle values,1191and1200, or1195.5. The mode is 1000.**1000 + 1000 + 1181 + 1191 + 1200 + 1268 + 1328**= 7 8168 = x 7 for Example 1 GUIDED PRACTICE 1. What If ?In Example 1, suppose you eliminate the greatest data value, 2584. Which measure of central tendency best represents the remaining data? Explain your reasoning. SOLUTION = 1166 The median is 1191 The mode is 1000.**ANSWER**The median best represents the data. The mode is significantly less than most of the data, and the mean is lower than most of the data. for Example 1 GUIDED PRACTICE**RUNNING**EXAMPLE 2 Compare measures of dispersion The top 10 finishing times (in seconds) for runners in two men’s races are given. The times in a 100 meter dash are in set A, and the times in a 200 meter dash are in set B. Compare the spread of the data for the two sets using (a) the range and (b) the mean absolute deviation. A: 10.62, 10.94, 10.94, 10.98, 11.05, 11.13, 11.15, 11.28, 11.29, 11.32**ANSWER**The range of set Bis greater than the range of set A. So, the data in Bcover a wider interval than the data in A. EXAMPLE 2 Compare measures of dispersion B: 21.37, 21.40, 22.23, 22.23, 22.34, 22.34, 22.36, 22.60, 22.66, 22.73 SOLUTION a. A: 11.32 – 10.62 = 0.7 B: 22.73 –21.37 = 1.36**| 21.37 – 22.226 | + | 21.40 – 22.226 | + … + | 22.73**– 22.226 | 10 | 10.62 – 11.07 | + | 10.94 – 11.07 | + … + | 11.32 – 11.07 | 10 EXAMPLE 2 Compare measures of dispersion b. The mean of setAis11.07, so the mean absolute deviation is: = 0.164 The mean of setBis22.226, so the mean absolute deviation is: = 0.3364**ANSWER**The mean absolute deviation of set Bis greater, so the average variation from the mean is greater for the data in B than for the data in A. EXAMPLE 2 Compare measures of dispersion**RUNNING**2. for Example 2 GUIDED PRACTICE The top 10 finishing times (in seconds) for runners in a men’s 400 meter dash are 46.89, 47.65, 48.15, 49.05, 49.19, 49.50, 49.68, 51.09, 53.31, and 53.68. Compare the spread of the data with that of set Ain Example 2 using (a) the range and (b) the mean absolute deviation. a. A: 11.32 – 10.62 = 0.7 400 meter – 53.68 – 46.89 = 6.79 The range of finishing time for the men’s 400m dash (6.79) is greater than the range for set A(0.7)**| 46.89 – 49.819 | + | 47.65 – 49.819| + | 48.15 –**49.819| +…+ | 53.68 – 49.819| | 10.62 – 11.07 | + | 10.94 – 11.07 | + … + | 11.32 – 11.07 | 10 10 for Example 2 GUIDED PRACTICE b. The mean of setAis11.07, so the mean absolute deviation is: = 0.164 The mean of 400m men’sis 49.819, so the mean absolute deviation is : = 1.7246**How do you compare measures of central tendency and**dispersion? Essential Question:**Textbook p. 877-878**Independent Practice