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6.7 Use Measures of Central Tendency and Dispersion

6.7 Use Measures of Central Tendency and Dispersion. Vocabulary. Mean: average, “x-bar” Median: middle # when values written in numerical order; avg. of 2 numbers if there is an even set of data Mode: value that occurs most often Measure of dispersion: describes the spread of data.

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6.7 Use Measures of Central Tendency and Dispersion

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  1. 6.7 Use Measures of Central Tendency and Dispersion

  2. Vocabulary • Mean: average, “x-bar” • Median: middle # when values written in numerical order; avg. of 2 numbers if there is an even set of data • Mode: value that occurs most often • Measure of dispersion: describes the spread of data

  3. Vocabulary • Range: difference of greatest value and least value • Deviation from the mean: difference of the data value and the mean • Mean absolute deviation: avg. deviation of the data from the mean

  4. Compare measures of central tendency 1000 + 1000 + 1181 + 1191 + 1200 + 1268 + 1328 + 2584 8 = 10,752 = x 8 The heights (in feet) of 8 waterfalls in the state of Washington are listed below. Find mean, median, & mode. Which measure of central tendency best represents the data? 1000, 1000, 1181, 1191, 1200, 1268, 1328, 2584 SOLUTION = 1344

  5. 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. Compare measures of central tendency 1000, 1000, 1181, 1191, 1200, 1268, 1328, 2584 The median is the mean of the two middle values,1191and1200, or1195.5. The mode is 1000.

  6. Compare measures of dispersion RUNNING 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 B: 21.37, 21.40, 22.23, 22.23, 22.34, 22.34, 22.36, 22.60, 22.66, 22.73

  7. 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. A: 10.62, 10.94, 10.94, 10.98, 11.05, 11.13, 11.15, 11.28, 11.29, 11.32 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

  8. | 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 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

  9. 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.

  10. RUNNING 2. 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. ANSWER a. The range of finishing times for the men’s 400m dash (6.79) is greater than the range for set A (0.7). b. The mean absolute deviation is greater for the men’s 400m dash (1.7246) than it is for set A (0.164).

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