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Statistics Chapter 12

Statistics Chapter 12. 1. Measures of Central Tendency Section 12.2. 2. Measures of Central Tendency. The mean of a set of n numbers is the sum of the numbers divided by n .

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Statistics Chapter 12

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  1. StatisticsChapter 12 1

  2. Measures of Central TendencySection 12.2 2

  3. Measures of Central Tendency • The mean of a set of n numbers is the sum of the numbers divided by n. • The median is the middle number when the numbers are arranged in order of magnitude. If there is an even set of numbers, the median is the average of the two middle numbers. The • median is in the (n+1)/2 position. • The mode is the number of the set that occurs most often. 3

  4. Examples During a certain week, the following maximum temperatures(in °F) were recorded in a large eastern city: 78, 82, 82, 71, 69, 73, and 70. Find the mean, median, and mode of these temperatures. 4

  5. Solution Mean = (78 + 82 + 82 + 71 + 69 + 73 + 70) / 7 = 75 69, 70 ,71, 73, 78, 82, 82 Median = 73 Mode = 82 5

  6. Example A mathematics professor lost a paper belonging to one of her students. She remembered that the mean score for the class of 20 was 82, and that the sum of the 19 other scores was 1560. What was the grade on the lost paper? Grade = 20x82 – 1560 = 80 6

  7. Example • Tom has a mean score of 60 on four tests • taken. What score must he obtain on his • next test to have a mean(average) score of • a. 70 on all five tests? • b. What is the highest mean Tom can get • on all five tests? a. T5 5x70 – 4x60 = 110 b. HM (4x60 + 100)/5 = 68 7

  8. Example A professor grades students on two tests, three quizzes, and a final examination. Each test counts the same as three quizzes and the final examination counts the same as two tests. Mike has test scores of 88 and 96. Mike’s quiz scores are 88, 77, and 85. His final examination score is 94. Find Mike’s weighted average for the course. The professor rounds averages to the nearest integer. Avg = (388 + 396 + 88 + 77 + 85 + 694)/15 = 91 8

  9. Example Find the mean, the median, and the mode for the pay rates in the given frequency distribution. Hourly pay rates for employees Frequency $ 8.50 8 $11.00 28 $17.00 10 $20.00 5 Total 51 9

  10. Solution Mean = (8$8.50 + 28$11 +10$17 + 5$20)/51 Mean = $646/51 = $12.67 Because there are 51 employees, the median hourly rate would be that of the 26th employee. (51+1)/2 = 26 Median = $11.00 Mode = $11.00 END 10

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