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DESCRIPTIVE STATISTICS FOR ALGEBRA I

DESCRIPTIVE STATISTICS FOR ALGEBRA I. Mean, Standard Deviation, Mean Absolute Deviation and Z-scores. Mean.

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DESCRIPTIVE STATISTICS FOR ALGEBRA I

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  1. DESCRIPTIVE STATISTICS FOR ALGEBRA I Mean, Standard Deviation, Mean Absolute Deviation and Z-scores

  2. Mean Mean = the average of the numbers. To find the mean of a set of numbers, you must add up the numbers then divide by the number of numbers. The µ symbol is used to represent mean in a formula. Example: {18 23 10 39 22 17 16 15} 18+23+10+39+22+17+16+15 = 160 160 = 20 8

  3. Mean Practice • Find the mean of the following sets of numbers. {20 30 15 40 20 80 60 50} {34 20 56 82 78 30}

  4. Mean Practice 20 + 30 + 15 + 40 + 20 + 80 + 60 + 50 = 320 320 34 + 20 + 56 +82 + 78 + 30 = 300 300 = 40 8 = 50 6

  5. Standard Deviation The standard deviation is the positive square root of the variance of the data set. The greater the value of the standard deviation, the more spread out the data are about the mean. The lesser (closer to 0) the value of the standard deviation, the closer the data are clustered about the mean. The symbol is used to represent standard deviation in a formula. σ

  6. Mean and Standard Deviation Computation of descriptive statistics using the graphing calculator. • From the menu screen select STAT and press EXE. {40, 30, 50, 30, 50, 50, 60, 50, 30, 40, 50, 40, 60, 50} • In List 1, enter the data set. Enter the first element and press EXE to move to the next line and continue until all the elements have been entered into List 1. • Press F2 (CALC) • Press F1 (1VAR) ͞x = Mean of the data = 45 x n-1 = Standard deviation = 10.2 σ

  7. Mean Absolute Deviation The mean absolute deviation is the arithmetic mean of the absolute values of the deviations of elements from the mean of a data set.

  8. Mean Absolute Deviation Computation Method 1 Continuing from the calculation of standard deviation, press EXIT until the function button menus read “GRAPH CALC TEST INTR DIST” • Press ► to move the cursor to List 2. • Press ▲ until “List 2” is highlighted. • Press OPTN F4 (NUM) F1 (ABS) • Press ( OPTN F1 F1 (to get “List”) 1 - VARS F3 F5 F2 ) EXE (List 2 will automatically fill with data). • Continue to press EXIT until the function button menus read “GRAPH CALC TEST INTR DIST” • Press F2 (CALC) F6 (SET) F2 (List 2) • Press EXE EXIT • Press F2 (CALC) F1 (1 VAR) ͞ X = mean absolute deviation

  9. Mean Absolute Deviation Computation Method 2 Continuing from the calculation of standard deviation, press EXIT until the function button menus read “GRAPH CALC TEST INTR DIST”. • Press ► to move the cursor to List 2. • Enter the absolute value of the difference between the data value and the mean for all the data values in List 1. • Press F2 (CALC) F6 (SET) F2 (List 2) • Press EXE EXIT • Press F2 (CALC) F1 (1 VAR) ͞ X = mean absolute deviation

  10. Mean Absolute Deviation The mean absolute deviation for the data {40, 30, 50, 30, 50, 50, 60, 50, 30, 40, 50, 40, 60, 50} is 8.57142857.

  11. Mean and Standard Deviation Here are 20 test scores from Mr. Barnes algebra class. • 82 79 72 84 73 94 77 81 74 68 87 • 89 80 90 84 83 81 75 78 What is the mean score? What is the standard deviation?

  12. Mean and Standard Deviation Here are 20 test scores from Mr. Barnes algebra class. • 82 79 72 84 73 94 77 81 74 68 87 • 89 80 90 84 83 81 75 78 What is the mean score? 80.7 What is the standard deviation? 6.5

  13. Mean Absolute Deviation Here are 20 test scores from Mr. Barnes algebra class. • 82 79 72 84 73 94 77 81 74 68 87 • 89 80 90 84 83 81 75 78 What is the mean absolute deviation?

  14. Mean Absolute Deviation Here are 20 test scores from Mr. Barnes algebra class. • 82 79 72 84 73 94 77 81 74 68 87 • 89 80 90 84 83 81 75 78 What is the mean absolute deviation? 5.03

  15. WHERE DO I STAND? Z-Scores

  16. Z-score • Position of a data value relative to the mean. • Tells you how many standard deviations above or below the mean a particular data point is. z-score =describes the location of a data value within a distribution referred to as a standardized value Population = Z-SCORE

  17. Z-score In order to calculate a z-score you must know: • a data value = X i • the mean = µ σ • the standard deviation =

  18. Z-scores Here are 18 test scores from Mr. Barnes algebra class. • 84 89 83 79 83 77 73 67 75 • 80 78 93 83 77 80 81 79 The bold score is Michael’s. How did he perform relative to his classmates?

  19. Z-scores Here are 18 test scores from Mr. Barnes stat class. • 84 89 83 79 83 77 73 67 75 • 80 78 9383 77 80 81 79 The bold score is Michael’s. How did he perform relative to his classmates? Michael’s score is “above average”, but how much above average is it?

  20. Z-scores If we convert Michael’s score to a standardized value, then we can determine how many standard deviations his score is away from the mean. • What we need: • Michele’s score • mean of test scores • standard deviation 93 80.2 93 – 80.2 5.8 = 2.2 Z= 5.8 Therefore, Michael’s standardized test score is 2.2, over two standard deviations above the class mean.

  21. 62.8 68.6 74.4 80.2 86.0 91.8 97.6 Z-score Number Line Michael’s Score = 93 Michael’s z-score = 2.2 3 -2 -1 -3 0 2 1

  22. Z-scores Here are 18 test scores from Mr. Barnes algebra class. • 84 89 83 79 83 77 73 67 75 • 80 78 93 83 77 80 81 79 The bold score is Bonnie’s. How did she perform relative to her classmates?

  23. Z-scores Here are 18 test scores from Mr. Barnes stat class. • 84 89 83 79 83 77 73 67 75 • 80 78 93 83 77 80 81 79 The bold score is Bonnie’s. How did she perform relative to her classmates? Bonnie’sscore is “below average”, but how much above average is it?

  24. Z-scores If we convert Michael’s score to a standardized value, then we can determine how many standard deviations his score is away from the mean. • What we need: • Michele’s score • mean of test scores • standard deviation 67 80.2 67 – 80.2 5.8 = -2.3 Z= 5.8 Therefore, Bonnie’sstandardized test score is -2.3, over two standard deviations below the class mean.

  25. 62.8 68.6 74.4 80.2 86.0 91.8 97.6 Z-score Number Line Bonnie’sScore = 67 Bonnie’sz-score = -2.3 1 -3 0 2 -2 -1 3

  26. Z-scores Here are 18 test scores from Mr. Barnes algebra class. • 84 89 83 79 83 77 73 67 75 • 80 78 93 83 77 80 81 79 The bold score is Ray’s. How did he perform relative to his classmates?

  27. Z-scores Here are 18 test scores from Mr. Barnes stat class. • 84 89 83 79 83 77 73 67 75 • 80 78 9383 77 80 81 79 The bold score is Ray’s. How did he perform relative to his classmates? Ray’s score is slightly “below average”, but how much below average is it?

  28. Z-scores If we convert Ray’s score to a standardized value, then we can determine how many standard deviations his score is away from the mean. • What we need: • Ray’s score • mean of test scores • standard deviation 80 80.2 80 – 80.2 5.8 = -0.03 Z= 5.8 Therefore, Ray’s standardized test score is -0.03, slightly below the class mean.

  29. 62.8 68.6 74.4 80.2 86.0 91.8 97.6 Z-score Number Line Ray’s Score = 80 Ray’s z-score = -0.03 0 -2 -1 1 3 -3 2

  30. Practice Problem 1 Which students’ heights have a z-score greater than 1? A) All of them B) Ann, Bill, Caryn, Ken and Gary C) Ken D) Don and Fred

  31. Practice Problem 1 Which students’ heights have a z-score greater than 1? Mean = 53 Standard Deviation = 4.7 A) All of them B) Ann, Bill, Caryn, Ken and Gary C) Ken X – 53 4.7 > 1 D) Don and Fred X > 57.7

  32. Practice Problem 2 Which students have a z-score less than -1? A) All of them B) Bill, Don and Fred C) Only Bill D) Ann, Caryn, Ken and Gary

  33. Practice Problem 2 Which students have a z-score less than -1? Mean = 51 Standard Deviation = 4.6 A) All of them B) Bill, Don and Fred C) Only Bill X – 51 4.6 < -1 D) Ann, Caryn, Ken and Gary X < 46

  34. Practice Problem 3 Which student’s height has a z-score of zero? A) Bill B) Fred C) Caryn D) None of them

  35. Practice Problem 3 Which student’s height has a z-score of zero? A) Bill Mean = 50 Standard Deviation = 5.3 B) Fred X – 50 5.3 C) Caryn = 0 D) None of them X = 50

  36. Practice Problem 4 Given a data set with a mean of 135 and a standard deviation of 10, describe the z-score of a data value of 100? A) Less than -5 B) Between -1 and 0 C) Between -5 and -1 D) Greater than 0

  37. Practice Problem 4 Given a data set with a mean of 135 and a standard deviation of 10, describe the z-score of a data value of 100? Mean = 135 Standard Deviation = 10 A) Less than -5 B) Between -1 and 0 100 – 135 10 Z = C) Between -5 and -1 Z = - 3.5 D) Greater than 0

  38. Practice Problem 5 Given a data set with a mean of 35 and a standard deviation of 2.5, find the data value associated with a z-score of 3? A) 46 B) 45 C) 44.5 D) 42.5

  39. Practice Problem 5 Given a data set with a mean of 35 and a standard deviation of 2.5, find the data value associated with a z-score of 3? Mean = 35 Standard Deviation = 2.5 A) 46 B) 45 X – 35 2.5 = 3 C) 44.5 X = 42.5 D) 42.5

  40. Practice Problem 6 Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were: {2.5, 0, -3.2, -2.1, 1.75, 0.4} 1) List the z-scores of students that were above the mean.

  41. Practice Problem 6 Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were: {2.5, 0, -3.2, -2.1, 1.75, 0.4} 1) List the z-scores of students that were above the mean. 2.5, 1.75, and 0.4

  42. Practice Problem 7 Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were: {2.5, 0, -3.2, -2.1, 1.75, 0.4} 2) If the mean of the exam is 85, did any of the students selected have an exam score of 85? Explain.

  43. Practice Problem 7 Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were: {2.5, 0, -3.2, -2.1, 1.75, 0.4} 2) If the mean of the exam is 85, did any of the students selected have an exam score of 85? Explain. One student with a z-score of 0.

  44. Practice Problem 8 Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were: {2.5, 0, -3.2, -2.1, 1.75, 0.4} 3) If the standard deviation of the exam was 4 and the mean is 85, what was the actual test score for the student having a z-score of 1.75?

  45. Practice Problem 8 Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were: {2.5, 0, -3.2, -2.1, 1.75, 0.4} 3) If the standard deviation of the exam was 4 and the mean is 85, what was the actual test score for the student having a z-score of 1.75? X – 85 = 1.75 4 X = 92

  46. Practice Problem 9 A set of mathematics exam scores has a mean of 70 and a standard deviation of 8. A set of English exam scores has a mean of 74 and a standard deviation of 16. For which exam would a score of 78 have a higher standing?

  47. Practice Problem 9 A set of mathematics exam scores has a mean of 70 and a standard deviation of 8. A set of English exam scores has a mean of 74 and a standard deviation of 16. For which exam would a score of 78 have a higher standing? Math English 1 4 78 – 74 16 78 – 70 8 = = 1 Math because it has a higher z-score.

  48. Practice Problem 10 Maya represented the heights of boys in Mrs. Constantine’s and Mr. Kluge’s classes on a line plot and calculated the mean and standard deviation. Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches) ________________________________________________________________ 64 65 66 67 68 69 70 71 72 73 x x x x x x x x x x x x x x x x x x x x Mean = 68.4 Standard Deviation = 2.3 1. How many elements are above the mean?

  49. Practice Problem 10 Maya represented the heights of boys in Mrs. Constantine’s and Mr. Kluge’s classes on a line plot and calculated the mean and standard deviation. Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches) ________________________________________________________________ 64 65 66 67 68 69 70 71 72 73 x x x x x x x x x x x x x x x x x x x x Mean = 68.4 Standard Deviation = 2.3 • How many elements are above the mean? There are 9 elements above • the mean value of 68.4.

  50. Practice Problem 11 Maya represented the heights of boys in Mrs. Constantine’s and Mr. Kluge’s classes on a line plot and calculated the mean and standard deviation. Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches) ________________________________________________________________ 64 65 66 67 68 69 70 71 72 73 x x x x x x x x x x x x x x x x x x x x Mean = 68.4 Standard Deviation = 2.3 1. How many elements are below the mean?

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