MEASURE OF CENTRAL TENDENCY

Measure of central tendency provides a very convenient way of describing a set of scores with a single number that describes the PERFORMANCE of the group.
It is also defined as a single value that is used to describe the “center” of the data.

1. MEASURE OF CENTRAL TENDENCY AB.RAJAR. ASSOCIATE PROFESSOR. COMMUNITY HEALTH SCIEMCES

2. MEASURE OF CENTRAL TENDENCY In statistics, a central tendency is a central value or a typical value for a probability distribution. It is occasionally called an average or just the center of the distribution . Measure of central tendency provides a very convenient way of describing a set of scores with a single number that describes the PERFORMANCE of the group. It is also defined as a single value that is used to describe the “center” of the data.

3. OBJECTIVES. The main objectives of averages are: To present diversified and complex data in a Summarized form . To facilitate Comparisons. To facilitate D ecision making . To establish Relationship. To facilitate Statistical Analysis.

4. Common Measures. There are three commonly used measures of central tendency. These are the following: MEAN. MODE. MEDIAN. Quartile. Deciles. Percentiles.

5. MEAN (Arithmetic Mean ) The MEAN of a set of values or measurements is the sum of all the measurements divided by the number of measurements in the set. Among the three measures of central tendency, the mean is the most popular and widely used. It is sometimes called the arithmetic mean. If we compute the mean of the population, we call it the parametric or population mean , denoted by μ ( read “mu ”). If we get the mean of the sample , we call it the sample mean and it is denoted by ( read “x bar ”).  

6. Equations . For ungrouped or raw data , the mean has the following formula. Simply µ = Simply µ Where: = ( read X bar ) for sample. = Sum of the measurements or values. ∑𝑓𝑥 = Sum of all frequencies n = Number of measurement. µ = Mu ( for population)   SAMPLE POPULATION

7. MEAN It is the most commonly used measure of the center of data. It is also referred as the “arithmetic average”. Computation of Sample Mean: X = Σ = ̅ ̅  

8. Exercise. Example: Scores of 15 students in Community Medicine quiz consist of 25 items .(MCQS) The highest score is 25 and the lowest score is 10 . Here are the scores: 25, 20, 18, 18, 17, 15, 15, 15, 14, 14, 13, 12, 12, 10, 10. Find the mean in the following scores .? X score 25 14 20 14 18 13 18 12 17 12 15 10 15 10 15   = = 15.2  

9. MEAN Result: = 15.2 Analysis: The average performance of 15 students who participated in Community Medicine quiz consisting of 25 items is 15.20 . The implication of this is that student who got scores below 15.2 did not perform well in the said examination . Students who got scores higher than 15.2 performed well in the examination compared to the performance of the whole class.  

10. Exercise. The diastolic blood pressure of 10 individuals is as: 83,75,81,79,71,95,75,77,84 and 90. Find the average (mean) diastolic blood pressure ? Solution: Hence. = = 81.   = 81. The average diastolic blood pressure is 81.  

11. Exercise. Example. Mr Zain collects the data on the ages of Pathology teachers in Muhammad Medical College and His study yields the following: 38,35,28,36,35,33 and 40 . Find the average age of teachers ? =     = 35.   = 35   Based on the computed mean, 35 is the average age of Pathology teachers in MMC.

12. Exercise. A unit test wise performance (marks) of 8 students in Community Medicine are 400, 600,360,450 and 250. Find Arithmetic mean ??

13. Exercise. A Pathology class of 8 students obtained the following marks in an examination. Find the Arithmetic mean ?? 60,80,75,35,80,70,78,44.

14. Weighted Mean Weighted mean is the mean of a set of values wherein each value or measurement has a different weight or degree of importance. The following is its formula: Where = = mean.( read as x bar ) = sum of each measurements multiplied by respective values. x = measurement or value w = number of measurements.  

15. Weighted Mean Sometimes, we want to find the average of certain values which are not of equal importance, therefore we assign certain numerical values ( i.e,called weights) according to their relative importance. Weights may frequencies or quantities etc. Thus if ……… are n- values & are corresponding weights, then weighted mean is defined as: = =  

16. Exercise Example: Marks obtained by a student in three semesters of different duration are recorded below. Calculate the mean marks per test ?? Test 1 2 3 Marks 40 90 80 Duration (Hours) 0.5 2.0 0.5

17. Solution: = = = 80 marks .   Marks (x) Duration (w) wx 40 0.5 20 90 2.0 180 80 o.5 40 3.0 240 Result: Mean marks /test are = 80

18. Exercise . Below are Shazia’s subjects and the corresponding number of units and grades she got for the previous grading period. Compute her grade point average??   Subject Unit Grade Anatomy 0.9 86 Bio-chemistry 1.5 85 Physiology 1.5 88 Pharmacology 1.8 87 Forensic Medicine 0.9 86 Pathology 1.2 83 Community Medicine 1.2 87 =86.1 R esult = Shazia’s average point score is 86.1  

19. Arithmetic Mean From a Frequency Distribution ( Ungrouped Data)

20. Arithmetic Mean From a Frequency Distribution ( Ungrouped Data) If ……… be a set of k-values with corresponding frequencies then arithmetic mean is computed as: = Where Σ f = the sum of the frequencies Σ fx = the sum of each observation multiplied by its frequency  

21. Exercise. From the following data of sales ( rounded to the nearest hundred rupees) on 25 days, find the mean sales per day.? Sales ( in rupees) 1650 1900 2050 2100 2350 No of days 3 5 8 7 2

22. Solution: x f fx 1650 3 4950 1900 5 9500 2050 8 16400 2100 7 14700 2350 2 4700 Total 25 50250 X = ------------ = 2010 Rs 50250 25

23. Exercise. Find the average income from the following data ? Daily income in Rs ( x ) 200 300 350 700 840 950 No of workers ( f ) 6 2 2 1 1 1

24. Solution: x f fx 200 6 1200 300 2 600 350 2 700 700 1 700 840 1 840 950 1 950 Total 13 4990 = ------------ = 383.85 Average Income/worker   4990 13

25. Exercise. If the mean of the following numbers is 30 find the value of k. 22,28,31,k,23.??

26. Arithmetic Mean From a Frequency Distribution ( Grouped Data) If ……… are the mid points of different class intervals and ,are the corresponding frequencies then the arithmetic mean is computed as : =   Where Σ f = the sum of the frequencies Σ fx = the sum of each observation multiplied by its frequency  

27. Exercise. Compute arithmetic mean from the following frequency distribution. C.I 2-6 7-11 12-16 17-21 22-26 Total f 2 5 11 9 3 30

28. Solution: C.I f x fx 2-6 2 4 8 7-11 5 9 45 12-16 11 14 154 17-21 9 19 171 22-26 3 24 72 Total 30 450 X = ------------ = 15. 450 30

29. Exercise. The following frequency table gives the height ( in inches ) of 100 students in a college. Calculate the average height of these students. Heights ( inches) 60-62 62-64 64-66 66-68 68-70 70-72 No of students 5 18 42 20 8 7

30. Solution: Height (inches) f x fx 60-62 5 61 350 62-64 18 63 1134 64-66 42 65 2730 66-68 20 67 1340 68-70 8 69 552 70-72 7 71 497 Total 100 6558 X = ------------ = 65.58ʺ [inches] 6558 100

31. Properties of the Mean. It measures stability . Mean is the most stable among other measures of central tendency because every score contributes to the value of the mean. The sum of each score’s distance from the mean is zero. It may be easily affected by extreme values . It can be applied to interval level of measurement. It is very easy to compute .

32. MEDIAN

33. The Median The median is simply another name for the 50th percentile . The most middle value of an arranged data is called Median. In other words, median is that quantity which divides the data into equal parts.i.e, 50% of the data lie below the median and 50% lie above the median. OR It is the score in the middle; half of the scores are larger than the median and half of the scores are smaller than the median.

34. The Median If ꭕ 1 , ꭕ 2 , ꭕ 3 , ……….. ꭕ n are the positive values and are arranged in ascending order of magnitudes, then. th value.   Median = th value .  

35. Computing the Median First Step Arrange the observations in an ordered array Second Step For an array with an odd number of terms, the median is the middle number. Third Step For an array with an even number of terms, the median is the average of the two middle numbers . Locating the Median The median’s location in an ordered array is found by (n+1)/2

36. Median Example with an Odd Number of Data Let X be an ordered array such that X has the following values: 3, 4, 5, 7, 8, 9, 11, 14, 15, 16, 16, 17, 19, 19, 20, 21, 22 There are 17 values in the ordered array Position of median = (n+1)/2 = (17+1)/2 = 9 th position Counting from left to right to the 9 th position, the median is 15 Advantage - extreme values do not distort the median Note that if 22 (maximum value) is replaced by 100, the median is still 15 If 3 (minimum value) is replaced by -103, the median is still 15

37. Median Example with an Even Number of Data Let X be an ordered array such that X assumes the following values: 3, 4, 5, 7, 8, 9, 11, 14, 15, 16, 16, 17, 19, 19, 20, 21 There are 16 values in the ordered array Position of median = (n+1)/2 = (16+1)/2 = 8.5 th position The median is a value between the 8th and 9th observations in the ordered array. The median is 14 + 0.5(15-14) = 14.5 or simply, (14+15)/2 =14.5 Advantage - extreme values do not distort the median If 21 (maximum value) is replaced by 100, the median is still 14.5 If 3 (minimum value) is replaced by -88, the median is still 14.5

38. Exercise. Example: Find median for the following data:, 7,7,2,3,4,2,7,9,31. Solution: First we arrange the values in ascending order i.e,2,2,3,4,7,7,7,9,31. Where n= 9 Than, Median = 7 Median = th value.   th value = 5 th value = 7  

39. Exercise Example 2. For the data given below, find the Median ? 36,41,27,32,29,38,39,43. Solution: Arranging in order of magnitude i.e,27,29,32,36,38,39,41,43. Where n = 8. Then, = = 4.5 th value. The median is a value between the 4th and 5th observations in the ordered array. = = 37 Median = 37.   Median = th value.  

40. Exercise #1 Find the Median for the following data set ? 12,16,08,17,19,35,39,40,7,54,5.

41. Exercise # 2 Find the Median for the following data set ? 55,65,45,11,22,36,47,51.

42. Median from the frequency distribution ( ungrouped data) In this case, the Median can be found directly by forming the column of cumulative frequency. Therfore,the value of the variable corresponding to the cumulative frequency gives the Median , where N is the total frequency i.e., Example 1. Find the Median for the following frequency distribution.   x 2 3 4 5 6 7 f 7 12 17 19 21 24

43. Solution: First we form a column of Cumulative frequency, then Where Since 50.5 lies in the cumulative frequency corresponding to 5,therefore the Median = 5.   Median = th value.   th value = 5 th (value = 50.5 )   x f C.f 2 7 7 3 12 19 4 17 36 5 19 55 6 21 76 7 24 100 Total 100= N 50.5 th value

44. Exercise Example 2. The table given below shows the number of children in the family for 47 families in a certain town. Find the median number of children per family. No of children 1 2 3 4 5 Frequency 5 17 14 11 6 4

45. Solution: Form a column of frequency. Since 24 th value lies in the cumulative frequency corresponding to 2. Therefore the median number of children per family is 2. Median = th value.   th value = 24 th   No of Children f C.f 5 5 1 7 12 2 14 26 3 11 37 4 6 43 5 4 47 Total 47 = N No of children 1 2 3 4 5 Frequency 5 17 14 11 6 4

46. Exercise The following table gives the marks of 20 students in a class test of 100 marks. Find the Median marks. Marks 50 55 58 60 63 66 75 No of Students 1 2 2 5 4 5 1

47. Solution Since. = = 61.5 marks. Therfore,10.5 th value is the mean of two middle values i.e., ( 10 th and 11 th ) in the given data. So median is 61.5 marks.   x f C.f 50 1 1 55 2 3 58 2 5 60 5 10 63 4 14 66 5 19 75 1 20 Total 20 = N Median = th value.   th value = 10.5 th value.  

48. When To Use the Median The median is often used when the distribution of scores is either positively or negatively skewed The few really large scores (positively skewed) or really small scores (negatively skewed) will not overly influence the median

49. MODE

50. MODE. The mode is the value of the observation that appears most frequently.

51. MODE. Mode is the value which repeats maximum number of times in a set of data. Mode is that point about which there is maximum concentration of frequency. Mode is the value which has a maximum frequency than others in its neighborhood. The mode or the modal score is a score or scores that occurred most in the distribution . The mode applies to all levels of data measurement: N ominal, O rdinal, I nterval , and ratio

52. Example 1 . Exercise 1. Find the Mode for the following data. 9,10,12,13,14,11,15,16,11,9,11. Mode. Solution: First we Array the data: 9,9,10,10,11,11,11,12,13,14,15,16. Since 11 is the most repeated value, therefore, Mode = 11

53. Example 2. Exercise 2. Find modal height (in inches) from the heights of 20 students. 60,65,64,58,69,72,64,64,65,60,61,67,64,63,65,64,68,63,64 and 66 . Solution : First we Array the data: Mode 58,60,60,63,63,64,64,64,64,64,64,65,65,65,66,67,68,69,72. Since 64 is the most repeated value, therefore: Mode = 64ʺ inches.

54. Example 3. Exercise 3. Find mode in the following data. 5,4,3,1,2,6,8,17,12. Solution: First Array the data: 1,2,3,4,5,6,7,8,12,17. Since each value is repeating the same number of times, therefore there is: N o Mode.

55. Example 4. Exercise 4. Determine Mode in the following data ? 2,5,5,5,6,7,7,8,8,8,8 and 9 . Solution: Since 5 and 8 are the most repeated values, therefore there are two Modes i.e., 5 and 8 Mode = 5 and 8. Bimodal.

56. Mode from a Frequency Distribution ( Ungrouped Data) Shoe size 2 3 4 5 6 Frequency 8 15 23 20 14 Shoe size Frequency 2 3 4 5 6 8 15 23 20 14 Solution: Since shoe size 4 has maximum frequency, therefore Modal size is 4 Example 4 . Find Mode in the following frequency distribution. Modal size is 4

57. TYPES OF MODE It is classified as: U nimodal. Bimodal. T rimodal or mulitimodal . Unimoda : is a distribution of scores that consists of only one mode. Bimodal : is a distribution of scores that consists of two modes. Trimodal : is a distribution of scores that consists of three modes or Multimodal is a distribution of scores that consists of more than two modes.

58. MODE. Example: Scores of 10 students in Section A, Section B and Section C. Scores of Section A Scores of Section B Scores of Section C 25 25 25 24 24 25 24 24 25 20 20 22 20 18 21 20 18 21 16 17 21 12 10 18 10 9 18 7 7 18

59. MODE The score that appeared most in Section A is 20 , hence, the mode of Section A is 20. There is only one mode, therefore, score distribution is called unimodal distribution. The modes of Section B are 18 and 24 , since both 18 and 24 appeared twice. There are two modes in Section B, hence, the distribution is a bimodal distribution. The modes of Section C are 18,21 & 25 .There are three modes for section C, Therefore it is called as trimodal or multimodal distribution.

60. Bimodal Distributions When a distribution has two “modes,” it is called bimodal

61. Multimodal Distributions If a distribution has more than 2 “modes,” it is called multimodal

62. Any peak is considered a mode, even if all peaks do not have the same height. A distribution with a single peak is called a single-peaked, or unimodal, distribution. A distribution with two peaks, even though not the same size, is a bimodal distribution.

63. When To Use the Mode The mode is not a very useful measure of central tendency It is insensitive to large changes in the data set That is, two data sets that are very different from each other can have the same mode

64. When To Use the Mode The mode is primarily used with nominally scaled data It is the only measure of central tendency that is appropriate for nominally scaled data

65. EMPIRICAL RELATION BETWEEN MEAN,MEDIAN & MODE.

66. Symmetrical distribution. A distribution is symmetric if its left half is a mirror image of its right half A symmetric distribution with a single peak and a bell shape is known as a normal distribution

67. Symmetrical distribution. A symmetric distribution with a single peak and a bell shape is known as a normal distribution Where: Mode = Mean = Median .

68. Asymmetry or Skewness A distribution is left-skewed (or negatively skewed) if the values are more spread out on the left, meaning that some low values are likely to be outliers . In this case: Mode > Median > Mean.

69. Asymmetry or Skewness A distribution is right skewed or positively skewed if the values are more spread out on the right. It has a tail pulled toward the right. In this case: Mean > Median > Mode.

70. What is the relationship between mean, median and mode of a left-skewed distribution? Find the mean, median and mode of: , 5, 10, 20, 40, 45, 45, 50, 50, 50, 60, 60, 60, 60, 60, 60, 70, 70, 70, 70, 70, 70, 70, 70. The mean is 51.5. The median is 60. The mode is 70.

71. Left-skewed Mode > Median > Mean.

72. What is the relationship between mean, median and mode of a right-skewed distribution? Find the mean, median, and mode of: 20 , 20, 20, 20, 20, 20, 20, 20, 30, 30, 30, 30, 30, 30, 45, 45, 45, 50, 50, 60, 70, 90 . The mean is 36.1. The median is 30. The mode is 20.

73. Right-Skewed Mean > Median > Mode.

74. THANKS FOR YOU ATTENTION.