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### Measures of Central Tendency - Biostatstics

• 1. Measures of Central Tendency Harshit Jadav M.S.Pharm, PGDRA
• 2. • Measures of central tendency are also usually called as the averages. • They give us an idea about the concentration of the values in the central part of the distribution. • The following are the five measures of central tendency that are in common use: • (i) Arithmetic mean, (ii) Median, (iii) Mode, (iv) Geometric mean, and (v) Harmonic mean (vi) weighted mean
• 3. MEASURE OF CENTRAL TENDANCY MEAN MEDIAN MODE The average of the data The middle value of the data most commonly occurring value
• 4. Mean (Average) Mean locate the centre of distribution. Also known as arithmetic mean Most Common Measure The mean is simply the sum of the values divided by the total number of items in the set. Affected by Extreme Values XX XX nn XX XX XX nn ii ii nn nn == == ++ ++ ++== ∑∑ 11 11 22 
• 5. Merits: • It is easy to understand and easy to calculate • It is based upon all the observations • It is familiar to common man and rigidly defined • It is capable of further mathematical treatment. • It is affected by sampling fluctuations. Hence it is more stable.
• 6. Demerits • It cannot be determined by inspection. • Arithmetic mean cannot be used if we are dealing with qualitative characteristics, which cannot be measured quantitatively like caste, religion, sex. • Arithmetic mean cannot be obtained if a single observation is missing or lost • Arithmetic mean is very much affected by extreme values.
• 7. MEAN UNGROUPED DATA GROUPED DATA n x n i ∑= = 1 ix n x n i ∑= = 1 ii xf MEAN = nsobservatioofno.total nsobservatioofSum
• 8. Birth weight of new borns are : 3.3, 6.1, 5.8, 3.8, 2.7, 4.1, 3.4, 3.9, 5.1, 3 n x n i ∑= = 1 ix =41.2/10 =4.12 kg
• 9. Q. A Survey of 100 families each having five children, revealed the following distribution No. of male children 0, 1, 2, 3, 4, 5 No. of Families 9, 24, 35, 24, 6, 2 Find the Mean of male children. x f f . X 0 9 0 1 24 24 2 35 70 3 24 72 4 6 24 5 2 10 N=100 Σ f.x = 200 Mean = X =Σ f.x / Σf X =200/ 100 = 2
• 10. Find mean days of confinement after delivery in the following series:- Day of confinement No. of patients 6 5 7 4 8 4 9 3 10 2
• 11. Day of confinement (x) No. of patients (f) X*f 6 5 30 7 4 28 8 4 32 9 3 27 10 2 20 Total 18 137 Solution:- n x n i ∑= = 1 ii xf =137/18 =7.61
• 12. Median 1.Measure of Central Tendency. 2. The median is determined by sorting the data set from lowest to highest values and taking the data point in the middle of the sequence. 3.Middle Value In Ordered Sequence • If Odd n, Middle Value of Sequence • If Even n, Average of 2 Middle Value 4.Not Affected by Extreme Values
• 13. Merits: • It is rigidly defined • It is easy to understand and easy to calculate. • It is not at all affected by extreme values. • It can be calculated for distributions with open-end classes. • Median is the only average to be used while dealing with qualitative data. • Can be determined graphically.
• 14. Demerits: • In case of even number of observations median cannot be determined exactly. • It is not based on all the observations. • It is not capable of further mathematical treatment
• 15. For ungrouped data:- Step-1 Arranged data in ascending or descending order. Step:-2 If total no. of observations ‘n’ is odd then used the following formula for median Step:-3 If total no. of observations ‘n’ is even then used the following formula for median = arithmetic mean of two middle observations. . 2 1 nobservatioth n + =
• 16. Median If X1, X2, X3,... …. ,Xn are n observations arranged in ascending order of magnitude. X(n+1)/2 if n is odd Median = { Xn/2 + X(n/2)+1 if n is even ------------------ 2
• 17. Calculate the median for the following series :- 2,3,5,1,4,5,8 1,2,3,4,5,5,8. Median . 2 1 nobservatioth n + = = 7+1/2 =4th number
• 18. 74+75 Median = ---------- = 74.5 2 The data on pulse rate per minute of 10 healthy individuals are 82, 79, 60, 76, 63,81, 68, 74, 60, 75. n= 10 60, 60, 63, 68, 74,75, 76, 79, 81, 82 Xn/2 + X(n/2)+1 / 2
• 19. Find out the median for number of sports injuries happened in cricket in all teams 37, 57, 65, 46, 12, 14, 19, 23, 56, 78, 5, 33
• 20. Median:- f hc n l       − + 2 l = lower limit of class interval where the median occurs f = Frequency of the class where median occurs h = Width of the median class C= Cumulative frequency of the class preceding the median class (PCF) For Grouped data:-
• 21. Weight of infant in kg No of infants 2.0-2.5 37 2.5-3.0 117 3.0-3.5 207 3.5-4.0 155 4.0-4.5 48 4.5 and above 26 Find the median weight of 590 infants born in a hospital in one year from the following table.
• 22. Weight of infants in kg No of infants Cumulative frequency 2.0-2.5 37 37 2.5-3.0 117 37+117=154 3.0-3.5 207 154+207=361 3.5-4.0 155 361+155=516 4.0-4.5 48 516+48=564 4.5 and above 26 564+26=590 N/2 =590/2 =295 Median Class = 3.0-3.5, so L=3.0, f= 207 Cf = 154 h=0.5
• 23. f hc n l       − + 2 = 3.0 + (295-154) * 0.5 207 = 3.0 + 0.34 = 3.34 Median Weight
• 24. For grouped Data:- Class interval Frequency 5-9 2 10-14 11 15-19 26 20-24 17 25-29 8 30-34 6 35-39 3 40-44 2 45-49 1 Calculate the median for the following data series:-
• 25. Solution:- Class interval Frequency cumulative frequency 5-9 2 2 10-14 11 13 15-19 26 39 20-24 17 56 25-29 8 64 30-34 6 70 35-39 3 73 40-44 2 75 45-49 1 76
• 26. n=76 l = lower limit of class interval where the median occurs = 15 h = Width of the median class = 4 f = Frequency of the class where median occurs = 26 C = Cumulative frequency of the class preceding the median class =13 f hc n l       − + 2
• 27. Mode 1.Measure of Central Tendency 2.The mode is the most frequently occurring value in the data set. 3.May Be No Mode or Several Modes
• 28. Merits: • Mode is readily comprehensible and easy to calculate. • Mode is not at all affected by extreme values. • Mode can be conveniently located even if the frequency distribution has class intervals of unequal magnitude • Open-end classes also do not pose any problem in the location of mode. • Mode is the average to be used to find the ideal size.
• 29. Demerits: • Mode is ill defined. • It is not based upon all the observations. • It is not capable of further mathematical treatment. • As compared with mean, mode is affected to a great extent by fluctuations of sampling.
• 30. Mode Example No Mode Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7 One Mode Raw Data: 6.3 4.9 8.9 6.3 4.9 4.9 More Than 1 Mode Raw Data: 21 28 28 41 43 43
• 31. Mode for ungrouped data:- 2,2,3,4,6,7,4,4,4,4,8,9,0 mode is 4 10,10,3,3,4,2,1,6,7 mode is 10 and 3 10,34,23,12,11,3,4 no mode
• 32. Mode for Group Data Fm-F1 Mode = L1 + ------------------ * c 2Fm– F1- F2 Where •Fm is mode freq. • F1 is freq. just lower mode class. •F2 is freq. after mode class. • L1 is lower limit of mode class •C is class difference.
• 33. C I FREQU. (F) 20 - 30 3 30 - 40 20 40 - 50 27 50 - 60 15 60 - 70 9 Q. Find the Mode for group data Age group 20-30 30-40 40-50 50-60 60-70 No. of persons 3 20 27 15 9
• 34. Q. Find the Mode for group data Age group 20-30 30-40 40-50 50-60 60-70 No. of persons 3 20 27 15 9 C I FREQU. (F) 20 - 30 3 30 - 40 20 40 - 50 27 50 - 60 15 60 - 70 9 L1 Fm F1 F2
• 35. FFmm -- FF11 Mode = LMode = L11 + ------------------ * c+ ------------------ * c 2F2Fmm – F– F11- F- F22 2727 –– 20 7020 70 Mode = 40 + ------------------ * 10 = 40 + -----Mode = 40 + ------------------ * 10 = 40 + ----- 2*272*27 – 20- 15 19– 20- 15 19 Mode = 43.68Mode = 43.68
• 36. Calculate the mode for the following frequency distribution:- IQ Range Frequency 90-100 11 100-110 27 110-120 36 120-130 38 130-140 43 140-150 28 150-160 16 160-170 1
• 37. Modal class by inspection is 130-140 fm= 43 f1= 38 f2= 28 C=10 l = 130 c fff ff l m m       −− − + )(2 )( 21 1 =130.6579
• 38. Relationship between Mean, Median and Mode Mode = 3 Median – 2 Mean
• 39. Summary of Central Tendency Measures MeasureMeasure DescriptionDescription MeanMean Balance PointBalance Point MedianMedian Middle ValueMiddle Value When OrderedWhen Ordered ModeMode Most FrequentMost Frequent
• 40. Ex. Calculate Mean, Median, Mode. Age Group No. of Patients 25-30 4 30-35 3 35-40 2 40-45 3 45-50 4 50-55 8 55-60 6
• 41. Age Group No. of Patients (F) X F*X C.F 25-30 4 27.5 110 4 30-35 3 32.5 97.5 7 35-40 2 37.5 75 9 40-45 3 42.5 127.5 12 45-50 4 47.5 190 16 50-55 8 52.5 420 24 55-60 6 57.5 345 30 30 1365 MEAN =1365/30 = 45.5 MEDIAN =45+ (15-12)*5/4 = 48.75 MODE=50 + [(8-4)*5/(2*8-4-6)]=53.34
• 42. Ex. The following table gives the frequency distribution of marks obtained by 2300 medical students of Gujarat in MCQ of PSM exam. Find Mean, Median and Mode. Marks No. of students 11-20 141 21-30 221 31-40 439 41-50 529 51-60 495 61-70 322 71-80 153
• 43. Geometric Mean • Geometric mean is defined as the positive root of the product of observations. Symbolically, • It is also often used for a set of numbers whose values are meant to be multiplied together or are exponential in nature, such as data on the growth of the human population or interest rates of a financial investment. • Find geometric mean of rate of growth: 34, 27, 45, 55, 22, 34 n nxxxxG /1 321 )( =
• 44. Geometric mean of Group data • If the “n” non-zero and positive variate-values occur times, respectively, then the geometric mean of the set of observations is defined by: [ ] Nn i f i Nf n ff in xxxxG 1 1 1 21 21       == ∏=  ∑= = n i ifN 1 Where nxxx ,........,, 21 nfff ,.......,, 21
• 45. Geometric Mean (Revised Eqn.) )( 321 nxxxxG =         = ∑= n i ixLog N AntiLogG 1 1         = ∑= n i ii xLogf N AntiLogG 1 1 )( 321 321 n fff xxxxG = Ungroup Data Group Data
• 47. What is the geometric mean of 4,8.3,9 and 17?
• 48. First, multiply the numbers together and then take the 5th root (because there are 5 numbers) G = (4*8*3*9*17)(1/5) G = 6.81
• 49. Harmonic Mean • Harmonic mean (formerly sometimes called the subcontrary mean) is one of several kinds of average. • The harmonic mean is a very specific type of average. • It’s generally used when dealing with averages of units, like speed or other rates and ratios.
• 50. Harmonic Mean Group Data • The harmonic mean H of the positive real numbers x1,x2, ..., xn is defined to be ∑= = n i i i x f n H 1 ∑= = n i ix n H 1 1 Ungroup Data Group Data
• 51. What is the harmonic mean of 1,5,8,10? Here, N=4 H = 4 / (1/1) +(1/5) + (1/8) + (1/10) H = 4/ 1.425 H = 2.80 ∑= = n i ix n H 1 1
• 52. Rahul drives a car at 20 mph for the first hour and 30 mph for the second. What’s his average speed?
• 53. We need the harmonic mean: = 2/(1/20 + 1/30) = 2(0.05 + 0.033) = 2 / 0.083 = 24.09624 mph.
• 54. Weighted Mean • The Weighted mean of the positive real numbers x1,x2, ..., xn with their weight w1,w2, ..., wn is defined to be ∑ ∑ = = = n i i n i ii w xw x 1 1 Σ = the sum of (in other words…add them up!). w = the weights. x = the value.
• 55. A weighted mean is a kind of average. Instead of each data point contributing equally to the final mean, some data points contribute more “weight” than others. If all the weights are equal, then the weighted mean equals the arithmetic mean (the regular “average” you’re used to). Weighted means are very common in statistics, especially when studying populations.
• 56. Steps: 1.Multiply the numbers in your data set by the weights. 2.Add the numbers in Step 1 up. Set this number aside for a moment. 3.Add up all of the weights. 4.Divide the numbers you found in Step 2 by the number you found in Step 3.
• 57. You take three 100-point exams in your statistics class and score 80, 80 and 95. The last exam is much easier than the first two, so your professor has given it less weight. The weights for the three exams are: •Exam 1: 40 % of your grade. (Note: 40% as a decimal is .4.) •Exam 2: 40 % of your grade. •Exam 3: 20 % of your grade
• 58. 1.Multiply the numbers in your data set by the weights: .4(80) = 32 .4(80) = 32 .2(95) = 19 2. Add the numbers up. 32 + 32 + 19 = 83. 3. (0.4+0.4+0.2) = 1 4. 83/1= 83
• 59. The arithmetic mean is best used when the sum of the values is significant. For example, your grade in your statistics class. If you were to get 85 on the first test, 95 on the second test, and 90 on the third test, your average grade would be 90. Why don't we use the geometric mean here? What about the harmonic mean?
• 60. What if you got a 0 on your first test and 100 on the other two? The arithmetic mean would give you a grade of 66.6. The geometric mean would give you a grade of 0!!! The harmonic mean can't even be applied at all because 1/0 is undefined.
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