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Central tendency

This document discusses various measures of central tendency including the mean, median, and mode. It defines each measure and provides examples of how to calculate each from both ungrouped and grouped data. The mean is the average value and is calculated by summing all values and dividing by the total number of values. The median is the middle value when values are arranged in order and divides the data set in half. The mode is the most frequently occurring value.

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MEASURES OF CENTRAL TENDENCY BY: Mrs.Keerthi Samuel Asst.Professor,
INTRODUCTION AND DEFINITION • Measures of central tendency is a typical value of the entire data. • It describes the characteristics of the entire data. • It is the value around which other values are distributed. • It gives us the mental picture of the central value which is generally called average or central tendency. • This value always lies between maximum and minimum values and generally , it is located in the center or middle of the distribution. • Mean, median and mode are commonly used averages.
CHARACTERISTICS • It should be easy to understand • It should be rigidly defined • It should be easy to calculate and simple to follow • It should be based on all the observations of the series. • It should not be affected by fluctuations of sampling
CLASSIFICATION AVERAGES MATHEMATICAL AVERAGE ARTHEMATIC MEAN POSITIONAL AVERAGE MEDIAN MODE
MERITS • Easy to understand • Easy to calculate • It takes into account each and every observation • Sampling fluctuations do not affect the value of mean • It is easy for further calculation
DEMERITS • It is influenced by very big or very small numbers. • example: the mean of 10,80,90,1,4 is 37 . This mean 37 neither represents big number like 80,90 nor represent small numbers like 10,1,4 properly. • Sometimes it looks ridiculous when we calculate average number of children born to the women. We may get a mean value of 2.5 children. This id evidently absurd. • It is not a positional average like median and mode. • Sometimes it may give false conclusions.
ARTHEMATIC MEAN • Mean is the simplest measurement of central tendency and is widely used measure. • It is relatively a stable measure of central tendency. • Mean is defined as sum of the values divided by the number of values in the case of ungrouped data. • Ungrouped data: 𝒙 = 𝒙 𝒏 ∑= symbol of summation n= number of observations • This formula is used in case of individual observations where frequencies are not given.
ARTHEMATIC MEAN • The duration of survival of eight patients after the diagnosis of blood cancer is as follows . Find the mean duration of survival 5,3,7,4,30,2,3,2 𝒙 = 𝒙 𝒏 =5+3+7+4+30+2+3+2 =7 8
GROUPED DATA-WITHOUT CLASS INTERVAL • If the number of observations are many it is a laborious process to calculate mean without grouping them into frequency tables. • In this method the values of the variables are multiplied by their respective frequencies and the products so obtained are totaled • The total is divided by the total number of frequencies • STEPS: • Multiply each variable by its frequency (fx) • Add all the (fx) = (∑fx) • Divide (∑fx) by the total frequency (∑f) • The formula is 𝑓𝑥 𝑓
GROUPED DATA-WITHOUT CLASS INTERVAL • 𝑥 = 𝐴𝑟𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐 𝑀𝑒𝑎𝑛 • 𝑓𝑥 = sum of fx • 𝑓= total frequency • This method corresponds to the direct method of calculation of mean for grouped data. This also can be called as grouped data without class intervals
• The following data is the number of illness for students in a hostel. Find the mean number of illness per student NUMBER OF ILLNESS(X) NUMBER OF STUDENTS(F) 0 1 2 3 4 5 6 7 24 76 114 115 86 51 26 18
𝑥 = 𝑓𝑥 𝑓 =1530/510=3 Mean number of illness per student = 3 NUMBER OF ILLNESS NUMBER OF fx 0 1 2 3 4 5 6 7 24 76 114 115 86 51 26 18 0 76 228 345 344 255 156 126
• The following data gives the distribution of students according to their age in college. find the mean age of students of the class NUMBER OF ILLNESS NUMBER OF STUDENTS 18 40 19 30 20 20 21 10 22 10
𝑥 = 𝑓𝑥 𝑓 =2120/110=19.2 Mean age of the students of the class is= 19.2 NUMBER OF ILLNESS NUMBER OF fx 18 40 720 19 30 570 20 20 400 21 10 210 22 10 220 𝑓=110 𝑓𝑥=2120
GROUPED DATA WITH CLASS INTERVAL • When the range of Values of the variable is large, the values of the variables are grouped into appropriate class intervals and the corresponding frequencies will be grouped into frequency distribution. • Mean 𝑥 = 𝛴𝑓𝑚 𝑓
• Protein intake of 400 females is given in the following table. Find the mean. PROTEIN INTAKE NO. OF FEMALES 15-25 30 25-35 40 35-45 100 45-55 110 55-65 80 65-75 30 75-85 10
𝑥 = 𝛴𝑓𝑚 𝑓 =19000/400=47.50 PROTEIN INTAKE NO. OF FEMALES Mid point of CI(m) F(m) 15-25 30 20 600 25-35 40 30 1200 35-45 100 40 4000 45-55 110 50 5500 55-65 80 60 4800 65-75 30 70 2100 75-85 10 80 800 𝛴𝑓 = 400 𝛴𝑓𝑚 = 19000
• Calculate the weighted mean of the grouped data weights of 50 students given , calculate mean Weight in kgs students 65-69 6 60-64 5 55-59 10 50-54 9 45-49 9 40-44 8 35-39 7
𝑥 = 𝛴𝑓𝑚 𝑓 =2768/54=51.25 Weight in kgs students Mid point of CI F(x) 65-69 6 67 402 60-64 5 62 310 55-59 10 57 570 50-54 9 52 468 45-49 9 47 423 40-44 8 42 336 35-39 7 37 259 𝛴𝑓 = 54 𝛴𝑓𝑥 = 2768
MEDIAN
MEDIAN • A value that divides a distribution into two equal halves or central or middle value of a series of observations, when the observed values are arranged in ascending or descending order of magnitude. • Median is just the 50th perceive value. Median is denoted by’Md’ • Like mean, median can also be calculated. • Ungrouped data • Discrete series • Continous series
MEDIAN-UNGROUPED DATA • STEP-I: • Arrange the data in ascending or descending order of magnitude • STEP-2: • If the number of items are odd, n+1/2 th item gives the serial number of the median. • If the number of items are even, the average of n/2th and n/2+1 th item gives the serial number of the median.
MEDIAN-PROBLEM 1. The diastolic blood pressure of 10 individuals is as follows. find the median 83,75,81,79,71,95,75,77,84,90 Arrange in ascending order n/2= 10/2 = 5th item in the serial =10/2+1=6th item in serial Ascending order: 71,75,75,77,79,81,83,84,90,95 5th item is 79 and 6th item is 81 Average of 5th and 6th values is : 79+81/2=160/2=80 Median = 80 𝑛 2 + 1
MEDIAN-PROBLEM 2. In a hospital ward the following are the number of days of stay of patients. Find the median days of stay in the hospital 13,42,8,9,7,3,6,52,82,11,11,10,2 Ascending order: • 2,3,6,7,8,8,9,10,11,11,13,42,52 • Patients are in odd number • 𝑛+1 2 th value =13+1/2=14/2= 7th value in the series Median=9
MEDIAN-PROBLEM(GROUPED) Find the median from the following data Step-I : find the cumulative frequencies Step-II : value Step-III : check the obtained value in relation to cumulative frequencie and the corresponding x value will be the median Size of shoe 4 5 6 7 8 9 frequency 10 15 22 16 12 5 𝑁 + 1th 2
𝑁+1th 2 = 80+1/2=81/2=40.5th value Here 40.5th value lies between 25 and 47 of cumulative frequency and falls on 47.So, we take the corresponding x value of cf 47. The corresponding ‘x’ value of cf is 6. Median is 6 . SIZE OF SHOE F cf 4 10 10 5 15 25 6 22 47 7 16 63 8 12 75 9 5 80 80
MEDIAN-PROBLEM(GROUPED) Find the median from the following data Step-I : find the cumulative frequencies Step-II : value Step-III : check the obtained value in relation to cumulative frequencies and the corresponding x value will be the median Size of shoe 4 5 6 7 8 9 frequency 10 15 22 16 12 5 𝑁 + 1th 2
MEDIAN-PROBLEM(GROUPED WITH CI) 𝑴𝒆𝒅𝒊𝒂𝒏 = 𝒍 + 𝑵 𝟐 − 𝒎 𝒇 × 𝒄 l= lower limit of the median class N= total frequency m= the cumulative frequency value above the median class F= the frequency corresponding to the median class C=class Interval
MEDIAN-PROBLEM(GROUPED WITH CI) 𝑴𝒆𝒅𝒊𝒂𝒏 = 𝒍 + 𝑵 𝟐 − 𝒎 𝒇 × 𝒄 Protein intake Number of families 15-25 30 25-35 40 35-45 100 45-55 110 55-65 80 65-75 30 75-85 10
MEDIAN-PROBLEM(GROUPED WITH CI) 𝑴𝒆𝒅𝒊𝒂𝒏 = 𝒍 + 𝑵 𝟐 − 𝒎 𝒇 × 𝒄 Protein intake Number of families cf 15-25 30 30 25-35 40 70 35-45 100 170-m or cf 45 (l) -55 110 (f) 280-Median class 55-65 80 360 65-75 30 390 75-85 10 400 N=400
MEDIAN-PROBLEM(GROUPED WITH CI) • N=Total frequency; N/2= 400/2=200 • The value 200 falls on 280 in the cumulative frequency. • Median class value is 280. the cumulative frequency value above the median class m=170. • The frequency corresponding to the median class is f=110, the lower limit corresponding to the median class is l=45;c=10. 𝑴𝒆𝒅𝒊𝒂𝒏 = 𝒍 + 𝑵 𝟐 − 𝒎 𝒇 × 𝒄 =45+ 200−170 110 x10 =45+ 300 110 =47.73grams
MEDIAN ADVANTAGES: • It is simple to calculate and easy to understand • It can be located by inspection • The value of median is unaffected by extreme values • Its value generally lies in distribution. DISADVANTAGES: • Median is not familiar average like mean. • If the data is large , it is tedious to arrange the data in ascending order. • Since, it is a positional average, the value is not influenced by each and every observation. • The median value is affected by sampling fluctuations.
MODE • Mode is the most frequently occurring value in a series of observations. • Mode is also a positional average. This is the another measure of central tendency which is least influenced by the size of the individual observations. • Mode is more useful in certain type od observations , where it is required to know the value of observation, which has high influence in the series. • Ex: when studying the age of attack , or onset of a disease it is advisable to know the maximum number of persons are affected rather than the mean age of onset or median age of onset. • Mode also can be calculated from the following relationship, if median and mean are known. Mode=3(median)-2(mean)
MODE-TYPES • The most frequently occurring value in the series of observations. • Ex: 4,5,8,6,7,5,9,5 • Mode is 5 • BIMODEL: • If two sets of observations are same or distribution with two modes is bimodal • Ex: 4,8,9,10,4,6,12,10 • Bimodel:4 and 10 • MULTIMODEL: • If more than two sets of observations are same or distribution with more than two modes is multimodel. • Ex: 5,8,9,10,5,9,16,8 • Multimodel:5,8,9
MODE-UNGROUPED DATA • The diastolic BP of 20 individuals is 85,75,81,79,71,95,75,77,75,90,71,75,79,95,75,77,84,75,81,75 The mode is 75 • If all sets of values are same there is no mode • Ex:3,7,3,4,7,4,9,9 • If different values are there is no mode • Ex:31,33,34,36,37,39,40
MODE-GROUPED DATA(discreete) Wages employees 145 3 170 16 180 8 190 20 200 6 210 2 Since the highest frequency is 20 the mode will be 190
MODE-GROUPED DATA(Continous) 𝑴𝑶𝑫𝑬 = 𝑙 + 𝑓2 𝑓1 + 𝑓2 𝑥𝐶 • f= modal class frequency • l= lower limit of the modal class • 𝑓1 = frequency preceding the model class • 𝑓2= frequency succeeding the model class • h= height of the class (or) class interval of model class
MODE-GROUPED DATA(Continous) Protein No.of families 15-25 30 25-35 40 35-45 100-f1 45(l)-55 110 -f 55-65 80-f2 65-75 30 75-85 10 𝑴𝑶𝑫𝑬 = 𝑙 + 𝑓2 𝑓1 + 𝑓2 𝑥𝐶
𝑴𝑶𝑫𝑬 = 𝑙 + 𝑓2 𝑓1 + 𝑓2 𝑥𝐶 𝑴𝑶𝑫𝑬 = 𝟒𝟓 + 𝟖𝟎 𝟏𝟎𝟎 + 𝟖𝟎 𝑥𝟏𝟎 = 𝟒𝟓 + 𝟖𝟎𝟎 𝟏𝟖𝟎 =45+4.4 =49.4 Mode = 49.4
MERITS • It is easy to understand and not affected by the extreme values • It is a positional average and can be located easily by inspection
DEMERITS • It is an average, which is ill defined and indeterminate • In the case of bimodal class, the calculation is difficult as it involves grouping and analysis • It is not based on all observations • The exact location is uncertain • Mode is not used in biological and medical sciences