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Measures of Central Tendency

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Measures of Central Tendency 1 UNIT-1 FREQUENCY DISTRIBUTION Structure: 1.0 Introduction 1.1 Objectives 1.2 Measures of Central Tendency 1.2.1 Arithmetic mean 1.2.2 Median 1.2.3 Mode 1.2.4 Empirical relation among mode, median and mode 1.2.5 Geometric mean 1.2.6 Harmonic mean 1.3 Partition values 1.3.1 Quartiles 1.3.2 Deciles 1.3.3 Percentiles 1.4 Measures of dispersion 1.4.1 Range 1.4.2 Semi-interquartile range 1.4.3 Mean deviation 1.4.4 Standard deviation 1.2.5 Geometric mean 1.5 Absolute and relative measure of dispersion 1.6 Moments 1.7 Karl Pearson’s β and γ coefficients 1.8 Skewness 1.9 Kurtosis 1.10 Let us sum up 1.11 Check your progress : The key. 2 1.0 INTRODUCTION According to Simpson and Kafka a measure of central tendency is typical value around which other figures aggregate‘. According to Croxton and Cowden ‗An average is a single value within the range of the data that is used to represent all the values in the series. Since an average is somewhere within the range of data, it is sometimes called a measure of central value‘. 1.1 OBJECTIVES The main aim of this unit is to study the frequency distribution. After going through this unit you should be able to : describe measures of central tendency ; calculate mean, mode, median, G.M., H.M. ; find out partition values like quartiles, deciles, percentiles etc; know about measures of dispersion like range, semi-inter-quartile range, mean deviation, standard deviation; calculate moments, Karls Pearsion’s β and γ coefficients, skewness, kurtosis. 1.2 MEASUIRES OF CENTRAL TENDENCY The following are the five measures of average or central tendency that are in common use : (i) Arithmetic average or arithmetic mean or simple mean (ii) Median (iii) Mode (iv) Geometric mean (v) Harmonic mean Arithmetic mean, Geometric mean and Harmonic means are usually called Mathematical averages while Mode and Median are called Positional averages. 3 1.2.1 ARITHMETIC MEAN To find the arithmetic mean, add the values of all terms and them divide sum by the number of terms, the quotient is the arithmetic mean. There are three methods to find the mean : (i) Direct method: In individual series of observations x1, x2,… xn the arithmetic mean is obtained by following formula. x x x x............. x x AM.. 1 2 3 4nn 1 n (ii) Short-cut method: This method is used to make the calculations simpler. Let A be any assumed mean (or any assumed number), d the deviation of the arithmetic mean, then we have fd MA. ( d=(x-A)) N (iii)Step deviation method: If in a frequency table the class intervals have equal width, say i than it is convenient to use the following formula. fu M A i n where u=(x-A)/ i ,and i is length of the interval, A is the assumed mean. Example 1. Compute the arithmetic mean of the following by direct and short -cut methods both: Class 20-30 30-40 40-50 50-60 60-70 Freqyebcy 8 26 30 20 16 Solution. Class Mid Value f fx d= x-A f d x A = 45 20-30 25 8 200 -20 -160 30-40 35 26 910 -10 -260 40-50 45 30 1350 0 0 50-60 55 20 1100 10 200 6070 65 16 1040 20 320 Total N = 100 ∑ fx = 4600 ∑f d = 100 By direct method M = (∑fx)/N = 4600/100 = 46. By short cut method. Let assumed mean A= 45. 4 M = A + (∑ fd )/N = 45+100/100 = 46. Example 2 Compute the mean of the following frequency distribution using step deviation method. : Class 0-11 11-22 22-33 33-44 44-55 55-66 Frequency 9 17 28 26 15 8 Solution. Class Mid-Value f d=x-A u = (x-A)/i fu (A=38.5) i=11 0-11 5.5 9 -33 -3 -27 11-22 16.5 17 -22 -2 -34 22-33 27.5 28 -11 -1 -28 33-44 38.5 26 0 0 0 44-55 49.5 15 11 1 15 55-66 60.5 8 22 2 16 Total N = 103 ∑fu = -58 Let the assumed mean A= 38.5, then M = A + i(∑fu )/N = 38.5 + 11(-58)/103 = 38.5 - 638/103 = 38.5 - 6.194 = 32.306 PROPERTIES OF ARITHMETIC MEAN Property 1 The algebraic sum of the deviations of all the variates from their arithmetic mean is zero. Proof . Let X1, X2,… Xn be the values of the variates and their corresponding frequencies be f1, f2, …, fn respectively. Let xi be the deviation of the variate Xi from the mean M, where i = 1,2, …, n. Then Xi = Xi –M, i = 1,2,…, n. nn fixi fii() X M ii11 nn = M fii M f ii11 5 = 0 Exercise 1(a) Q.1) Marks obtained by 9 students in statistics are given below. 52 75 40 70 43 65 40 35 48 calculate the arithmetic mean. Q.2) Calculate the arithmetic mean of the following distribution Variate : 6 7 8 9 10 11 12 Frequency: 20 43 57 61 72 45 39 Q.3) Find the mean of the following distribution Variate : 0-10 10-20 20-30 30-40 40-50 Frequency: 31 44 39 58 12 1.2.2 MEDIAN The median is defined as the measure of the central term, when the given terms (i.e., values of the variate) are arranged in the ascending or descending order of magnitudes. In other words the median is value of the variate for which total of the frequencies above this value is equal to the total of the frequencies below this value. Due to Corner, ―The median is the value of the variable which divides the group into two equal parts one part comprising all values greater, and the other all values less then the median‖. For example. The marks obtained, by seven students in a paper of Statistics are 15, 20, 23, 32, 34, 39, 48 the maximum marks being 50, then the median is 32 since it is the value of the 4th term, which is situated such that the marks of 1st, 2nd and 3rd students are less than this value and those of 5th, 6th and 7th students are greater then this value. COMPUTATION OF MEDIAN (a)Median in individual series. Let n be the number of values of a variate (i.e. total of all frequencies). First of all we write the values of the variate (i.e., the terms) in ascending or descending order of magnitudes Here two cases arise: 6 Case 1. If n is odd then value of (n+1)/2th term gives the median. th th n 1 Case2. If n is even then there are two central terms i.e., n/2 and The mean of 2 these two values gives the median. (b) Median in continuous series (or grouped series). In this case, the median (Md) is computed by the following formula n cf M l 2 i d f Where Md = median l = lower limit of median class cf = total of all frequencies before median class f = frequency of median class i = class width of median class. Example 1 – According to the census of 1991, following are the population figure, in thousands, of 10 cities : 1400, 1250, 1670, 1800, 700, 650, 570, 488, 2100, 1700. Find the median. Solution. Arranging the terms in ascending order. 488, 570, 650, 700, 1250, 1400, 1670, 1800, 2100. Here n=10, therefore the median is the mean of the measure of the 5th and 6th terms. Here 5th term is 1250 and 6th term is 1400. Median (Md) = (1250+14000)/2 Thousands = 1325 Thousands Examples 2. Find the median for the following distribution: Wages in Rs. 0-10 10-20 20-30 30-40 40-50 No. of workers 22 38 46 35 20 Solution . We shall calculate the cumulative frequencies. 7 Wages in Rs. No. of Workers f Cumulative Frequencies (c.f.) 0-10 22 22 10-20 38 60 20-30 46 106 30-40 35 141 40-50 20 161 n th st st Here N = 161. Therefore median iscf the measure of (N + 1)/2 term i.e 81 term. Clearly 81 term is situated in theM class l20 -230. Thus i 20-30 is the median class. Consequently. d f Median = 20 + (½ 161 – 60) / 46 10 = 20 + 205/46 = 20 + 4.46 = 24.46. Example 3. Find the median of the following frequency distribution: Marks No. of students Marks No. of students Less than 10 15 Less than 50 106 Less than 20 35 Less than 60 120 Less than 30 60 Less than 70 125 Less than 40 84 Solution . The cumulative frequency distribution table : Class (Marks) Frequency f Cumulative (No. of students) Frequency (C. F.) 0-10 15 15 10-20 20 35 20-30 25 60 30-40 24 84 40-50 22 106 50-60 14 120 60-70 5 125 Total N = 125 th 125 1 Median = measure of term 2 8 = 63rd term. Clearly 63rd term is situated in the class 30-40. Thus median class = 30 - 40 Median n =30 + (125/2 – 60) / 24cf 10 M l 2 i = 30 + 25/24d f = 30+1.04 = 31.04 1.2.3 MODE The word ‗mode is formed from the French word ‗La mode‘ which means ‗in fashion‘. According to Dr.
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