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2 Measures of Central Tendency MEASURES OF CENTRAL 2 TENDENCY INTRODUCTION A frequency distribution, in general, shows clustering of the data around some central value. Finding of this central value or the average is of importance, as it gives the most representative value of the whole group. Different methods give different averages which are known as the measures of central tendency. The commonly used measures of central values are mean, median, mode, geometric mean and harmonic mean. There are two main objectives of the study of averages. (i) To get single value that describes the characteristic of the entire group. Measures of central value, by condensing the mass of data in one single value, enable us to get a bird’s-eye view of the the entire data. (ii) To facilitate comparison, measures of central value, by reducing the mass of data to one single figure, enable comparison to be made. Comparison can be made either at a point of time or over a period of time. For example, we can compare the percentage results of the students of different colleges in a certain examination. Since an average is a single value representing a group of values, it is desired that the value satisfies the following properties: (i) Easy to understand (ii) Simple to compute (iii) Based on all the items (iv) Not be undully affected by extreme observations (v) Capable of further algebraic treatment (vi) Sampling stability Definition A numerical expression which is used to present a whole series should neither have the lowest value nor the heighest value in the series, but a value some- where between these two limits, possibly in the centre, where most of the items of the series cluster, such figures are called measures of central tendency or measures of locations or averages. 40 Statistics for Managers __1 f1 f2 fn n GM = [ (x1) (x2) ... (xn ) ] __1 \ log GM = n [f1 log x1 + f2 log x2 + ... fn log xn] S________ fi log xi \ GM = Antilog ( n ) Harmonic Mean If x1, x2, x3 ... xn be a set of n observations then the harmonic mean is defined as the reciprocal of the (arithmetic) mean of the reciproculs of the quantities. 1 Thus, HM = __________________ __1 __1 __1 __1 + + ... + n ( x1 x2 xn ) In a frequency distribution harmonic mean is given by 1 HM = __________________ 1 f1 f2 fn __ __ + __ + ... + __ N ( x1 x2 xn ) where N = Â fi MEDIAN If the values of the variable are arranged in the ascending or descending order of magnitude, the median is the middle item if the number is odd and is the mean of the two middle items if the number is even. Thus, the median is equal to the mid-value i.e. the value which divides the total frequency into two equal parts. For computation of median, it is necessary that the items be arranged in ascending order. 1. Ungrouped frequency distribution: Arrange the n values of the variate in ascending or descending order. n + 1 (a) When n is odd, the middle value, i.e. _____ th value gives the median. ( 2 ) n n (b) When n is even, there are two middle values __ th and __ + 1 th. The arithmetic mean 2 ( 2 ) of these two values gives the median. 2. For grouped frequency distribution N __ – C ( 2 ) Median = l + _______ × h f where l is the lower limit of the median class N is the total frequency f is the frequency of this class h is the width of the class and C is the cumulative frequency upto the class preceding the median class. 44 Statistics for Managers 52.5-60.5 5 56.5 3 15 60.5-68.5 8 64.5 4 32 68.5-76.5 2 72.5 5 10 Sfi = 90 Sfiui = –13 Here h = 8 A = 32.5 hSf u _____i i Mean = A + Sfi 8 × –13 = 32.5 + _______ 90 = 31.35 Example 4 The mean of 200 items was 50. Later on, it was found that two items were read as 92 and 8 instead of 192 and 88. Find the correct mean. _ Solution Here n = 200 and incorrect value of x = 50 _ ___Sx _ Since x = n \ Sx = n x \ Incorrect total = 200 × 50 = 10,000 \ Correct total = 10,000 – (92 + 8) + (192 + 88) = 10180 10180 \ Correct mean = ______ = 50.9 200 Example 5 Find the missing frequency from the following data. Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 5 15 20 – 20 10 Given AM is 34 marks. Solution Let f1 be the missing frequency. _ Now, AM = x = 34 _ Sf x ____i i x = Sfi 2200 + 35f1 \ 34 = __________ 70 + f1 \ 34 (70 + f1) = 2200 + 35 f1 \ fi = 180 Example 6 Calculate the missing frequency from the following data. Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 4 7 12 – 22 11 3 Measures of Central Tendency 45 Given AM is 37 Solution Let fi be the missing frequency. _ Now AM x = 37 _ Sfixi x = ____ Sfi 2955 + 35f __________1 37 = 59 + f1 \ 37(59 + fi) = 2955 + 35fi \ 37fi – 35fi = 2955 – 37 × 59 \ 2fi = 772 \ fi = 386 Example 7 Daily income of ten families of a particular place is given below. Find geometric mean. 85, 70, 15, 75, 500, 8, 45, 250, 40, 60. Solution To find GM first, we calculate log x as follows. x 85 70 15 75 500 2 45 250 40 36 log x 1.9294 1.8457 1.1761 1.8757 2.6990 0.9030 1.6532 2.3979 1.6021 1.5563 \ S log x = 17.6373 S_______ log x \ GM = Antilog ( n ) 17.6373 = Antilog _______ = 58.03. ( 10 ) Example 8 Calculate the geometric mean from the following data. x 125, 1462, 38, 7, 0.22, 0.08, 12.75, 0.5 Solution First find log x x 125 1462 38 7 0.22 0.08 12.75 0.5 _ _ _ log x 2.0969 3.1650 1.5798 0.8451 1 .3424 2 .9031 1.1055 1 .6990 \ Sx = 10.7368 and n = 8 S log x 10.7368 \ GM = Antilog _______ = Antilog _______ ( n ) ( 10 ) = 6.952 Example 9 Find the geometric mean for the following data given below. Marks 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 36-40 Frequency 6 10 18 30 15 12 10 6 2 Solution To find GM, we form the table as follows. 46 Statistics for Managers Class Midpoint m f log m f × log m 4-8 6 6 0.7782 4.6692 8-12 10 10 1.0000 10.0000 12-16 14 18 1.1461 20.6298 16-20 18 30 1.2553 37.6590 20-24 22 15 1.3224 20.1360 24-28 26 12 1.4150 16.9800 28-32 30 10 1.4771 14.7710 32-36 34 6 1.5315 9.1890 36-40 38 2 1.5798 3.1596 N = 109 Sf × log m = 137.1936 Sf × log m \ GM = Antilog __________ ( N ) 137.1936 = Antilog ________ = Antilog (1.2587) ( 109 ) = 18.14 Example 10 Find the harmonic mean from the following data. x 2574 475 75 5 0.8 0.08 0.005 0.0009 __1 Solution To find harmonic mean, calculate x . x 2574 475 75 5 0.8 0.08 0.005 0.0009 __1 x 0.0004 0.0021 0.0133 0.2000 1.2500 12.500 200.00 1111.1111 N 8 Here n = 8 HM = ___ = _________ __1 1325.0769 S x = 0.006. Example 11 From the following data, compare the harmonic mean. Marks 10 20 25 40 50 Students 20 30 50 15 5 Solution To find HM we form the table as follows. Marks (x) 10 20 25 40 50 Students (f) 20 30 50 15 5 f/x 2.00 1.500 2.00 0.375 0.100 Here N = 120 and Sf/x = 5.975 N 120 \ HM = ____ = _____ = 20.08. Sf/x 5.975 Measures of Central Tendency 47 Example 12 Compute the harmonic mean from the following data. Class 10-20 20-30 30-40 40-50 50-60 Frequency 4 6 10 7 3 Solution To find the harmonic mean, we form the table as follows. Class Freq. Midpoint f/m f m 10-20 4 15 0.267 20-30 6 25 0.240 30-40 10 35 0.286 40-50 7 45 0.158 50-60 3 55 0.055 N = 30 Sf/m = 1.004 N 30 \ Harmonic mean = ____ = _____ Sf/m 1.004 = 29.88. Example 13 An incomplete frequency distribution is given below Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 12 30 – 65 – 25 18 Given that the total frequency is 229 and median is 46. Find the missing frequencies. Solution Let f1 and f2 be the missing frequencies of the classes 30-40 and 50-60 respectively. Since the median lies in the class 40-50 N __ – C ( 2 ) Median = l + _______ × h f 229 ____ – (12 + 30 + f ) [ 2 1 ] 46 = 40 + __________________ × 10 65 229 ____ – (42 + f ) [ 2 1 ] \ 46 – 40 = ______________ × 10 65 6 × 65 229 \ ______ = ____ – 42 – f 10 2 1 229 390 \ f = ____ – ____ – 42 1 2 10 = 33.5 34 \ f1 = 34 And f2 = 229 – (12 + 30 + 34 + 65 + 25 + 18) = 45 \ Missing frequencies are f1 = 34 and f2 = 45 48 Statistics for Managers Example 14 Calculate median, the lower quartile, upper quartile, semi-interquartile range and the mode for the following data.
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