MEASURES OF CENTRAL TENDENCY INTRODUCTION

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1 2 MEASURES OF CENTRAL TENDENCY INTRODUCTION A requency distribution, in general, shows clustering o the data around some central value. Finding o this central value or the average is o importance, as it gives the most representative value o the whole group. Dierent methods give dierent averages which are known as the measures o central tendency. The commonly used measures o central values are mean, median, mode, geometric mean and harmonic mean. There are two main objectives o the study o averages. (i) To get single value that describes the characteristic o the entire group. Measures o central value, by condensing the mass o data in one single value, enable us to get a bird s-eye view o the the entire data. (ii) To acilitate comparison, measures o central value, by reducing the mass o data to one single igure, enable comparison to be made. Comparison can be made either at a point o time or over a period o time. For example, we can compare the percentage results o the students o dierent colleges in a certain examination. Since an average is a single value representing a group o values, it is desired that the value satisies the ollowing properties: (i) Easy to understand (ii) Simple to compute (iii) Based on all the items (iv) Not be undully aected by extreme observations (v) Capable o urther algebraic treatment (vi) Sampling stability Deinition 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 somewhere between these two limits, possibly in the centre, where most o the items o the series cluster, such igures are called measures o central tendency or measures o locations or averages.

2 40 Statistics or Managers GM = [(x 1 ) 1 (x 2 ) (xn) n ] n \ log GM = n 1 [ 1 log x log x n log x n ] \ GM = Antilog ( S i log x i n ) Harmonic Mean I x 1, x 2, x 3... x n be a set o n observations then the harmonic mean is deined as the reciprocal o the (arithmetic) mean o the reciproculs o the quantities. 1 Thus, HM = 1 n ( x x x n ) where In a requency distribution harmonic mean is given by 1 HM = 1 N ( 1 x x n 2 x n ) MEDIAN N = Â i I the values o the variable are arranged in the ascending or descending order o magnitude, the median is the middle item i the number is odd and is the mean o the two middle items i the number is even. Thus, the median is equal to the mid-value i.e. the value which divides the total requency into two equal parts. For computation o median, it is necessary that the items be arranged in ascending order. 1. Ungrouped requency distribution: Arrange the n values o the variate in ascending or descending order. (a) When n is odd, the middle value, i.e. ( n ) th value gives the median. (b) When n is even, there are two middle values n 2 th and ( n ) th. The arithmetic mean o these two values gives the median. 2. For grouped requency distribution ( N 2 C ) Median = l + h where l is the lower limit o the median class N is the total requency is the requency o this class h is the width o the class and C is the cumulative requency upto the class preceding the median class.

3 44 Statistics or Managers S i = 90 S i u i = 13 Here h = 8 A = 32.5 Mean = A + hs i u i S i = = Example 4 The mean o 200 items was 50. Later on, it was ound that two items were read as 92 and 8 instead o 192 and 88. Find the correct mean. Solution Here n = 200 and incorrect value o _ x = 50 Since _ Sx x = n \ Sx = n _ x \ Incorrect total = = 10,000 \ Correct total = 10,000 (92 + 8) + ( ) = \ Correct mean = = 50.9 Example 5 Find the missing requency rom the ollowing data. Marks No. o students Given AM is 34 marks. Solution Let 1 be the missing requency. Now, AM = _ x = 34 _ S i x i x = S i \ 34 = \ 34 ( ) = \ i = 180 Example 6 Calculate the missing requency rom the ollowing data. Class Frequency

4 Measures o Central Tendency 45 Given AM is 37 Solution Let i be the missing requency. Now AM _ x = 37 _ S i x i x = S i = \ 37(59 + i ) = i \ 37 i 35 i = \ 2 i = 772 \ i = 386 Example 7 Daily income o ten amilies o a particular place is given below. Find geometric mean. 85, 70, 15, 75, 500, 8, 45, 250, 40, 60. Solution To ind GM irst, we calculate log x as ollows. x log x \ S log x = \ GM = Antilog ( S log x n ) = Antilog ( ) = Example 8 Calculate the geometric mean rom the ollowing data. x 125, 1462, 38, 7, 0.22, 0.08, 12.75, 0.5 Solution First ind log x x _ log x \ Sx = and n = 8 \ GM = Antilog ( S log x = n ) = Antilog ( ) Example 9 Find the geometric mean or the ollowing data given below. Marks Frequency Solution To ind GM, we orm the table as ollows.

5 46 Statistics or Managers Class Midpoint m log m log m N = 109 S log m = S \ GM = Antilog log m ( N ) = Antilog ( ) = Antilog (1.2587) = Example 10 Find the harmonic mean rom the ollowing data. x Solution To ind harmonic mean, calculate 1 x. x x Here n = 8 HM = N S 1 x = = Example 11 From the ollowing data, compare the harmonic mean. Marks Students Solution To ind HM we orm the table as ollows. Marks (x) Students () /x Here N = 120 and S/x = \ N HM = S/x = =

6 Example 12 Compute the harmonic mean rom the ollowing data. Measures o Central Tendency 47 Class Frequency Solution To ind the harmonic mean, we orm the table as ollows. Class Freq. Midpoint /m m N = 30 S/m = \ Harmonic mean = N S/m = = Example 13 An incomplete requency distribution is given below Class Frequency Given that the total requency is 229 and median is 46. Find the missing requencies. Solution Let 1 and 2 be the missing requencies o the classes and respectively. Since the median lies in the class ( N 2 C ) Median = l + h [ ( ] ) 46 = [ ( ] ) \ = \ = \ 1 = = \ 1 = 34 And 2 = 229 ( ) = 45 \ Missing requencies are 1 = 34 and 2 = 45

7 48 Statistics or Managers Example 14 Calculate median, the lower quartile, upper quartile, semi-interquartile range and the mode or the ollowing data. Class Marks Frequency Solution We orm the table as ollows. Class Frequency Cumulative requency N = 49 (1) Median ( or N 2 = 49 2 ) = 24.5 alls in the class and is given by N 2 C ) Median = l + h where l = 15, N = 49, C = 11, h = 5 and = 15 ( 49 2 ) 11 \ Median = = 19.5 marks. 15 (2) Lower quartile (Q 1 ) ( or N 4 = ) alls in the class ( N 4 C ) \ Q 1 = l + h where l = 15, N = 49, C = 11, h = 5 and = 15 ( 49 4 ) 11 \ Q 1 = = 15.4 marks 15 3N (3) Upper quartile (Q 3 ) ( or 4 ) = alls in the class N 4 ) C Q 3 = l + h

8 Measures o Central Tendency 49 where l = 25, N = 49, C = 36, h = 5 and = 5 \ Q 3 = 25 + ( ) 36 5 = marks (4) Semi-interquartile range Q = 1 2 (Q 3 Q 1 ) = 1 ( ) = (5) Mode: It is seen that the mode value alls in the class m 1 Mode = l + h 2 m 1 2 where l = 15, m = 15, 1 = 6, 2 = 10 and h= = = 18.2 marks Example 15 From the ollowing marks ind Q 1, Q 2 and Q 3 marks. 23, 48, 34, 68, 15, 36, 24, 54, 65, 75, 92, 10, 70, 61, 20, 47, 83, 19, 77 Solution Let us irst arrange the given data in the ascending order. SN x SN x Here n = 19 (i) Q 1 is the size o ( n ) ( th item = = 5th item = 23 (ii) Q 2 is the size o ( n + 1 (iii) Q 3 is size o 3 ( n ) ( th item ) th item 2 ) th item = 10th item = 48 4 ) = 15th item = 70 4 ) th item = 3 ( Example 16 Find the median value o the ollwing data. 43, 62, 15, 80, 56, 72, 34, 8, 25 Solution Arrange the data in an array orm with serial numbers S. No x

9 50 Statistics or Managers Median is the size o ( n ) th item \ median = 43 = ( ) th Example 17 Find the median value o the ollowing. 36, 5, 19, 26, 6, 28, 56, 18, 63, 4 Solution Arrange the data in an array orm with serial numbers S. No x Median is the size o ( n ) th item = = 5.5th item 2 The item lies in between 5th and 6th item. \ Median = = = 22.5 Example 18 Locate median and quartiles rom the ollowing data. Size o shoes Frequency Solution x CF (1) Q 1 is the size o ( n ) th item = = 80.25th item (lies in 100 CF). 4 Against 100 CF, the value is 5.0 \ Q 1 = 5.

10 (2) Median Q 2 = ( n ) th item = th item (lies in 230 CF). Against 230 CF the value is 6 \ Q 2 = 6 3(n + 1) (3) Q 3 is the size o th item = th item (lies in 260 CF) 4 Against 260 CF, the value is 6.5 \ Q 3 = 6.5 Measures o Central Tendency 51 Example 19 The mean height o 25 male workers in a actory is 61 inches and the mean height o 35 emale workers in the same actory is 58. Find the combined mean o 60 workers in the actory. Solution Given n 1 = 25, _ x 1 = 61, n 2 = 35 and _ x 2 = 58 The combined mean is given by _ x = n 1 _ x 1 + n 2 x 2 n 1 + n = = 3555/ = Thus, the combined mean height is inches. Example 20 Compute mean, median and mode or the ollowing data. Class Freq Solution Form the table as ollows Class Freq Mid values d i = x i A h i d i CF x i A Here S = 90 = N S i d i = 75 h = 10 (1) Mean = A + h S id i S i = = = 33.83

11 52 Statistics or Managers (2) Median ( N 2 = 90 2 ) = 45 alls in the class ( N 2 C ) \ Median = l + h ( 90 2 ) 35 = = = (3) Mode: The maximum requuency is in the class Mode = l + m 1 2 m 1 2 h = = Example 21 Find the numbers whose arithmetic mean is 12.5 and geometric mean is 10. Solution Let the two numbers be a and b Then, AM = a + b and GM = ab 2 a + b \ = 12.5 ab = 10 2 \ a + b = 25 or, ab = 100 Consider (a + b) 2 (a b) 2 = 4ab \ (25) 2 (a b) 2 = \ (a b) = = 225 \ a b = 15. \ a + b = 25 Solving, a = 20, b = 5 a b = 15 Example 22 An aeroplane covers the our sides o a square at speeds o 1,000, 2,000, 3000 and 4000 km per hour respectively. What is the average speed o the plane in its light around the square? Solution For arithmetic man, we compute as _ x = = 2,500 km per hour. 4

12 Measures o Central Tendency 53 However, that is not the correct answer in such a problem harmonic mean is an appropriate average. 4 HM = = 25 12, ,000 = = 1,920 km per hour. 25 Note: The harmonic mean is a measure o central tendency or data expressed as rates such as km per hour, km per litre, hours per semester tonnes per month, etc. Example 23 The mean annual salaries paid to 100 employees o a company was Rs. 5,000. The mean annual salaries paid to male and emale employees were Rs. 5,200 and Rs. 4,200 respectively. Determine the percentage o males and emales employed by the company. Solution Let n 1 represent percentage o males and n 2 represent precentage o emales so that n 1 + n 2 = 100 Given, combined mean _ x = 5,000, _ x 1 = 5,200 and _ x 2 = 4200 Consider, _ x = \ 5000 = n _ 1 x 1 + n 2 x 2 n 1 + n 2 n n n 1 + n 2 \ (n 1 + n 2 ) 5000 = 5200 n n 2 \ 5000 [n 1 + (100 n 1 )] = 5200 n (100 n 1 ) 5000 n n 1 = 5200 n n 1 \ 1,000 n 1 = 80,000 \ n 1 = 80 and n 2 = 20 Thus, the percentage o males and emales employed is 80 and 20 respectively. General merits and uses o Averages So ar we have discussed the methods o computing the various types o averages and also their distinctive eatures. At this point, we have a problem which o these is the best average to be used. To have a solution to this we must know the ollowing: (i) The purpose which the average is designed to serve. (ii) Would the average be used or urther computations? (iii) The type o data available. I they are badly skewed (avoid the mean). gappy around the middle (avoid the median) and unequal in class interval (avoid the mode). (iv) The typical value required in the particular problem. Within the ramework o descriptive statistics the main requirement is to know what each average means and then select one that uluils the purpose. (v) In some cases it is required to work out more than one averages and present them.

13 54 Statistics or Managers Median The median is the best average in open-end grouped distribution especially where i plotted as a requency curve. For example, in case o price distribution or income distribution. In such cases very high or very low values would cause the mean to be higher to lower than the most common values. In such cases the median or middle value o the series may be a more representative igure to use in describing the mass o data. Mode Mode can be used to describe quantitative data. The mode can be used in problems involving the expression o preerences where quantitative measurements are not possible. Thus, the preerred type o package design among a number o alternative design would be the model design. I we want to compare consumer preerences or dierent kinds o products or dierent kinds o advertising we can compare the model preerences expressed by dierent groups o people but we cannot calculate the median or mean. Mode is a particularly useul average or discrete, series and best suited or large requency. Geometric mean Geometric mean is useul in averaging ratio and percentages and in computing average rates o increase or decrease. It is particularly important in economics and business statistics in index number construction. Harmonic mean Harmonic mean is useul in problems in which values o a variable are compared with a constant quantity o another variable, i.e. rates distance covered within certain time and quantity purchased or sold per unit, etc. In the ollowing cases arithmetic mean should not be used. (a) In highly skewed distributions. (b) In distributions with open end intervals. (c) When the distribution is unevenly spread. (d) The arithmetic mean should not be used to average ratios and rates o change. In such cases, the geometric mean is more suitable. (e) When there are very large and very small items, arithmetic mean would be seriously misleading on account o undue inluence rom extreme items. General demerits o averages: (a) Since an average is a single value representing a group o values. Otherwise there is every possibility o jumping to wrong conclusions. For example, a person had to cross the river rom one bank to another. He was not aware o the depth o the river so he enquired o another man who told him that the average depth o water is 5 t. 4 inches. The man was 5 t. and 8 inches and he thought that he can very easily cross the river because at all times the would be above the level o water so he started. In the beginning, the level o water was very low but it as he reached the middle the water was 15 t. deep and he lost his lie. The man was drowned because he had a misconception that average depth means uniorm depth throughout but it is not so. An average represents a group o values and lies somewhere in between the two extremes, i.e. the largest and the smallest items o the series. (b) An average may give us a value that does not exist in the data. For example, the arithmetic mean o 100, 300, 250, 50 and 100 is 160 a value that does not exist in the data. (c) At times the average may give a very absurd result. For example. It we calculate size o a amily we may get a value 5.6. But this is impossible as persons cannot be in ractions. However, we should remember that it is an average value representing the entire group.

14 Measures o Central Tendency 55 (d) Measures o central tendency ail to give an idea about the ormation o the series. Two or more series may have the same central value but may dier widely in composition. (e) We must remember that an average is a measure o central tendency. Hence, unless the data show a clear single concentration o observations an average may not be meaningul at all. This evidently precludes the use o any average to typiy a bimodal, a U-shaped or J-shaped distribution. QUESTION BANK A. Choose the correct answer rom the given alternatives. 1. Dierent methods give dierent averages which are known as the (a) measures o central tendency (b) statistics (c) measures o dispersion (d) skewness 2. I x 1, x 2, x 3 x n are a set o n values o a variate, then the mean is given by (a) n Sx i (b) Sx i n n 1 x 1 + n 2 x 2 n 1 + n 2 (c) nsx i (d) 3. I x 1, x 2 x 3 x n are a set o n observations then the geometric mean is n (a) n Slog x i (b) Slog x i (c) Slog x i n (d) Antilog Slog x i n 4. I the values o the variables are arranged in ascending order o magnitude, the middle term is (a) mean (b) mode (c) median (d) quartile 5. In a grouped distribution, the value o the variable o maximum requency is called (a) mode (b) mean (c) median (d) quartile 6. In a symmetrical distribution, the mean, median and mode (a) dier (b) coincide (c) mean-median = mode (d) dier by In case o individual observations and discrete series, the size can be determined by using the ( n ) th item or (a) D 10 (b) D 9 (c) D 1 (d) D 2 8. For grouped requency distribution the mode is calculated by using the ormula m 1 (a) l + h (b) l 2 m 1 2 n + m 1 1 2