9. Illustration 4.1 For the above data, the A.M. is 2717 + 2796 +…… 4645+….. + 5424 + ….+ 6874 = -------------------------------------------------------------------------- 20 = Rs. 4565.4 Millions x
11. Illustration 4.2 The calculation is illustrated with the data relating to equity holdings of the group of 20 billionaires given in Table 3.1 Class Interval ( 1 ) Frequency ( f i ) ( 2 ) Mid Value of Class Interval ( x i ) ( 3 ) f i x i Col.(4) = Col.(2) x Col.(3) 2000 – 3000 2 2500 5000 3000 – 4000 5 3500 17500 4000 – 5000 6 4500 27000 5000 – 6000 4 5500 22000 6000 – 7000 3 6500 19500 Sum f i = 20 f i x i = 91000
12. Illustration 4.2 values of f i and f i x i , in formula = 9100 ÷ 20 = 4550 i i f f i x x
13. Weighted Arithmetic Mean if the values x 1 , x 2 x 3 , …. x i , ….x n have weights w 1 , w 2 w 3 , …. w i , ….,w n then the weighted mean of x is given as i i i w x w x
14. Illustration 4.3 Item Monthly Consumption Weight (w i ) Rise in Price (Percentage) (p i ) w i p i Sugar 5 5 20 100 Rice 20 20 10 200
15. Illustration 4.3 Therefore, the average price rise could be evaluated as = = = = = 12. Thus the average price rise is 12 % . 20 5 200 100 25 300 i i i w p w p
16. Geometric Mean The Geometric Mean ( G. M.) of a series of observations with x 1 , x 2 , x 3 , ……..,x n is defined as the n th root of the product of these values . Mathematically G.M. = { ( x 1 )( x 2 )( x 3 )…………….(x n ) } (1/ n ) It may be noted that the G.M. cannot be defined if any value of x is zero as the whole product of various values becomes zero.
17. Illustration 4.5 For the data with values, 2,4, and 8, G.M. = (2 x 4 x 8 ) (1/3) = (64) 1/3 = 4
18. Average Rate of Growth of Production/Business or Increase in Prices If P 1 is the production in the first year and P n is the production in the nth year, then the average rate of growth is given by ( G – 100) % where, G = 100 (P n / P 1 ) 1/(n-1) or log G = log 100 + { 1/(n–1) } (log P n – log P 1 )
19. Example 4.4 The wholesale price index in the year 2000-01 was 145.3. It increased to 195.5 in the year 2005-06. What has been the average rate of increase in the index during the last 5 years. Solution: By using the formula ( 4.8), we have log G = 2 +{ (1/5) ( log 195.5 – log145.3 ) } = 2.02578 Therefore, G = Anti log (2.02578) = 106.11 Thus the average rate of increase = 106.11 100 = 6.11%
20. Combined G.M. of Two Sets of Data If G 1 & G 2 are the Geometric means of two sets of data, then the combined Geometric mean, say G, of the combined data is given by : n 1 log G 1 + n 2 log G 2 log G = ------------------------------- n 1 + n 2
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22. Combined G.M. of Two Sets of Data 5 log 120 + 5 log 115 5 x 2.07918 + 5 x 2.06070 log G = ------------------------------- = ---------------------------------- 5 + 5 10 20.6994 = ------------ = 2.06994 10 Therefore, G = antilog 2.06994 = 117.47 Thus the combined average rate of growth for the period of 10 years is 17.47%.
23. Weighted Geometric Mean Just like weighted arithmetic mean, we also have weighted Geometric mean If x 1 , x 2 ,….,x i, ….,x n are n observations with weights w 1 , w 2 , …w i ,.., w n , then their G.M. is defined as: w i log x i G.M. = ---------------------- w i
24. Harmonic Mean The harmonic mean (H.M.) is defined as the reciprocal of the arithmetic mean of the reciprocals of the observations. For example, if x 1 and x 2 are two observations, then the arithmetic means of their reciprocals viz 1/x 1 and 1/ x 2 is {(1 / x 1 ) + (1 / x 2 )} / 2 = (x 2 + x 1 ) / 2 x 1 x 2 The reciprocal of this arithmetic mean is 2 x 1 x 2 / (x 2 + x 1 ). This is called the harmonic mean. Thus the harmonic mean of two observations x 1 and x 2 is 2 x 1 x 2 ----------------- x 1 + x 2
25. Relationship Among A.M. G.M. and H.M. The relationships among the magnitudes of the three types of Means calculated from the same data are as follows: (i) H.M. ≤ G.M. ≤ A.M. i.e. the arithmetic mean is greater than or equal to the geometric which is greater than or equal to the harmonic mean. ( ii ) G.M. = i.e. geometric mean is the square root of the product of arithmetic mean and harmonic mean. ( iii) H.M. = ( G.M.) 2 / A .M. . . . M H M A
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27. Median - Ungrouped Data First the data is arranged in ascending/descending order. In the earlier example relating to equity holdings data of 20 billionaires given in Table 4.1, the data is arranged as per ascending order as follows 2717 2796 3098 3144 3527 3534 3862 4187 4310 4506 4745 4784 4923 5034 5071 5424 5561 6505 6707 6874 Here, the number of observations is 20, and therefore there is no middle observation. However, the two middle most observations are 10 th and 11 th . The values are 4506 and 4745. Therefore, the median is their average. 4506 + 4745 9251 Median = ----------------- = ----------- 2 2 = 4625.5 Thus, the median equity holdings of the 20 billionaires is Rs.4625.5 Millions.
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29. Illustration 4.2 Class Interval Frequency Cumulative frequency 2000-3000 2 2 3000-4000 5 7 4000-5000 6 13 5000-6000 4 17 6000-70000 3 20
30. Illustration 4.2 Here, n = 20, the median class interval is from 4000 to 5000 as the 10 th observation lies in this interval. Further, L m = 4000 f m = 6 f c = 7 w m = 1000 Therefore, 20/2 –7 x 1000 Median = 4000 + ------------------------- 6 = 4000 + 3/6 x 1000 = 4000 + 500 = 4500
35. Quartiles data Q 1 and Q 3 are defined as values corresponding to an observation given below : Ungrouped Data Grouped Data (arranged in ascending or descending order) Lower Quartile Q 1 {( n + 1 ) / 4 } th ( n / 4 ) th Median Q 2 { ( n + 1 ) / 2 } th ( n / 2 ) th Upper Quartile Q 3 {3 ( n + 1 ) / 4 } th (3 n / 4 ) th
36. Quartiles 1 1 1 ) 4 / ( 1 Q Q c Q w f f n L Q 3 3 3 ) 4 / 3 ( 3 Q Q c Q w f f n L Q
37. Equity Holding Data Class Interval Frequency Cumulative frequency 2000-3000 2 2 3000-4000 5 7 4000-5000 6 13 5000-6000 4 17 6000-70000 3 20
38. ( (20/4) – 2 ) Q 1 = 3000 + --------------- 1000 5 ( 5 – 2 ) = 3000 + -------------------- 1000 5 3000 = 3000 + ------------- 5 = 3000 + 600 = 3600 The interpretation of this value of Q 1 is that 25 % billionaires have equity holdings less than Rs.
39. (15 – 13) Q 3 = ------------- 1000 +5000 4 2 = ------- 1000 +5000 4 = 5500 The interpretation of this value of Q 3 is that 75 % billionaires have equity holdings less than Rs. 5500 Millions.
40. Percentiles (95/100) n – f c P 95 = L P95 + ------------------- x w P95 f P95 where, L P95 is the lower point of the class interval containing 95 th percent of total frequency, f c is the cumulative frequency up to the 95 th percentile interval, f P95 is the frequency of the 95 th percentile interval and w P95 is the width of the 95 th percentile interval.
52. Geometric Mean Advantages Disadvantages (i) Makes use of full data (ii) Extreme large values have lesser impacts (ii) Useful for data relating to rations and percentage (iv) Useful for rate of change/growth (i) Cannot be calculated if any observation has the value zero (ii) Difficult to calculate and interpret