Measures Of Central Tendancy cont Geometric Mean Geometric

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Measures Of Central Tendancy (cont. ) Geometric Mean: Geometric mean is defined at the

Measures Of Central Tendancy (cont. ) Geometric Mean: Geometric mean is defined at the nth root of the product of n observations of a distribution. Symbolically, GM =. . . n√x 1……x 2……x 3…… If we have only two observations, say, 4 and 16 then GM = √ 4× 16 = √ 64 = 8. Similarly, if there are three observations, then we have to calculate the cube root of the product of these three observations; and so on. When the number of items is large, it becomes extremely difficult to multiply the numbers and to calculate the root. To simplify calculations, logarithms are used.

Example: If we have to find out the geometric mean of 2, 4 and

Example: If we have to find out the geometric mean of 2, 4 and 8, then we find Log GM = (∑ log xi)N = (Log 2 + Log 4 + Log 8)3 = (0. 3010 + 0. 6021+ 0. 9031)3 = (1. 8062)/3 = 0. 60206 GM = Antilog (0. 60206) =4

Harmonic Mean The harmonic mean is defined as the reciprocal of the arithmetic mean

Harmonic Mean The harmonic mean is defined as the reciprocal of the arithmetic mean of the reciprocals of individual observations. Symbolically, HM= n / (1/xi+1/x 2+1/x 3+………… 1/xn) =Reciprocal (∑ 1/x)/N The calculation of harmonic mean becomes very tedious when a distribution has a large number of observations. In the case of grouped data, the harmonic mean is calculated by using the following formula: HM =N/∑[f/x]

Example: Suppose we have three observations 4, 8 and 16. We are required to

Example: Suppose we have three observations 4, 8 and 16. We are required to calculate the harmonic mean. Reciprocals of 4, 8 and 16 are: 1/4 , 1/8 , 16 1 respectively Since HM = N/(1/ x 1+ 1/ x 2+1/ x 3) = 3(1/ 4 +1/ 8 +1/ 16) = 3/(0. 25 +0. 125 +0. 0625 )3 = 6. 857 (approx. )

Example: Consider the following series: Class-interval Frequency 2 -4 20 4 -6 40 6

Example: Consider the following series: Class-interval Frequency 2 -4 20 4 -6 40 6 -8 30 8 -10 10 Calculate Harmonic Mean.

Solution: Class-interval 2 -4 -6 8 10 Mid-value 3 5 7 9 =∑[f*1/x]/N =

Solution: Class-interval 2 -4 -6 8 10 Mid-value 3 5 7 9 =∑[f*1/x]/N = 100/20. 0641 = 4. 984 approx. Frequency Reciprocal of MV f /x 20 0. 3333 6. 6660 40 0. 2000 8. 0000 30 0. 1429 4. 2870 10 0. 1111 1. 1111 Total 20. 0641 4 68 -