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• 1. Session – 8 Measures of Dispersions Mean Deviation Mean deviation is the average differences among the items in a series from the mean itself or median or mode of that series. It is concerned with the extent of which the values are dispersed about the mean or median or the mode. It is found by averaging all the deviations from control tendency. These deviations are taken into computations with regard to negative sign. Theoretically the deviations of item are taken preferably from median instead than from the mean and mode. Merits of Mean Deviation • It is rigidly defined and easy to compute. • It takes all items in to considerations and gives weight to deviation according to these sign. • It is less affected by extreme values. • It removes all irregularities by obtaining deviation and provides correct measures. Demerits of Mean Deviation • It is not suitable for algebraic treatments. • It is positive which is not justified mathematically. • It is not satisfactory measure when the deviations are taken from mode. • It is not suitable when class intervals are open end. 1
• 2. Formula to compute Mean Deviation If xi is variant and takes the values x1, x2, x3, …….. xn with average. A (mean, median, mode), then mean deviation from the average – A is defined by ∑ xi − A MD = N For the grouped data ∑f xi − A MD = N MD Coefficient of MD = Mean 1. Compute MD and CMD from mean for the given data below. X d = xi − x 21 26.55 32 15.55 38 9.55 41 6.55 49 1.45 54 6.45 59 11.45 66 18.45 68 20.45 Σx = 428 Σ x i − x = Σd= 116.45 x= ∑ n xi x= 428 = 47.35 i =1 9 ∑ xi − x 116.45 MD = = N 9 MD = δ = 12.938 MD Coefficient of MD = Avg 12.938 = = 0.272 47.55 2
• 3. 2. Following are the wages of workers. Find mean deviation from median and its coefficient. x Wages x i − Me = x i − 47 59 17 30 32 22 25 67 25 22 43 32 15 22 43 4 17 47  M 0 64 55 8 55 59 12 47 64 17 80 67 20 25 80 33 25 Σ x i − M = 186 Σ x i − M = 186  11 + 1  th Median =   item  2   11 + 1  =  = 6th item  2  Me = 47 ∑ x i − Me MD = N 186 = = 16.91 11 MD Coefficient of MD = Median 16.91 = = 0.359 47 3. Compute MD about its mode and its coefficient. 3
• 4. x f d = x i − Mode fd 20 6 100 600 40 19 80 1520 60 40 60 2400 80 23 40 920 100 65 20 1300 120  Mode 83  Modal 0 0 class 140 55 20 1100 160 20 40 800 180 9 60 5401 Σf = 320 Σf x i − Mode = 9180 the highest frequency is 83 and hence Z = 120 ∑ x i − Mode MD= N  9180  Median =    320  = 28.68 28.68 Coefficient of MD = 120 = 0.239 4
• 5. 4. Find out the mean deviation from the data given below about its median. Salaries 40 50 50-100 100-200 200-400 No. of Employees 22 18 10 8 2 No. of x x(mv) cf d = x i − Me fd Employees 40 22 40 22 10 220 50 18 50 40 0 0 50-100 10 75 50 25 250 100-200 8 150 58 100 800 200-400 2 300 60 250 500 Σf = 60 Σf x i − Me = 1770 th  N + 1 Median =   item  2  60 + 1 = 2 61 = = 30.5 It lies in 40 cf and against 40 cf 2 discrete value is 50 ∑ x i − Median MD = N  1770  =   60  MD = 29.5 MD Coefficient of MD = Median 29.5 = 50 = 0.59 5