stands for the mean deviation from median My for the total of the values above

# Stands for the mean deviation from median my for the

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stands for the mean deviation from median, My for the total of the values above the actual median, and Mx for the values below it, and N for the number of items. δ ´ X = 1 N ( ´ X y ´ X x ) Where δ ´ X stand for the mean deviation from mean, ´ X y stands for the total of the values above the actual arithmetic average and ´ X x for the values below it. Example 8 The following are the marks obtained by a batch of 9 students in a certain test: Sl. No. Marks (out of 100) Sl. No. Marks (out of 100) 1 68 6 38 2 49 7 59 3 32 8 66 4 21 9 41 5 54 Calculate mean deviation of the series from median.
(Answer: Mean Deviation = 12.8 marks) Solution Direct Method Calculation of mean deviation of the series of marks of 9 students (arranged in ascending order of magnitude) Marks (out of 100) Deviations from Median (49) (X) (+, - sign ignored) 21 28 32 17 38 11 41 8 49 0 54 5 59 10 66 17 68 19 ∣dM∣ = 115 Median = the value of ( N + 1 2 ) th item = 49 Mean Deviation, δM = dM N Where, dM represents the summation of he deviations from the median and N, the number of items δM = 115 9 = 12.8 Marks Short cut Method Marks arranged in ascending order of magnitude
Sum of items above median (with values less than median) = 21+32+38+41 = 132 (Mx) Sum of items below median (with values more than median) = 54+59+66+68 = 247 (My) δM = 1 N ( My Mx ) δM = 1 9 ( 247 132 ) = 12.8 Marks Practice Example 9 Calculate Mean Deviation (from arithmetic Average) for the following values. Also, calculate its Coefficient. 4800, 4600, 4400, 4200, 4000. (Answer: Mean Deviation = 240, Co-efficient of Mean Deviation = 0.54) Practice Example 10 Calculate mean deviation (from arithmetic Average) for the following values. 100.500 100.250 100.375 100.625 100.750 100.125 100.375 100.625 100.500 100.125 (Answer: Mean Deviation = 0.175) Calculation of mean deviation in Discrete Series Example 11 Find the mean deviation of the distribution given below No. of Accidents Persons having said no. of Accidents No. of Accidents Persons having said no. of Accidents 0 15 7 2 1 16 8 1 2 21 9 2 3 10 10 2 4 17 11 0 5 8 12 2 6 4 (Answer: Mean deviation from Median is 1.96 accidents)
Solution No. of Accidents Persons having said no. of Accidents Cumulative Number of Accidents Deviation from Median (2) (+, - signs ignored) Total Deviations (X) (f) (dM) (fdM) 0 15 15 2 30 1 16 31 1 16 2 21 52 0 0 3 10 62 1 10 4 17 79 2 34 5 8 87 3 24 6 4 91 4 16 7 2 93 5 10 8 1 94 6 6 9 2 96 7 14 10 2 98 8 16 11 0 98 9 0 12 2 100 10 20 N = 100 fdM =19 6 Mean Deviation δM = dM N = 196 100 = 1.96 accidents Practice Example 12 Calculate the mean deviation (from mean) of the following series: Marks No. of Students 5 5 15 8 25 15 35 16 45 6 (Answer: mean deviation 9.44 marks) (Hint: calculate mean by step deviation, further find mean deviation from mean) Calculation of Mean Deviation in Continuous Series
Example 13 Calculate the mean deviation (from median) from the following data Class Interval Frequency Class Interval Frequency 1-3 6 9-11 21 3-5 53 11-13 16 5-7 85 13-15 4 7-9 56 15-17 4 (Answer: mean deviation is 2.1) (Hint: Refer Example 25 from Measures of Central Tendency for Mean) Solution Class Interval Mid- Point Deviation from Actual Median Frequency Dev * f Dev from Assumed Dev * f 1-3 6 3-5 53 5-7 85 7-9 56 9-11 21 11-13 16 13-15 4 15-17 4 Total Direct Method ___________________________________________________________________________________ ___________________________________________________________________________________