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Ex 15.1, 4 - Find mean deviation about median 36, 72, 46 - Mean deviation about median - Ungrouped

Ex 15.1, 4 - Chapter 15 Class 11 Statistics - Part 2


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Ex15.1, 4 Find the mean deviation about the median for the data 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 Arranging data in ascending order, 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 Here, n = number of observations = 10 (even) Since n is even Median = (( /2)^ + ( /2 + 1)^ )/2 M = ((10/2)^ + (10/2 + 1)^ )/2 M = (5^ + 6^ )/2 M = (46 + 49)/2 = 95/2 = 47.5 Thus, for 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 Median = 47.5 Mean deviation about median = ( 128 | M| )/10 M.D.(M) = (( (|36 47.5| + |42 47.5| + |45 47.5| + |46 47.5| + |46 47.5| @+ |49 47.5| + |51 47.5| + |53 47.5|+ |60 47.5| + |72 47.5| )))/10 = (( (| 11.5| + | 5.5| + | 2.5| + | 1.5| + | 1.5| @+ |1.5| + |3.5|+ |5.5| + |12.5| + |24.5| )))/10 = (11.5 + 5.5 +2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5)/10 = 70/10 = 7

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