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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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo