Ex 15.1, 12 - Chapter 15 Class 11 Statistics (Term 1)
Last updated at May 29, 2018 by Teachoo
Last updated at May 29, 2018 by Teachoo
Transcript
Ex15.1, 12 Calculate the mean deviation about median age for the age distribution of 100 persons given below: Converting the given data in continuous frequency by subtracting 0.5 from lower age limit adding and 0.5 in upper limit N = 128 = 100 Median Class = ( /2)^ term = (100/2)^ term = 50th term In above data, cumulative frequency of class 35.5 40.5 is 63 which is greater than 50. Median class = 35.5 40.5 Median = + ( /2 )/ Where, = lower limits of median class N = sum of frequencies = frequency of median class C = Cumulative frequency of class before median class Here, = 35.5, N = 100, C = 37, = 5, = 26 Median = 35.5 + (100/2 37)/26 5 = 35.5 + (50 37)/26 5 = 20 + 13/26 5 = 35.5 + 2.5 = 38 Now, 128 = 100 128 | | = 735 Mean Deviation (M) = ( _ | M| )/ _ = 735/100 = 7.35
Ex 15.1
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