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Find the mean deviation about the median for the data given below. 45, 36, 50, 60, 53, 46, 51, 48, 72, 42. |
Answer» Arranging the given data in an ascending order, we get : 36, 42, 45, 46, 48, 50, 51, 53, 60, 72. Here n=10, which is even. `therefore" median"=(1)/(2).{(n)/(2)"th observation"+((n)/(2)+1)"th observation"}` `=(1)/(2)("5th observation + 6th observation")` `=(1)/(2)(48+50)=(98)/(2)=49.` Thus, M=49. The values of `(x_(i)-M)` are `-13,-7,-4,-3,-1,1,2,4,11,23.` `therefore" "overset(10)underset(i=1)Sigma|x_(i)-M|=(13+7+4+3+1+1+2+4+11+23)=69` `rArr" "MD(M)=(overset(10)underset(i=1)Sigma|x_(i)-M|)/(10)=(69)/(10)=6.9.` |
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