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If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations isA. 50B. 30C. 51D. 31 |
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Answer» Correct Answer - D Let the 50 observations are `x_(1),x_(2),x_(3), ..., x_(50).` Now, deviations of these observations from 30 are `(x_(1)-30),(x_(2)-30), (x_(3)-30), ..., (x_(50)-30)` `because sum_(i=1)^(50)(x_(i)-30)=50 " "` (given) `rArr sum_(i=1)^(50)x_(i)-(30xx50)=50` `rArr sum_(i=1)^(50)x_(i)-50(30+1)=50xx31` `therefore` mean of 50 observations `=(bar(x))=(sum_(i=1)^(50)x_(i))/(50)=31` |
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