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if `x_(1),x_(2)….x_(10)` are 10 observations, in which mean of `x_(1),x_(2),x_(3),x_(4)` is 11 while mean of `x_(5),x_(6)….x_(10)` is 16. also `x_(1)^(2)+x_(2)^(2)+….x_(10)^(2)=2000` then value of standard devition isA. `2sqrt(2)`B. 2C. 4D. `sqrt(2)` |
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Answer» Correct Answer - B key idea Formula of standard deviation (sigma), for n observations `=sqrt((Sigmax_(1)^(2))/(n)-((Sigmax_(1))/(n))^(2))` Given 10 observations are `x_(1),x_(2),x_(3), … , x_(10)` ` therefore (x_(1)+x_(2)+x_(3)+x_(4))/(4) =11` `rArr x_(1)+x_(2)+x_(3)+x_(4)=44 " ...(i)" ` `and (x_(5)+x_(6)+x_(7)+x_(8)+x_(9)+x_(10))/(6)=16` ` implies x_(5)+x_(6)+x_(7)+x_(8)+x_(9)+x_(10)=96 " ...(ii)" ` So, mean of given 10 observations `=(44+96)/(10)=(140)/(10)=14` Since, the sum of squares of all the observations = 2000 `x_(1)^(2)+x_(2)^(2)+x_(3)^(2)+ ... +x_(10)^(2)=2000 " ...(iii)' ` Now, `sigma^(2)=("standard deviation")^(2)=(Sigma x_(i)^(2))/(10)-((Sigma x_(i)^(2))/(10))^(2)` `=(2000)/(10)-(14)^(2)=200-196=4` So, `sigma =2` |
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