JEE Main 20206 Sep, Morning shiftMathematicsEasyStatistics and Probability258× since 2002If ∑i=1n(xi−a)=n\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = ni=1∑n(xi−a)=n and ∑i=1n(xi−a)2=na\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = nai=1∑n(xi−a)2=na (n, a > 1) then the standard deviation of n observations x₁ , x₂ , ..., x_n is :AOption A. aaa – 1BOption B. na−1n\sqrt {a - 1}na−1COption C. n(a−1)\sqrt {n\left( {a - 1} \right)}n(a−1)DOption D. a−1\sqrt {a - 1}a−1CheckSkip · show solution
