MathematicsEasy89×since 2002Q5452If ∑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 :Aaaa – 1Bna−1n\sqrt {a - 1}na−1Cn(a−1)\sqrt {n\left( {a - 1} \right)}n(a−1)Da−1\sqrt {a - 1}a−1Check answerSkip