JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 12)
If $$\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$$ and $$\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$$
(n, a > 1) then the standard deviation of n
observations x1 , x2 , ..., xn is :
(n, a > 1) then the standard deviation of n
observations x1 , x2 , ..., xn is :
$$a$$ – 1
$$n\sqrt {a - 1} $$
$$\sqrt {n\left( {a - 1} \right)} $$
$$\sqrt {a - 1} $$
Explanation
S.D = $$\sqrt {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n} - {{\left( {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n}} \right)}^2}} $$
= $$\sqrt {{{na} \over n} - {{\left( {{n \over n}} \right)}^2}} $$
= $$\sqrt {a - 1} $$
= $$\sqrt {{{na} \over n} - {{\left( {{n \over n}} \right)}^2}} $$
= $$\sqrt {a - 1} $$
Comments (0)
