JEE MAIN - Mathematics (2018 (Offline) - No. 9)
If $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9$$ and
$$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$$, then the standard deviation of the 9 items
$${x_1},{x_2},.......,{x_9}$$ is
$$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$$, then the standard deviation of the 9 items
$${x_1},{x_2},.......,{x_9}$$ is
3
9
4
2
Explanation
IMPORTANT POINT :-
When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged.
so let $${x_i} - 5 = {y_i}$$
So, new equation is $$\sum\limits_{i = 1}^9 {{y_i}} = 9$$
and $$\sum\limits_{i = 1}^9 {y_i^2} = 45$$
As, we know. Standard Deviation (S.D)
$$ = \sqrt {{{\sum\limits_{i = 1}^9 {y_i^2} } \over 9} - {{\left( {{{\sum\limits_{i = 1}^9 {yi} } \over 9}} \right)}^2}} $$
$$ = \sqrt {{{45} \over 9} - {{\left( {{9 \over 9}} \right)}^2}} $$
$$ = \sqrt {5 - 1} $$
$$=$$ $$\sqrt 4 $$
$$=2$$
When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged.
so let $${x_i} - 5 = {y_i}$$
So, new equation is $$\sum\limits_{i = 1}^9 {{y_i}} = 9$$
and $$\sum\limits_{i = 1}^9 {y_i^2} = 45$$
As, we know. Standard Deviation (S.D)
$$ = \sqrt {{{\sum\limits_{i = 1}^9 {y_i^2} } \over 9} - {{\left( {{{\sum\limits_{i = 1}^9 {yi} } \over 9}} \right)}^2}} $$
$$ = \sqrt {{{45} \over 9} - {{\left( {{9 \over 9}} \right)}^2}} $$
$$ = \sqrt {5 - 1} $$
$$=$$ $$\sqrt 4 $$
$$=2$$
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