JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 9)
A data consists of n observations : x1, x2, . . . . . . ., xn.
If $$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$$ and
$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$$
then the standard deviation of this data is :
If $$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$$ and
$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$$
then the standard deviation of this data is :
2
$$\sqrt 5 $$
5
$$\sqrt 7 $$
Explanation
$$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n $$
$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)$$
$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n $$
$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)$$
Performing (1) + (2), we get
$$2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n$$
$$\sum\limits_{i = 1}^n {x_i^2} = 6n$$
Performing (1) $$-$$ (2), we get
$$ \Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n$$
$$ \Rightarrow $$$$ \Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n$$
S.D($$\sigma $$)$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}} $$
$$\sigma $$ $$ = \sqrt {{{6n} \over n} - \left( 1 \right)} $$
$$\sigma $$ $$ = \sqrt 5 $$
$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)$$
$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n $$
$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)$$
Performing (1) + (2), we get
$$2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n$$
$$\sum\limits_{i = 1}^n {x_i^2} = 6n$$
Performing (1) $$-$$ (2), we get
$$ \Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n$$
$$ \Rightarrow $$$$ \Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n$$
S.D($$\sigma $$)$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}} $$
$$\sigma $$ $$ = \sqrt {{{6n} \over n} - \left( 1 \right)} $$
$$\sigma $$ $$ = \sqrt 5 $$
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