JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 15)
Let xi
(1 $$ \le $$ i $$ \le $$ 10) be ten observations of a
random variable X. If
$$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$$ and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$$
where 0 $$ \ne $$ p $$ \in $$ R, then the standard deviation of these observations is :
$$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$$ and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$$
where 0 $$ \ne $$ p $$ \in $$ R, then the standard deviation of these observations is :
$${7 \over {10}}$$
$${9 \over {10}}$$
$${4 \over 5}$$
$$\sqrt {{3 \over 5}} $$
Explanation
Standard deviation = $$\sqrt {Variance} $$
$$ = \sqrt {{{\sum {x_1^2} } \over n} - {{(\overline x )}^2}} $$
$$ = \sqrt {{{\sum\limits_{i = 1}^{10} {{{({x_i} - p)}^2}} } \over {10}} - {{\left( {{{\sum\limits_{i = 1}^{10} {({x_i} - p)} } \over {10}}} \right)}^2}} $$
[ Standard deviation is free from shifting of origin.]
$$ = \sqrt {{9 \over {10}} - {{\left( {{3 \over {10}}} \right)}^2}} $$
$$ = \sqrt {{9 \over {10}} - {9 \over {100}}} $$
$$ = \sqrt {{{90 - 9} \over {100}}} $$
$$ = \sqrt {{{81} \over {100}}} $$
$$ = {9 \over {10}}$$
$$ = \sqrt {{{\sum {x_1^2} } \over n} - {{(\overline x )}^2}} $$
$$ = \sqrt {{{\sum\limits_{i = 1}^{10} {{{({x_i} - p)}^2}} } \over {10}} - {{\left( {{{\sum\limits_{i = 1}^{10} {({x_i} - p)} } \over {10}}} \right)}^2}} $$
[ Standard deviation is free from shifting of origin.]
$$ = \sqrt {{9 \over {10}} - {{\left( {{3 \over {10}}} \right)}^2}} $$
$$ = \sqrt {{9 \over {10}} - {9 \over {100}}} $$
$$ = \sqrt {{{90 - 9} \over {100}}} $$
$$ = \sqrt {{{81} \over {100}}} $$
$$ = {9 \over {10}}$$
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