JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 8)
The outcome of each of 30 items was observed; 10 items gave an outcome $${1 \over 2}$$ – d each, 10 items gave outcome $${1 \over 2}$$ each and the remaining 10 items gave outcome $${1 \over 2}$$+ d each. If the variance of this outcome data is $${4 \over 3}$$ then |d| equals :
$${2 \over 3}$$
$${{\sqrt 5 } \over 2}$$
$${\sqrt 2 }$$
2
Explanation
Variance is independent of region. So we shift the given data by $${1 \over 2}$$.
so, $${{10{d^2} + 10 \times {0^2} + 10{d^2}} \over {30}} - {\left( 0 \right)^2} = {4 \over 3}$$
$$ \Rightarrow $$ d2 $$=$$ 2 $$ \Rightarrow $$ $$\left| d \right| = \sqrt 2 $$
so, $${{10{d^2} + 10 \times {0^2} + 10{d^2}} \over {30}} - {\left( 0 \right)^2} = {4 \over 3}$$
$$ \Rightarrow $$ d2 $$=$$ 2 $$ \Rightarrow $$ $$\left| d \right| = \sqrt 2 $$
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