JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 1)
If the standard deviation of the numbers
–1, 0, 1, k is $$\sqrt 5$$ where k > 0, then k is equal to
2$$\sqrt 6 $$
$$\sqrt 6 $$
$$2\sqrt {{{5} \over 6}} $$
$$2\sqrt {{{10} \over 3}} $$
Explanation
standard deviation = $$\sqrt 5$$
$$ \therefore $$ Variance = $${\left( {\sqrt 5 } \right)^2}$$ = 5
Also variance = $${{\sum {x_i^2} } \over N} - {\mu ^2}$$
Where $$\mu $$ = Mean = $${{ - 1 + 0 + 1 + k} \over 4}$$ = $${k \over 4}$$
$$ \therefore $$ Variance = $${{{{\left( { - 1} \right)}^2} + 0 + {1^2} + {k^2}} \over 4}$$ - $${{{k^2}} \over {16}}$$
$$ \Rightarrow $$ 5 = $${{2 + {k^2}} \over 4}$$ - $${{{k^2}} \over {16}}$$
$$ \Rightarrow $$ 8 + 3k2 = 80
$$ \Rightarrow $$ k2 = 24 = 2$$\sqrt 6 $$
$$ \therefore $$ Variance = $${\left( {\sqrt 5 } \right)^2}$$ = 5
Also variance = $${{\sum {x_i^2} } \over N} - {\mu ^2}$$
Where $$\mu $$ = Mean = $${{ - 1 + 0 + 1 + k} \over 4}$$ = $${k \over 4}$$
$$ \therefore $$ Variance = $${{{{\left( { - 1} \right)}^2} + 0 + {1^2} + {k^2}} \over 4}$$ - $${{{k^2}} \over {16}}$$
$$ \Rightarrow $$ 5 = $${{2 + {k^2}} \over 4}$$ - $${{{k^2}} \over {16}}$$
$$ \Rightarrow $$ 8 + 3k2 = 80
$$ \Rightarrow $$ k2 = 24 = 2$$\sqrt 6 $$
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