JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 13)
If the variance of 10 natural numbers 1, 1, 1, ....., 1, k is less than 10, then the maximum possible value of k is ________.
Answer
11
Explanation
$${\sigma ^2} = {{\sum {{x^2}} } \over n} - {\left( {{{\sum x } \over n}} \right)^2}$$
$${\sigma ^2} = {{(9 + {k^2})} \over {10}} - {\left( {{{9 + k} \over {10}}} \right)^2} < 10$$
$$(90 + {k^2})10 - (81 + {k^2} + 8k) < 1000$$
$$90 + 10{k^2} - {k^2} - 18k - 81 < 1000$$
$$9{k^2} - 18k + 9 < 1000$$
$${(k - 1)^2} < {{1000} \over 9} \Rightarrow k - 1 < {{10\sqrt {10} } \over 3}$$
$$k < {{10\sqrt {10} } \over 3} + 1$$
k $$ \le $$ 11
Maximum integral value of k = 11.
$${\sigma ^2} = {{(9 + {k^2})} \over {10}} - {\left( {{{9 + k} \over {10}}} \right)^2} < 10$$
$$(90 + {k^2})10 - (81 + {k^2} + 8k) < 1000$$
$$90 + 10{k^2} - {k^2} - 18k - 81 < 1000$$
$$9{k^2} - 18k + 9 < 1000$$
$${(k - 1)^2} < {{1000} \over 9} \Rightarrow k - 1 < {{10\sqrt {10} } \over 3}$$
$$k < {{10\sqrt {10} } \over 3} + 1$$
k $$ \le $$ 11
Maximum integral value of k = 11.
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