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 (integer)
11
Solution
$${\sigma ^2} = {{\sum {{x^2}} } \over n} - {\left( {{{\sum x } \over n}} \right)^2}$$<br><br>$${\sigma ^2} = {{(9 + {k^2})} \over {10}} - {\left( {{{9 + k} \over {10}}} \right)^2} < 10$$<br><br>$(90 + {k^2})10 - (81 + {k^2} + 8k) < 1000$<br><br>$90 + 10{k^2} - {k^2} - 18k - 81 < 1000$<br><br>$9{k^2} - 18k + 9 < 1000$<br><br>$${(k - 1)^2} < {{1000} \over 9} \Rightarrow k - 1 < {{10\sqrt {10} } \over 3}$$<br><br>$k < {{10\sqrt {10} } \over 3} + 1$
<br><br>k $\le$ 11
<br><br>Maximum integral value of k = 11.
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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