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If the standard deviation of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 is K, then the standard deviation of 10, 11, 12, ......, 19 is

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$K$

$10 K$

$K + \sqrt{10}$

$K + 10$

answer is A.

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Detailed Solution

$K = \sqrt{(\frac{(n^{2} - 1)}{12}) l^{2}} = \sqrt{\frac{33}{4}}; SD of 10, 11, \dots . 19 is, k$

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