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Question
The standard deviation of first 10 natural numbers is
Options
5.5
3.87
2.97
2.87
Solution
We know that the standard deviation of first n natural number is \[\sqrt{\frac{n^2 - 1}{12}}\]
∴ Standard deviation of first 10 natural numbers
\[= \sqrt{\frac{{10}^2 - 1}{12}}\]
\[ = \sqrt{\frac{99}{12}}\]
\[ = \sqrt{8 . 25}\]
\[ = 2 . 87\]
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