Question 141

The standard deviation of first 13 natural numbers is

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Solution:

Formula used 

Here, n = 13

$$\sqrt{\frac{\left(\left(\ 13\right)^2\ -\ 1\right)}{12}}\ =\ \sqrt{\ \frac{\left(13+1\right)\left(13-1\right)}{12}}\ =\ \sqrt{\ \frac{14.12}{12}}\ =\ \sqrt{\ 14}$$

$$\sigma\ =\ \sqrt{\ 14}\ $$ Answer

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