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If the standard deviation of 0,1,2, 3............9 in K, the standard deviation of 10,11,12,13......19 is

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Using

Standard Deviation -

In case of discrete frequency distribution 

\sigma = \sqrt{\frac{\sum f_{i}x_{i}^{2}}{\sum f_{i}}-\left ( \frac{\sum f_{i}x_{i}}{\sum f_{i}} \right )^{2}}

-

 

 Since, \sigma ^{2} = \frac{\sum \left ( x_{i} - \bar{x}\right )^{2}}{n}

and each data increases by \lambda \left ( \lambda =10 \right ) then there will be no change in standard deviation because \left ( x_{i}-\bar{x} \right ) will be same.

So, new standard deviation is K.

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