If the sum of 10 observations is 12 and the sum of their squares is 18, then the standard deviation is: 

Key Points

Standard deviation = \(\sqrt{{{(x_i - \bar x)}^2} \over n} = \sqrt {Variance} \) 
Therefore, Variance V(X) = \({1 \over n} {{(x_i - \bar x)}^2} \)
or, \(\rm v(x)=\frac{1}{n}\left[\Sigma(x_i^2-2\times x_i \times \bar x+\bar x^2)\right]\)

\(\rm =\frac{\Sigma x_i^2}{n}-2 \bar x\left(\frac{\Sigma x_i}{n}\right)+\frac{\Sigma \bar x^2}{n}\)

\(=\rm \frac{\Sigma x_i^2}{n}-2\bar x^2+\frac{n.\bar x^2}{n}\)

\(\rm = \frac{\Sigma x_i^2}{n}-2\bar x^2+\bar x^2\)

\(\rm V(x)=\frac{\Sigma x_i^2}{n}-\bar x^2\)
Given, the sum of 10 observations is 12 and the sum of their squares is 18. 
Thus, n = 10, \(\sum x_i = 12\), \(\sum x_i^2 = 18\) 
Therefore, mean \(\bar x = 1.2\)
Plugging this information into the equation of variance V(X), we have

\(\rm V(x)=\frac{\Sigma x_i^2}{n}-\bar x^2\)

\(=\frac{1}{10}\times 18-(1.2)^2\)

= 1.8 - 1.44

=0.36
Hence, the required standard deviation = \(\sqrt {Variance} = \sqrt{(0.36)} = 0.6 = {3 \over 5}\)