Calculate the standard deviation of
(a) −1, 0, 1
(b) −10, 0, 10
(a) x_{1}=-1,x_{2}=0,x_{3}=1. Clearly {\bar{x}}=0.
x_{1}-{\overline{{x}}}=-1\qquad x_{2}-{\overline{{x}}}=0\qquad x_{3}-{\overline{{x}}}=1variance = {\frac{(-1)^{2}+0^{2}+1^{2}}{3}}={\frac{2}{3}}
standard deviation = {\sqrt{\frac{2}{3}}}=0.816
(b) x_{1}=-10,x_{2}=0,x_{3}=10. Again {\overline{{x}}}=0 and so x_{i}-{\overline{{x}}}=x_{i}, for i = 1,2,3.
variance = {\frac{(-10)^{2}+0^{2}+10^{2}}{3}}={\frac{200}{3}}
standard deviation = {\sqrt{\frac{200}{3}}}=8.165
As expected, the second set has a much higher standard deviation than the first.