Respuesta :

kate200468
[tex] the\ standard\ deviation\ (\sigma)= \sqrt{ \frac{\big{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}}{\big{n}}} \\\\\\and\ \ \ \overline{x}= \frac{\big{x_1+x_2+...+x_n}}{\big{n}} \\\\---------------------\\\\x_1=0.26;\ \ \ x_2=0.23\ \ \ \Rightarrow\ \ \ \overline{x}= \frac{0.26+0.23}{2} =0.245\\\\\\\sigma=\sqrt{ \frac{\big{(0.26-0.245)^2+(0.23-0.245)^2}}{\big{2}}} =\sqrt{ \frac{\big{(0.015)^2+(-0.015)^2}}{\big{2}}} =[/tex]


[tex]=\sqrt{ \frac{ \big{2\cdot (0.015)^2}}{\big{2}}} =\sqrt{ (0.015)^2} =0.015[/tex]


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