Question

What is the standard deviation of the following data set rounded to the nearest tenth?

Answer 1

$Formula\ for\ standard\ deviation\ :\\\\ x=\sqrt{y^2}\\ y-variance\\\\ y^2=\frac{(a_1-a)^2+(a_2-a)^2...+(a_n-a)^2}{n}\\ n-amount\ of\ numbers\\ a-mean\\\ a=\frac{11+5+12+8+5+12+10}{7} =\frac{63}{7}=9\\\\ y^2=\frac{(11-9)^2+(5-9)^2+(12-9)^2+(8-9)^2+(5-9)^2(12-9)^2+(10-9)^2}{7} \\y^2=\frac{(2)^2+(-4)^2+(3)^2+(-1)^2+(-4)^2+(3)^2+1^2}{7}=\\ \frac{4+16+9+1+16+9+1}{7}=\frac{56}{7}=8\\\\ x=\sqrt8=2\sqrt2$

Answer 2

$\overline{x}\ \rightarrow\ the\ arithmetic\ mean\ \ \ \Rightarrow\ \ \ \overline{x}= \frac{11+5+12+8+5+12+10}{7} = \frac{63}{7} =9\\\\\sigma\ \rightarrow\ the\ standard\ deviation\\\\\sigma^2= \frac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}{n} \\\\ \sigma^2= \frac{(11-9)^2+2\cdot(5-9)^2+2\cdot(12-9)^2+(8-9)^2+(10-9)^2}{7}=\\\\= \frac{2^2+2\cdot(-4)^2+2\cdot3^2+(-1)^2+1^2}{7} = \frac{4+32+18+1+1}{7} = \frac{56}{7} =8\ \ \Rightarrow\ \ \sigma= \sqrt{8} \approx2.8$