Question

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

11, 5, 12, 8, 5, 12, 10

Answer 1

`\overline{x}\ \rightarrow\ the\ arithmetic\ mean\ \ \ \Rightarrow\ \ \ \overline{x}= (11+5+12+8+5+12+10)/(7) = (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= ((11-9)^2+2\cdot(5-9)^2+2\cdot(12-9)^2+(8-9)^2+(10-9)^2)/(7)=\\\\= (2^2+2\cdot(-4)^2+2\cdot3^2+(-1)^2+1^2)/(7) = (4+32+18+1+1)/(7) = (56)/(7) =8\ \ \Rightarrow\ \ \sigma= √(8) \approx2.8`

Answer 2

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