Question

Hey can anybody help me with this question please

Answer 1

`9;\ 10;\ 10;\ 10;\ 10;\ 11\\\\\overline{x}=(9+10+10+10+10+11)/(6)=(60)/(6)=10\\\\\delta^2=\frac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+(x_3-\overline{x})^2+...+(x_n-\overline{x})^2}{n}\\\\\delta^2=((9-10)^2+(10-10)^2+(10-10)^2+(10-10)^2+)/(6)\\(+(10-10)^2+(11-10)^2)/(6)\\\delta^2=((-1)^2+0+0+0+0+1^2)/(6)=(1+1)/(6)=(2)/(6)=(1)/(3)`


`standard\ deviation:\\\\√(\delta^2)=\sqrt{(1)/(3)}=(\sqrt1)/(\sqrt3)=(1\cdot\sqrt3)/(\sqrt3\cdot\sqrt3)=\boxed{(\sqrt3)/(3)}\approx\boxed{0.58}`

Answer 2

`\frac { \Sigma x }{ n } =\frac { 9+10+10+10+10+11 }{ 6 } =10\\ \\ \frac { \Sigma { x }^( 2 ) }{ n } =\frac { { 9 }^( 2 )+{ 10 }^( 2 )+{ 10 }^( 2 )+{ 10 }^( 2 )+{ 10 }^( 2 )+{ 11 }^( 2 ) }{ 6 } =\frac { 301 }{ 3 }`

Standard deviation formula:


`\sigma =\sqrt { \frac { \Sigma { x }^( 2 ) }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^( 2 ) }`

This means that:


`\sigma =\sqrt { \frac { 301 }{ 3 } -{ 10 }^( 2 ) }`

Answer:

0.58 (to 2 decimal places)