Question

hey can anybody help me with this question please

Answer 1

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

$standard\ deviation:\\\\\sqrt{\delta^2}=\sqrt{\dfrac{1}{3}}=\dfrac{\sqrt1}{\sqrt3}=\dfrac{1\cdot\sqrt3}{\sqrt3\cdot\sqrt3}=\boxed{\dfrac{\sqrt3}{3}}\approx\boxed{0.58}$

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)