Question

Hi there can you please help me with this question please anyone

Answer 1

$11;\ 13;\ 13;\ 13;\ 13;\ 15\\\\\overline{x}=\dfrac{11+13+13+13+13+15}{6}=\dfrac{78}{6}=13\\\\\delta^2=\dfrac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+(x_3-\overline{x})^2+...+(x_n-\overline{x})^2}{n}\\\\\delta^2=\dfrac{(11-13)^2+(13-13)^2+(13-13)^2+(13-13)^2+}{6}$
$\dfrac{+(13-13)^2+(15-13)^2}{6}\\\\\delta^2=\dfrac{(-2)^2+0+0+0+0+2^2}{6}=\dfrac{4+4}{6}=\dfrac{8}{6}=\dfrac{4}{3}$

$standard\ deviation:\\\\\sqrt{\delta^2}=\sqrt{\dfrac{4}{3}}=\dfrac{\sqrt4}{\sqrt3}=\dfrac{2\cdot\sqrt3}{\sqrt3\cdot\sqrt3}=\boxed{\dfrac{2\sqrt3}{3}}\approx\boxed{1.15}$