Mrs. Jensen’s statistics unit includes six quizzes, each worth 55 points. Avery’s scores are $\{47, 49, 50, 50, 51, 53\}.$ Determine the standard deviation for Avery’s quizzes.

$\begin{array}{cc}\mu =\frac{47+49+50+50+51+53}{6}=50& \\ {\left(x–\mu \right)}^2& \\ {\left(47–50\right)}^2= 9& \sqrt{{\sigma }^2}=\sqrt{\frac{20}{6}}\\ {\left(49–50\right)}^2= 1& \sigma =1.8257\\ {\left(50–50\right)}^2= 0& \sigma =1.8\\ {\left(50–50\right)}^2= 0& \\ {\left(51–50\right)}^2= 1& \\ {\left(53–50\right)}^2=\underline{+9}& \\ \sum _{}{\left(x–\mu \right)}^2=20& \end{array}$