Answer :
[tex]\begin{gathered} \text{Solve for the mean }\bar{x}\text{ first} \\ \bar{x}=\frac{\Sigma x_i}{n}=\frac{38+36+28+35+43}{5} \\ \bar{x}=36 \\ \text{Then solve for the variance }s^2 \\ s^2=\frac{\Sigma(x_i-\bar{x})^2}{n-1} \\ s^2=\frac{(38-36)^2+(36-36)^2+(28-36)^2+(35-36)^2+(43-36)^2}{5-1} \\ s^2=\frac{(2)^2+(0)^2+(-8)^2+(-1)^2+(7)^2}{5-1} \\ s^2=\frac{118}{4} \\ s^2=29.5 \\ \text{The square root of the variance is the standard deviation} \\ s=\sqrt[]{29.5} \\ s=5.4313902456001 \\ \text{round off to two decimal place and we have} \\ s=5.43 \end{gathered}[/tex]