Given:
The data set is 52.1, 45.5, 51, 48.8, 43.6.
Here, number of points are n=5.
The mean of this data is,
[tex]\begin{gathered} \mu=\frac{\sum^{}_{}x_i}{n} \\ =\frac{52.1+45.5+51+48.8+43.6.}{5} \\ =48.2 \end{gathered}[/tex]The variance is calculated as,
[tex]\begin{gathered} \sigma^2=\frac{\sum^{}_{}(x_i-\mu)^2}{n} \\ =\frac{(52.1-48.2)^2+(45.5-48.2)^2+(51-48.2)^2+(48.8-48.2)^2+(43.6-48.2)^2}{5} \\ =\frac{51.86}{5} \\ =10.372 \end{gathered}[/tex]The standard deviation is,
[tex]S\mathrm{}D\mathrm{}=\sqrt[]{10.372}=3.22[/tex]