Given that the dataset constitutes a population, we will need to use the two formulas below
[tex]\begin{gathered} \mu=\frac{1}{N}\sum_ix_i\rightarrow\text{ population mean} \\ \sigma=\frac{\sqrt{\sum_i(x_i-\mu)^2}}{\sqrt{N}}\rightarrow\text{ population standard deviation} \\ \end{gathered}[/tex]
Thus, in our case,
[tex]\begin{gathered} \Rightarrow\mu=\frac{1}{5}(93+97+111+95+109)=\frac{505}{5}=101 \\ \Rightarrow\mu=101 \end{gathered}[/tex]
Finding the standard deviation,
[tex]\begin{gathered} \Rightarrow\sigma=\frac{\sqrt{(93-101)^2+(97-101)^2+...+(109-101)^2}}{\sqrt{5}} \\ \Rightarrow\sigma=\sqrt{\frac{280}{5}}=\sqrt{56}\approx7.48 \end{gathered}[/tex]
Thus, the standard deviation of the population is 7.48
