Solution:
The standard deviation is expressed as
where
[tex]\begin{gathered} x_i\Rightarrow sample\text{ values} \\ \mu\implies mean \\ N\implies number\text{ of observation} \\ \sigma\Rightarrow standard\text{ deviation} \end{gathered}[/tex]
Given the sample data below:
[tex]31,\text{ 29, 29, 24, 24, 31}[/tex]
Step 1: Evaluate the mean.
The mean is evaluated as
[tex]\begin{gathered} \mu=\frac{31+29+29+24+24+31}{6} \\ =\frac{168}{6} \\ \Rightarrow\mu=28 \end{gathered}[/tex]
Step 2: Evaluate the standard deviation.
Thus, we have
[tex]\begin{gathered} \sigma=\sqrt{\frac{(31-28)^2+(29-28)^2+(29-28)^2+(24-28)^2+(24-28)^2+(31-28)^2}{6}} \\ =\sqrt{\frac{52}{6}} \\ =2.943920289 \\ \Rightarrow\sigma\approx2.94\text{ \lparen2 decimal places\rparen} \end{gathered}[/tex]
Hence, the standard deviation of the population, to two decimal places, is
[tex]2.94[/tex]
