Answer :
The standard deviation of a sample of data can be computed using the equation
[tex]\sigma=\sqrt[]{\frac{1}{N}\sum^N_{n\mathop=1}(x_n-\mu)^2}[/tex]where μ stands for the mean of the data and N as the number of samples.
The first step in solving the standard deviation of the given sample is to find the mean. The mean of the data given is
[tex]\mu=\frac{109+105+112+110+115+106+108+109}{8}=109.25[/tex]The second step is to subtract each number to the mean of the data then get the square of the difference. We have the following
[tex]\begin{gathered} (x_1-\mu)^2=(109-109.25)^2=0.0625 \\ (x_2-\mu)^2=(105-109.25)^2=18.0625 \\ (x_3-\mu)^2=(112-109.25)^2=7.5625 \\ (x_4-\mu)^2=(110-109.25)^2=0.5625 \\ (x_5-\mu)^2=(115-109.25)^2=33.0625 \\ (x_6-\mu)^2=(106-109.25)^2=10.5625 \\ (x_7-\mu)^2=(1098109.25)^2=1.5625 \\ (x_8-\mu)^2=(109-109.25)^2=0.0625 \end{gathered}[/tex]The third step is to find the sum of these squared differences and get the mean. We get
[tex]\frac{1}{N}\sum ^N_{n\mathop{=}1}(x_n-\mu)^2=\frac{0.0625+18.0625+7.5625+0.5625+33.0625+10.5625+1.5625+0.0625}{8}=8.933[/tex]Finally, for the last step, get the square root of the result in step 3, we have
[tex]\sqrt[]{\frac{1}{N}\sum^N_{n\mathop{=}1}(x_n-\mu)^2}=\sqrt[]{8.933}=\pm2.99[/tex]Therefore, the standard deviation of the given data is equal to ± 2.99.