We have to find the value of sx for the table given.
sx represents the sample standard deviation of x.
Then, we start by calculating the mean of x:
[tex]\begin{gathered} \bar{x}=\frac{1}{n}\sum x_i \\ \\ \bar{x}=\frac{1}{5}(2+5+7+11+15)=\frac{1}{5}(40)=8 \end{gathered}[/tex]
We now can calculate the standard deviation as:
[tex]\begin{gathered} s^2=\frac{1}{n-1}\sum(x_i-\bar{x})^2 \\ s^2=\frac{1}{4}[(2-8)^2+(5-8)^2+(7-8)^2+(11-8)^2+(15-8)^2] \\ s^2=\frac{1}{4}[(-6)^2+(-3)^2+(-1)^2+3^2+7^2] \\ s^2=\frac{1}{4}[36+9+1+9+49] \\ s^2=\frac{1}{4}[104] \\ s^2=26 \\ s=\sqrt{26} \\ s\approx5.099 \end{gathered}[/tex]
We have obtained the approximate value of sx as 5.099.
Answer: 5.099 [Option C]
