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:
![\begin{gathered} \bar{x}=(1)/(n)\sum x_i \\ \\ \bar{x}=(1)/(5)(2+5+7+11+15)=(1)/(5)(40)=8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2dv3x0xwz73s579p8xdz78xetkdylrn6qr.png)
We now can calculate the standard deviation as:
![\begin{gathered} s^2=(1)/(n-1)\sum(x_i-\bar{x})^2 \\ s^2=(1)/(4)[(2-8)^2+(5-8)^2+(7-8)^2+(11-8)^2+(15-8)^2] \\ s^2=(1)/(4)[(-6)^2+(-3)^2+(-1)^2+3^2+7^2] \\ s^2=(1)/(4)[36+9+1+9+49] \\ s^2=(1)/(4)[104] \\ s^2=26 \\ s=√(26) \\ s\approx5.099 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/g3bfm4z1w0v799a97csqwikl3xvcmhlg5j.png)
We have obtained the approximate value of sx as 5.099.
Answer: 5.099 [Option C]