Take into account that the standard deviation of a probability distribution table is given by:
![\sigma=\sqrt[\placeholder{⬚}]{\Sigma\left(x-\mu\right)^2P\left(x\right)}](https://img.qammunity.org/2023/formulas/mathematics/college/211jlepf1y1jk7jis8wah9u05xyy2f7fjr.png)
where x is each element of the first column of the table, μ is the mean and P(x) is the corresponding values of P(x) for each value of x in the second column.
By replacing the values of the table you obtain:
![\begin{gathered} \sigma=\sqrt[\placeholder{⬚}]{\left(0-3.79\right)^2\lparen0.04)+\left(1-3.79\right)^2\left(0.23\right)+\left(3-3.79\right)^2\left(0.35\right)+\left(6-3.79\right)^2\left(0.15\right)+\left(7-3.79\right)^2\left(0.23\right)} \\ \sigma=\sqrt[\placeholder{⬚}]{5.6859} \\ \sigma\approx2.38 \end{gathered}]()
Hence, the standard deviation of the given data is approximately 2.38