By definition, the standard deviation is
[tex]\sigma = \sqrt{\frac{\sum_{i=1}^{n}\left(x_i-\bar{x}\right)^2}{n}}[/tex]It seems hard so let's do it step by step, first, let's find the mean of the data
[tex]\begin{gathered} \bar{x}=\frac{24+29+2+21+9}{5} \\ \\ \bar{x}=17 \end{gathered}[/tex]Now we have the mean value, let's do each value of the set minus the mean value
[tex]\begin{gathered} x_1-\bar{x}=24-17=7 \\ \\ x_2-\bar{x}=29-17=12 \\ \\ x_3-\bar{x}=2-17=-15 \\ \\ x_4-\bar{x}=29-17=4 \\ \\ x_4-\bar{x}=9-17=-8 \end{gathered}[/tex]Now we have the difference between each element and the mean value, let's do the square of all values
[tex]\begin{gathered} (x_1-\bar{x})^2=7^2=49 \\ \\ (x_2-\bar{x})^2=12^2=144 \\ \\ (x_3-\bar{x})^2=(-15)^2=225 \\ \\ (x_4-\bar{x})^2=4^2=16 \\ \\ (x_5-\bar{x})^2=(-8)^2=64 \end{gathered}[/tex]Now we have the square of the difference we sum them
[tex]\begin{gathered} \sum_{i=1}^5\left(x_i-\bar{x}\right)^2=\left(x_1-\bar{x}\right)^2+\left(x_2-\bar{x}\right)^2+\left(x_3-\bar{x}\right)^2+\left(x_4-\bar{x}\right)^2+\left(x_5-\bar{x}\right)^2 \\ \\ \sum_{i=1}^5\left(x_i-\bar{x}\right)^2=49+144+225+16+64 \\ \\ \sum_{i=1}^5\left(x_i-\bar{x}\right)^2=498 \end{gathered}[/tex]Now we have the sum we must divide by the number of elements, in that case, 5 elements
[tex]\frac{\sum_{i=1}^5\left(x_i-\bar{x}\right)^2}{5}=99.6[/tex]Now we take the square root of that value to have the standard deviation!
[tex]\sigma=\sqrt{99.6}=9.979[/tex]We write it using only one decimal the result would be
[tex]\sigma=9.9[/tex]With no rounding.
Final answer:
[tex]\sigma=9.9[/tex]