Rather than computing each deviation from the mean individually, we can use the shortcut variance formula \( \text{Var} = \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2 \), which avoids working with the decimal mean at every step.
The data is 1, 2, 3, 4, 4, 5, 5, 5, 8, so \( n=9 \), \( \sum x = 1+2+3+4+4+5+5+5+8=37 \), and \( \sum x^2=1+4+9+16+16+25+25+25+64=185 \). Then:
\[ \text{Var} = \frac{185}{9} - \left(\frac{37}{9}\right)^2 \approx 20.56 - 16.90 \approx 3.65 \]\[ \text{Standard deviation} = \sqrt{3.65} \approx 1.9 \]Using the shortcut sum-of-squares formula gives a standard deviation close to 2, clearly closer to this option than to any of the others.
Therefore, the correct answer is 2.