$$N=30 \\ \sum_{i=1}^{30}x_i=120 \\ \\ \sum_{i=1}^{30}x_i^{2}=750\\ $$
Find a standard deviation of x:
$$ \bar{x}=\frac{120}{30}=4 \\ sd=\sqrt[]{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^{2}}{N}}=\sqrt[]{\frac{\sum_{i=1}^{n}(x_i^{2}-2\bar{x}x_i+\bar{x}^{2})}{N}}=\sqrt[]{\frac{\sum_{i=1}^{n}x_i^{2}-2n\bar{x}\sum_{i=1}^{n}x_i+n\bar{x}^{2}}{N}}=\sqrt[]{\frac{750-2*30*4*120+30*{4}^{2}}{30}}=30.1$$
But in the textbook the answer is 4