# Find a stardard deviation

$$N=30 \\ \sum_{i=1}^{30}x_i=120 \\ \\ \sum_{i=1}^{30}x_i^{2}=750\\$$

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

$$Var(X)=E((X-\mu)^2)=E(X^2)-E(\mu)^2=E(X^2)-\mu^2$$.
• @PHPUseless Are you sure about that? Is that the result of the variance of $X$ or the standard deviation of $X$? – user2974951 Jan 19 at 11:59
• @PHPUseless $SD(X)=\sqrt{750/30-(120/30)^2}=3$ for me. – user2974951 Jan 20 at 9:50