Respuesta :
[tex]\begin{gathered} 13,15,12,10,4,16,17,22,9 \\ S\tan dard\text{ deviation}=\sigma=? \\ Variance\text{ }\sigma^2=? \\ \sigma^2=\frac{\sum ^N_{i\mathop{=}1}(x_i-\bar{x})^2)}{N} \\ N=\text{ total of data=9} \\ \bar{x}=\frac{13+15+12+10+4+16+17+22+9}{9}=\frac{118}{9} \\ For\text{ }x_i=13 \\ (x_i-\bar{x})^2=(13-\frac{118}{9})^2=(\frac{-1}{9})^2=\frac{1}{81} \\ For\text{ }x_i=15 \\ (x_i-\bar{x})^2=(15-\frac{118}{9})^2=(\frac{17}{9})^2=\frac{289}{81} \\ For\text{ }x_i=12 \\ (x_i-\bar{x})^2=(12-\frac{118}{9})^2=(-\frac{10}{9})^2=\frac{100}{81} \\ For\text{ }x_i=10 \\ (x_i-\bar{x})^2=(10-\frac{118}{9})^2=(-\frac{28}{9})^2=\frac{784}{81} \\ For\text{ }x_i=4 \\ (x_i-\bar{x})^2=(4-\frac{118}{9})^2=(-\frac{82}{9})^2=\frac{6724}{81} \\ For\text{ }x_i=16 \\ (x_i-\bar{x})^2=(16-\frac{118}{9})^2=(\frac{26}{9})^2=\frac{676}{81} \\ For\text{ }x_i=17 \\ (x_i-\bar{x})^2=(17-\frac{118}{9})^2=(\frac{35}{9})^2=\frac{1225}{81} \\ For\text{ }x_i=22 \\ (x_i-\bar{x})^2=(22-\frac{118}{9})^2=(\frac{80}{9})^2=\frac{6400}{81} \\ For\text{ }x_i=9 \\ (x_i-\bar{x})^2=(9-\frac{118}{9})^2=(-\frac{37}{9})^2=\frac{1369}{81} \\ \sum ^N_{i\mathop{=}1}(x_i-\bar{x})^2)=\frac{17568}{81}=\frac{1952}{9} \\ \sigma^2=\frac{\frac{1952}{9}}{9}=\frac{1952}{81}\approx24.1 \\ \text{The variance is }24.1 \\ For\text{ standard deviation } \\ \sigma=\sqrt{\frac{\sum^N_{i\mathop{=}1}(x_i-\bar{x})^2)}{N}} \\ \sigma=\sqrt{24.1} \\ \sigma\approx4.9 \\ \text{The standard deviation is 4.9} \end{gathered}[/tex]
