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describe the center and spread of the data using the more appropriate status either the mean median range interquartile range or standard division

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Respuesta :

[tex]\begin{gathered} \mleft\lbrace1,12,2,9,2,3,7,3,3,6\mright\rbrace \\ \operatorname{mean}=\frac{1+12+2+9+2+3+7+3+3+6}{10}=\frac{48}{10}=4.8 \end{gathered}[/tex][tex]\sigma=\sqrt[]{\frac{\sum ^N_{i\mathop=1}(x-\operatorname{mean})^2}{N}}=3.4[/tex]

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