Chọn công thức đúng:D. \[\sigma \left( {\rm{X}} \right){\rm{ = }}\sqrt {{\rm{E}}\left( {\rm{X}} \right)} \]Lời giảiTa có: \[\sigma \left( {\rm{X}} \right){\rm{ = }}\sqrt {{\rm{V}}\left(

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Chọn công thức đúng:

\[\sigma \left( {\rm{X}} \right){\rm{ = }}{{\rm{V}}^{\rm{2}}}\left( {\rm{X}} \right)\]

\[\sigma \left( {\rm{X}} \right) = \sqrt {{\rm{x}}_{\rm{1}}^{\rm{2}}{{\rm{p}}_{\rm{1}}}{\rm{ + x}}_{\rm{2}}^{\rm{2}}{{\rm{p}}_{\rm{2}}}{\rm{ + }}...{\rm{ + x}}_{\rm{n}}^{\rm{2}}{{\rm{p}}_{\rm{n}}} - {{\rm{E}}^{\rm{2}}}\left( {\rm{X}} \right)} \]

\[\sigma \left( {\rm{X}} \right) = \sqrt {{\rm{x}}_{\rm{1}}^{\rm{2}}{{\rm{p}}_{\rm{1}}}{\rm{ + x}}_{\rm{2}}^{\rm{2}}{{\rm{p}}_{\rm{2}}}{\rm{ + }}...{\rm{ + x}}_{\rm{n}}^{\rm{2}}{{\rm{p}}_{\rm{n}}}} - {\rm{E}}\left( {\rm{X}} \right)\]

\[\sigma \left( {\rm{X}} \right){\rm{ = }}\sqrt {{\rm{E}}\left( {\rm{X}} \right)} \]

Giải thích

Lời giải

Ta có: \[\sigma \left( {\rm{X}} \right){\rm{ = }}\sqrt {{\rm{V}}\left( {\rm{X}} \right)} \]

Mà \[{\rm{V}}\left( {\rm{X}} \right){\rm{ = x}}_{\rm{1}}^{\rm{2}}{{\rm{p}}_{\rm{1}}}{\rm{ + x}}_{\rm{2}}^{\rm{2}}{{\rm{p}}_{\rm{2}}}{\rm{ + }}...{\rm{ + x}}_{\rm{n}}^{\rm{2}}{{\rm{p}}_{\rm{n}}} - {{\rm{E}}^{\rm{2}}}\left( {\rm{X}} \right)\]

nên\[\sigma \left( {\rm{X}} \right) = \sqrt {{\rm{x}}_{\rm{1}}^{\rm{2}}{{\rm{p}}_{\rm{1}}}{\rm{ + x}}_{\rm{2}}^{\rm{2}}{{\rm{p}}_{\rm{2}}}{\rm{ + }}...{\rm{ + x}}_{\rm{n}}^{\rm{2}}{{\rm{p}}_{\rm{n}}} - {{\rm{E}}^{\rm{2}}}\left( {\rm{X}} \right)} \]

Chọn B