• # question_answer If mean and standard deviation of 5 observations ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}}$ are 10 and 3, respectively, then the variance of 6 observations ${{x}_{1}},{{x}_{2}},.....{{x}_{3}}$ and $-50$ is equal to: A) $509.5$ B) $586.5$ C) $582.5$       D) $507.5$

Solution :

 $\sum{x}=50$ ${{(3)}^{2}}=\frac{1}{5}\left( e{{x}^{2}}-\frac{{{(ex)}^{2}}}{5} \right)$ $9=\frac{1}{5}\left( \sum{{{x}^{2}}-\frac{2500}{5}} \right)$ $\therefore$      $\sum{{{x}^{2}}=545}$ New variable$=\frac{1}{6}\left( 3045-\frac{0}{6} \right)=507.5$

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