MCQ
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$582.5$
- ✓$507.5$
${\left( 3 \right)^2} = \frac{1}{5}\left( {e{x^2} - \frac{{{{\left( {ex} \right)}^2}}}{5}} \right)$
$9 = \frac{1}{5}\left( {\sum {{x^2} - \frac{{2500}}{5}} } \right)$
$\therefore \sum {{x^2} = 545} $
New variable $ = \frac{1}{6}\left( {3045 - \frac{0}{6}} \right) = 507.5$
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