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13.Statistics
hard
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
$507.5$
(JEE MAIN-2019)
Solution
$\sum {x = 50} $
${\left( 3 \right)^2} = \frac{1}{5}\left( {e{x^2} – \frac{{{{\left( {ex} \right)}^2}}}{5}} \right)$
$9 = \frac{1}{5}\left( {\sum {{x^2} – \frac{{2500}}{5}} } \right)$
$\therefore \sum {{x^2} = 545} $
New variable $ = \frac{1}{6}\left( {3045 – \frac{0}{6}} \right) = 507.5$
Standard 11
Mathematics
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