If mean and standard deviation of 5 observati | vedclass

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Question

$\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( {\su

Vedclass · Accepted Answer

$507.5$

Vedclass · Answer

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

class	0-10	10-20	20-30	30-40
Freq	1	3	4	2

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

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