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$

${x_i}$	$60$	$61$	$62$	$63$	$64$	$65$	$66$	$67$	$68$
${f_i}$	$2$	$1$	$12$	$29$	$25$	$12$	$10$	$4$	$5$

${x_i}$	$6$	$10$	$14$	$18$	$24$	$28$	$30$
${f_i}$	$2$	$4$	$7$	$12$	$8$	$4$	$3$

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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