If sumlimits_{i = 1}^{18} {({x_i} - 8) = | vedclass

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Question

Varriance of observation $\left(\mathrm{x}_{1}-8\right) \forall \mathrm{i}=1,2,3, \ldots .18$

$=\frac{45}{18}-\left(\frac{9}{18}\right)^{2}=\frac{5

Vedclass · Accepted Answer

$3/2$

Vedclass · Answer

Varriance of observation $\left(\mathrm{x}_{1}-8\right) \forall \mathrm{i}=1,2,3, \ldots .18$

$=\frac{45}{18}-\left(\frac{9}{18}\right)^{2}=\frac{5}{2}-\frac{1}{4}=\frac{9}{4}$

then $S.D.$ of $x_{1} \forall i=1,2,3, \ldots . .18$

$=\sqrt{\frac{9}{4}}=\frac{3}{2}$

$X$	$c$	$2c$	$3c$	$4c$	$5c$	$6c$
$f$	$2$	$1$	$1$	$1$	$1$	$1$

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$K^{2}-1$	$K^{2}-1$	$K^{2}-1$	$k-3$

If $\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} - 8)^2 = 45} $ then the standard deviation of $x_1, x_2, ...... x_{18}$ is :-

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