The variance of 20 observation is 5 . If each | vedclass

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Question

$\frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2} = 5} $

$\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$\frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} ight)}^2} = 5} $

$\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} ight)}^2} = 100} $

new observations are $2 \mathrm{x}_{1}, 2 \mathrm{x}_{2}, \ldots \ldots, 2 \mathrm{x}_{20}$

Their mean $=\overline{\mathrm{x}}_{1}=\frac{2\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{20}ight)}{20}=2 \overline{\mathrm{x}}$

Now, variance$ = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {2{x_i} - 2\bar x} ight)}^2}} $

$ = \frac{1}{{20}} 	imes 4\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} ight)}^2}} $

$ = \frac{1}{{20}} 	imes 4 	imes 100 = 20$

Subject	Mathematics	Physics	Chemistty
Mean	$42$	$32$	$40.9$
Standard deviation	$12$	$15$	$20$

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

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