The mean and the standard deviation of 10 obs | vedclass

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(A) Given: Original mean $\bar{x} = 20$,original standard deviation $\sigma = 2$. If each observation $x_i$ is transformed to $y_i = p x_i - q$,then t

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$-20$

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(A) Given: Original mean $\bar{x} = 20$,original standard deviation $\sigma = 2$. If each observation $x_i$ is transformed to $y_i = p x_i - q$,then the new mean $\bar{y} = p \bar{x} - q$ and the new standard deviation $\sigma_y = |p| \sigma$. Given that the new mean is half of the original mean: $\bar{y} = \frac{20}{2} = 10$. So,$p(20) - q = 10 \implies 20p - q = 10$ $(i)$. Given that the new standard deviation is half of the original standard deviation: $\sigma_y = \frac{2}{2} = 1$. So,$|p| 	imes 2 = 1 \implies |p| = \frac{1}{2} \implies p = \pm \frac{1}{2}$. Case $1$: If $p = \frac{1}{2}$,then $20(\frac{1}{2}) - q = 10 \implies 10 - q = 10 \implies q = 0$. This contradicts the condition $q 
eq 0$. Case $2$: If $p = -\frac{1}{2}$,then $20(-\frac{1}{2}) - q = 10 \implies -10 - q = 10 \implies q = -20$. Thus,$q = -20$.

$Country$	$Agriculture$	$Industries$	$Services$	$Others$
$India$	$45$	$19$	$28$	$4$
$U.K.$	$3$	$40$	$44$	$13$
$Japan$	$6$	$48$	$43$	$3$
$U.S.A.$	$3$	$35$	$61$	$1$

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