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(A) Given observations are $\{k \alpha\}$ for $k=1, 2, \ldots, 50$.These are $\{\alpha, 2 \alpha, 3 \alpha, \ldots, 50 \alpha\}$.Since the number of o

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$4$

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(A) Given observations are $\{k \alpha\}$ for $k=1, 2, \ldots, 50$.These are $\{\alpha, 2 \alpha, 3 \alpha, \ldots, 50 \alpha\}$.Since the number of observations $n=50$ is even,the median $M$ is the average of the $25^{th}$ and $26^{th}$ terms:$M = \frac{25 \alpha + 26 \alpha}{2} = 25.5 \alpha$.The mean deviation about the median is given by $\frac{1}{n} \sum_{k=1}^{50} |k \alpha - M| = 50$.Substituting $M = 25.5 \alpha$:$\frac{1}{50} \sum_{k=1}^{50} |k \alpha - 25.5 \alpha| = 50$.$|\alpha| \sum_{k=1}^{50} |k - 25.5| = 2500$.The sum $\sum_{k=1}^{50} |k - 25.5|$ consists of terms $|1-25.5| + |2-25.5| + \ldots + |50-25.5|$.This is $2 	imes (24.5 + 23.5 + \ldots + 0.5) = 2 	imes \frac{25}{2} (24.5 + 0.5) = 25 	imes 25 = 625$.Thus,$|\alpha| 	imes 625 = 2500$.$|\alpha| = \frac{2500}{625} = 4$.

$\text{Marks obtained}$	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
$\text{Number of Boys}$	$6$	$8$	$10$	$4$	$2$

If $\alpha \neq 0$ and the mean deviation of the observations $\{k \alpha\}$ for $k=1, 2, \ldots, 50$ about its median is $50$,then $|\alpha|=$

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