If alpha neq 0 and the mean deviation of the | vedclass

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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,$|\alpha| 	imes 625 = 2500$.$|\alpha| = \frac{2500}{625} = 4$.

$x$	$10$	$11$	$12$	$13$
$f$	$6$	$12$	$18$	$12$

Income per day	Number of persons
$0-100$	$4$
$100-200$	$8$
$200-300$	$9$
$300-400$	$10$
$400-500$	$7$
$500-600$	$5$
$600-700$	$4$
$700-800$	$3$

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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