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(B) The observations are $x_K = 1 + K\alpha$ for $K = 0, 1, 2, \ldots, 100$. There are $n = 101$ observations. The mean $\bar{x}$ is given by $\bar{x}

Vedclass · Accepted Answer

$10.1$

Vedclass · Answer

(B) The observations are $x_K = 1 + K\alpha$ for $K = 0, 1, 2, \ldots, 100$. There are $n = 101$ observations. The mean $\bar{x}$ is given by $\bar{x} = \frac{1}{101} \sum_{K=0}^{100} (1 + K\alpha) = 1 + \alpha \frac{100 	imes 101}{2 	imes 101} = 1 + 50\alpha$. The mean deviation from the mean is $\frac{1}{n} \sum |x_K - \bar{x}| = \frac{1}{101} \sum_{K=0}^{100} |1 + K\alpha - (1 + 50\alpha)| = \frac{\alpha}{101} \sum_{K=0}^{100} |K - 50|$. This sum is $\sum_{K=0}^{50} (50 - K) + \sum_{K=51}^{100} (K - 50) = (50 + 49 + \ldots + 0) + (1 + 2 + \ldots + 50) = 2 	imes \frac{50 	imes 51}{2} = 2550$. Thus,the mean deviation is $\frac{\alpha 	imes 2550}{101} = 255$. Solving for $\alpha$: $\alpha = \frac{255 	imes 101}{2550} = \frac{101}{10} = 10.1$.

Age	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$5$	$8$	$5$	$12$	$X$	$Y$

If the mean deviation from the mean of the observations $\{1+K \alpha\}$,where $K=0, 1, 2, \ldots, 100$ and $\alpha > 0$ is $255$,then $\alpha =$

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