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

Question

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

Aicharya · Accepted Answer

Given observations $=\{\mathrm{K} \alpha\}$ Here, $\mathrm{K}=1,2,3, \ldots \ldots, 50$ $ 	herefore \quad\{\mathrm{K} \alpha\}=\{\alpha, 2 \alpha, 3 \alpha, \ldots . ., 50 \alpha\} $ Number of observations $=50$ (Even) $ \begin{aligned} & 	ext { Median of given observation }=\frac{25 \alpha+26 \alpha}{2} \ & M=25.5 \alpha \end{aligned} $ Now, mean deviation about median $\begin{aligned} & 	ext { M,D. (M) }=\frac{\sum_{k=1}^n\left|x_i-might|}{n} \ & 50=\frac{\sum_{k=1}^{50}|k \alpha-25.5 \alpha|}{50} \ & \Rightarrow \quad 50 	imes 50=\sum_{k=1}^{50}|\alpha(K-25.5)| \ & \Rightarrow \quad|\alpha| \sum_{k=1}^{50}|k-25.5|=2500 \ & \Rightarrow \quad 2|\alpha|[24.5+23.5+. .+1.5+0.5]=2500 \ & \Rightarrow \quad 2|\alpha| 	imes \frac{25}{2} 	imes(24.5+0.5)=2500 \ & \Rightarrow \quad|\alpha| 	imes 25=1400 \Rightarrow|\alpha|=4\end{aligned}$

If $\alpha \neq 0$ and the mean deviation of the observations $\{k \alpha\}$, for $k=1,2, \ldots \ldots, 50$…

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