Statistics Part 2

Concept:Mean deviation about median is the average of the absolute deviations from the median.Explanation:The numbers $k, 2k, \ldots, 1000k$ form an arithmetic progression with $n=1000$ terms.For even $n$, median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th terms.Here median $M = \frac{500k + 501k}{2} = \frac{1001k}{2}$.The absolute deviations from $M$ are symmetric.For $i=1$ to $500$, the deviation is $M - ik = \frac{1001k}{2} - ik = \frac{(1001-2i)k}{2}$.Sum of these 500 deviations = $\frac{k}{2}\sum_{i=1}^{500}(1001-2i) = \frac{k}{2}\left[500\cdot1001 - 2\cdot\frac{500\cdot501}{2}\right] = \frac{k}{2}[500500 - 250500] = 125000k$.Total sum of absolute deviations = $2 \times 125000k = 250000k$.Mean deviation = $\frac{250000k}{1000} = 250k$.Given mean deviation = $500$, so $250k = 500 \implies k = 2$.Thus $k^2 = 4$.Answer:$k^2 = 4$. Option D.