The mean deviation about the median of the nu | vedclass

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(D) First,arrange the given numbers in ascending order: $2, 3, 5, 7, 9, 11, 13, 15, 17, 20$. Since the number of observations $n = 10$ (which is even)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) First,arrange the given numbers in ascending order: $2, 3, 5, 7, 9, 11, 13, 15, 17, 20$. Since the number of observations $n = 10$ (which is even),the median is the average of the $5^{th}$ and $6^{th}$ observations. Median $M = \frac{9 + 11}{2} = \frac{20}{2} = 10$. Now,calculate the absolute deviations from the median $|x_i - M|$: $|2 - 10| = 8, |3 - 10| = 7, |5 - 10| = 5, |7 - 10| = 3, |9 - 10| = 1, |11 - 10| = 1, |13 - 10| = 3, |15 - 10| = 5, |17 - 10| = 7, |20 - 10| = 10$. The sum of these absolute deviations is $8 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 10 = 50$. The mean deviation about the median is $\frac{1}{n} \sum |x_i - M| = \frac{50}{10} = 5$.

The mean deviation about the median of the numbers $2, 3, 5, 20, 17, 15, 13, 11, 9, 7$ is

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