What is the coefficient of mean deviation abo | vedclass

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(B) Arrange the observations in ascending order: $40, 54, 62, 68, 76, 90$.Number of terms $(n) = 6$ (even).Median $(M) = \frac{(\frac{n}{2})^{th} 	ex

Vedclass · Accepted Answer

$0.2$

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(B) Arrange the observations in ascending order: $40, 54, 62, 68, 76, 90$.Number of terms $(n) = 6$ (even).Median $(M) = \frac{(\frac{n}{2})^{th} 	ext{ term} + (\frac{n}{2} + 1)^{th} 	ext{ term}}{2} = \frac{62 + 68}{2} = 65$.$\Sigma |x_i - M| = |40-65| + |54-65| + |62-65| + |68-65| + |76-65| + |90-65| = 25 + 11 + 3 + 3 + 11 + 25 = 78$.Mean deviation about median $= \frac{\Sigma |x_i - M|}{n} = \frac{78}{6} = 13$.Coefficient of mean deviation $= \frac{	ext{Mean Deviation}}{	ext{Median}} = \frac{13}{65} = 0.2$.

What is the coefficient of mean deviation about the median for the observations $40, 62, 54, 90, 68, 76$?

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