What is the mean deviation about the mean for | vedclass

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(B) The given series is an arithmetic progression: $5, 10, 15, \dots, 85$.Here,the first term $a = 5$,common difference $d = 5$,and the last term $l =

Vedclass · Accepted Answer

$21.17$

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(B) The given series is an arithmetic progression: $5, 10, 15, \dots, 85$.Here,the first term $a = 5$,common difference $d = 5$,and the last term $l = 85$.Number of terms $n$ is given by $85 = 5 + (n - 1)5$,which implies $n = 17$.The mean $\bar{x} = \frac{5 + 85}{2} = 45$.The mean deviation about the mean is given by $M.D. = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.For an arithmetic progression with $n$ terms,the mean deviation about the mean is $M.D. = \frac{d(n^2 - 1)}{4n}$ if $n$ is odd.Here $n = 17$ (odd) and $d = 5$.$M.D. = \frac{5(17^2 - 1)}{4 	imes 17} = \frac{5(289 - 1)}{68} = \frac{5 	imes 288}{68} = \frac{1440}{68} \approx 21.176$.

Class Interval	$0-2$	$2-4$	$4-6$	$6-8$	$8-10$
Frequency	$1$	$2$	$3$	$2$	$1$

What is the mean deviation about the mean for the observations $5, 10, 15, 20, \dots, 85$?

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