Find the mean deviation about the median for | vedclass

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(A) The given observations are already in ascending order.Adding a column for cumulative frequencies $(c.f.)$:						$x_i$			$f_i$			$c.f.$							$15$

Vedclass · Accepted Answer

$5.10$

Vedclass · Answer

(A) The given observations are already in ascending order.Adding a column for cumulative frequencies $(c.f.)$:						$x_i$			$f_i$			$c.f.$							$15$			$3$			$3$							$21$			$5$			$8$							$27$			$6$			$14$							$30$			$7$			$21$							$35$			$8$			$29$			Here,$N = \sum f_i = 29$,which is odd.Median $= \left(\frac{N+1}{2}\right)^{th}$ observation $= \left(\frac{29+1}{2}\right)^{th} = 15^{th}$ observation.This observation lies in the cumulative frequency $21$,for which the corresponding value is $x_i = 30$.So,Median $(M) = 30$.Calculating $|x_i - M|$ and $f_i|x_i - M|$:						$x_i$			$f_i$			$|x_i - 30|$			$f_i|x_i - 30|$							$15$			$3$			$15$			$45$							$21$			$5$			$9$			$45$							$27$			$6$			$3$			$18$							$30$			$7$			$0$			$0$							$35$			$8$			$5$			$40$			Sum $\sum f_i|x_i - M| = 45 + 45 + 18 + 0 + 40 = 148$.Mean Deviation about Median $= \frac{1}{N} \sum f_i|x_i - M| = \frac{148}{29} \approx 5.10$.

Income per day	Number of persons
$0-100$	$4$
$100-200$	$8$
$200-300$	$9$
$300-400$	$10$
$400-500$	$7$
$500-600$	$5$
$600-700$	$4$
$700-800$	$3$

Find the mean deviation about the median for the given data:

$x_i$ $15$ $21$ $27$ $30$ $35$

$f_i$ $3$ $5$ $6$ $7$ $8$

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Find the mean deviation about the mean for the following data:
Income per day Number of persons
$0-100$ $4$
$100-200$ $8$
$200-300$ $9$
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$400-500$ $7$
$500-600$ $5$
$600-700$ $4$
$700-800$ $3$

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Find the mean deviation about the median for the given data: $x_i$ $15$ $21$ $27$ $30$ $35$ $f_i$ $3$ $5$ $6$ $7$ $8$