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(A) The given data is $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$.Here,the number of observations is $n = 12$,which is even.Arranging the data in

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$2.33$

Vedclass · Answer

(A) The given data is $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$.Here,the number of observations is $n = 12$,which is even.Arranging the data in ascending order,we obtain:$10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$.Median $M = \frac{(\frac{n}{2})^{th} 	ext{ observation} + (\frac{n}{2} + 1)^{th} 	ext{ observation}}{2} = \frac{6^{th} 	ext{ observation} + 7^{th} 	ext{ observation}}{2} = \frac{13 + 14}{2} = 13.5$.The absolute deviations $|x_i - M|$ are:$|10 - 13.5| = 3.5$$|11 - 13.5| = 2.5$$|11 - 13.5| = 2.5$$|12 - 13.5| = 1.5$$|13 - 13.5| = 0.5$$|13 - 13.5| = 0.5$$|14 - 13.5| = 0.5$$|16 - 13.5| = 2.5$$|16 - 13.5| = 2.5$$|17 - 13.5| = 3.5$$|17 - 13.5| = 3.5$$|18 - 13.5| = 4.5$Sum of absolute deviations = $3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 = 28$.Mean deviation about median = $\frac{\sum |x_i - M|}{n} = \frac{28}{12} \approx 2.33$.

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

Marks obtained	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Number of students	$10$	$8$	$12$	$9$	$11$

Find the mean deviation about the median for the data: $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$.

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