Find the mean deviation about the mean for th | vedclass

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(D) First,find the mid-values $(x_i)$ for each class interval:$0-20: 10$$20-40: 30$$40-60: 50$$60-80: 70$$80-100: 90$Calculate the mean $(\bar{x})$:$\

Vedclass · Accepted Answer

$23.84$

Vedclass · Answer

(D) First,find the mid-values $(x_i)$ for each class interval:$0-20: 10$$20-40: 30$$40-60: 50$$60-80: 70$$80-100: 90$Calculate the mean $(\bar{x})$:$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{(10 	imes 10) + (8 	imes 30) + (12 	imes 50) + (9 	imes 70) + (11 	imes 90)}{10+8+12+9+11} = \frac{100+240+600+630+990}{50} = \frac{2560}{50} = 51.2$Now,calculate the mean deviation about the mean using the formula: $	ext{M.D.}(\bar{x}) = \frac{\Sigma f_i |x_i - \bar{x}|}{\Sigma f_i}$$\Sigma f_i |x_i - 51.2| = 10|10-51.2| + 8|30-51.2| + 12|50-51.2| + 9|70-51.2| + 11|90-51.2|$$= 10(41.2) + 8(21.2) + 12(1.2) + 9(18.8) + 11(38.8)$$= 412 + 169.6 + 14.4 + 169.2 + 426.8 = 1192$$	ext{M.D.}(\bar{x}) = \frac{1192}{50} = 23.84$

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

$x_i$	$3$	$6$	$9$	$12$	$13$	$15$	$21$	$22$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$

Find the mean deviation about the mean for the following data:
Marks obtained $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
Number of students $10$ $8$ $12$ $9$ $11$

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Find the mean deviation about the mean for the following data:Marks obtained$0-20$$20-40$$40-60$$60-80$$80-100$Number of students$10$$8$$12$$9$$11$