The mean deviation of the following data abou | vedclass

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(C) First,calculate the mean $(\bar{x})$:$\sum f_i = 8 + 48 + 56 + 32 + 16 = 160$$\sum f_i x_i = (5 	imes 8) + (15 	imes 48) + (25 	imes 56) + (35

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$8$

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(C) First,calculate the mean $(\bar{x})$:$\sum f_i = 8 + 48 + 56 + 32 + 16 = 160$$\sum f_i x_i = (5 	imes 8) + (15 	imes 48) + (25 	imes 56) + (35 	imes 32) + (45 	imes 16) = 40 + 720 + 1400 + 1120 + 720 = 4000$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4000}{160} = 25$Now,calculate the mean deviation about the mean ($M$.$D$.$(\bar{x})$):$M$.$D$.$(\bar{x})$ = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$Values of $|x_i - 25|$ are: $|5-25|=20, |15-25|=10, |25-25|=0, |35-25|=10, |45-25|=20$$\sum f_i |x_i - \bar{x}| = (8 	imes 20) + (48 	imes 10) + (56 	imes 0) + (32 	imes 10) + (16 	imes 20) = 160 + 480 + 0 + 320 + 320 = 1280$$M$.$D$.$(\bar{x})$ = $\frac{1280}{160} = 8$

$x_i$	$5, 15, 25, 35, 45$
$f_i$	$8, 48, 56, 32, 16$

The mean deviation of the following data about its mean is
$x_i$ $5, 15, 25, 35, 45$
$f_i$ $8, 48, 56, 32, 16$

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The mean deviation of the following data about its mean is$x_i$$5, 15, 25, 35, 45$$f_i$$8, 48, 56, 32, 16$