The mean deviation about the mean of the foll | vedclass

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(D) First,we calculate the mean $(\bar{x})$ of the given data:$\bar{x} = \frac{\Sigma f x}{\Sigma f} = \frac{(1 	imes 3) + (3 	imes 3) + (5 	imes 4

Vedclass · Accepted Answer

$2.95$

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(D) First,we calculate the mean $(\bar{x})$ of the given data:$\bar{x} = \frac{\Sigma f x}{\Sigma f} = \frac{(1 	imes 3) + (3 	imes 3) + (5 	imes 4) + (7 	imes 14) + (9 	imes 7) + (11 	imes 4) + (13 	imes 3) + (15 	imes 4)}{3 + 3 + 4 + 14 + 7 + 4 + 3 + 4}$$\bar{x} = \frac{3 + 9 + 20 + 98 + 63 + 44 + 39 + 60}{42} = \frac{336}{42} = 8$Now,we calculate the mean deviation about the mean using the formula: $	ext{M.D.}(\bar{x}) = \frac{\Sigma f|x - \bar{x}|}{N}$Calculating $|x - \bar{x}|$ for each $x$:$|1-8|=7, |3-8|=5, |5-8|=3, |7-8|=1, |9-8|=1, |11-8|=3, |13-8|=5, |15-8|=7$Calculating $\Sigma f|x - \bar{x}|$:$(3 	imes 7) + (3 	imes 5) + (4 	imes 3) + (14 	imes 1) + (7 	imes 1) + (4 	imes 3) + (3 	imes 5) + (4 	imes 7)$$= 21 + 15 + 12 + 14 + 7 + 12 + 15 + 28 = 124$$	herefore 	ext{M.D.}(\bar{x}) = \frac{124}{42} \approx 2.95$

Size $(x)$	$1$	$3$	$5$	$7$	$9$	$11$	$13$	$15$
Frequency $(f)$	$3$	$3$	$4$	$14$	$7$	$4$	$3$	$4$

$\text{Marks obtained}$	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
$\text{Number of Boys}$	$6$	$8$	$10$	$4$	$2$

The mean deviation about the mean of the following data is nearly:
Size $(x)$ $1$ $3$ $5$ $7$ $9$ $11$ $13$ $15$
Frequency $(f)$ $3$ $3$ $4$ $14$ $7$ $4$ $3$ $4$

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The mean deviation about the mean for the following data is:
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$\text{Number of Boys}$ $6$ $8$ $10$ $4$ $2$

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The mean deviation about the mean of the following data is nearly:Size $(x)$$1$$3$$5$$7$$9$$11$$13$$15$Frequency $(f)$$3$$3$$4$$14$$7$$4$$3$$4$