The mean deviation about the mean for the fol | vedclass

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(D) We have the following data.First,calculate the mean $\bar{x}$:$\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{(2 	imes 2) + (4 	imes 4) + (5

Vedclass · Accepted Answer

$1.733$

Vedclass · Answer

(D) We have the following data.First,calculate the mean $\bar{x}$:$\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{(2 	imes 2) + (4 	imes 4) + (5 	imes 10) + (7 	imes 8) + (9 	imes 6)}{2 + 4 + 10 + 8 + 6}$$\bar{x} = \frac{4 + 16 + 50 + 56 + 54}{30} = \frac{180}{30} = 6$Now,calculate the mean deviation about the mean using the formula $	ext{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$:$	ext{M.D.}(\bar{x}) = \frac{2|2-6| + 4|4-6| + 10|5-6| + 8|7-6| + 6|9-6|}{30}$$	ext{M.D.}(\bar{x}) = \frac{2(4) + 4(2) + 10(1) + 8(1) + 6(3)}{30}$$	ext{M.D.}(\bar{x}) = \frac{8 + 8 + 10 + 8 + 18}{30} = \frac{52}{30} = 1.733$

$x_i$	$2$	$4$	$5$	$7$	$9$
$f_i$	$2$	$4$	$10$	$8$	$6$

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

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

The mean deviation about the mean for the following data is:
$x_i$ $2$ $4$ $5$ $7$ $9$
$f_i$ $2$ $4$ $10$ $8$ $6$

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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 $(f)$ $3$ $3$ $4$ $14$ $7$ $4$ $3$ $4$

If the mean deviation of the numbers $1, 1+d, 1+2d, \dots, 1+100d$ from their mean is $255$,then $d$ is equal to:

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 about the mean for the following data is:$x_i$$2$$4$$5$$7$$9$$f_i$$2$$4$$10$$8$$6$