If 10 is the mean of the data 2, 3, 5, 18, 17 | vedclass

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(B) Given the mean $\bar{x} = 10$ for the $10$ observations: $\frac{2+3+5+18+17+15+13+x+9+7}{10} = 10$ $\Rightarrow 89 + x = 100$ $\Rightarrow x = 11$

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$4.8$

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(B) Given the mean $\bar{x} = 10$ for the $10$ observations: $\frac{2+3+5+18+17+15+13+x+9+7}{10} = 10$ $\Rightarrow 89 + x = 100$ $\Rightarrow x = 11$ The data set is ${2, 3, 5, 18, 17, 15, 13, 11, 9, 7}$. The mean deviation about the mean is calculated as $\frac{1}{n} \sum |x_i - \bar{x}|$:   $x_i$ $|x_i - 10|$   $2$ $8$   $3$ $7$   $5$ $5$   $18$ $8$   $17$ $7$   $15$ $5$   $13$ $3$   $11$ $1$   $9$ $1$   $7$ $3$   Sum of absolute deviations $= 8+7+5+8+7+5+3+1+1+3 = 48$. Mean Deviation $= \frac{48}{10} = 4.8$.

Class Interval	$0-4$	$4-8$	$8-12$	$12-16$
Frequency	$4$	$3$	$2$	$1$

If $10$ is the mean of the data $2, 3, 5, 18, 17, 15, 13, x, 9, 7$,then the mean deviation of this data about its mean is

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