If alpha, beta, gamma are the mean deviations | vedclass

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(B) Given data: $1, 2, 2, 3, 3, 3, 4, 6$. Mean $= \frac{1+2+2+3+3+3+4+6}{8} = \frac{24}{8} = 3$. Median is the average of the $4^{th}$ and $5^{th}$ te

Vedclass · Accepted Answer

$\alpha = \beta = \gamma$

Vedclass · Answer

(B) Given data: $1, 2, 2, 3, 3, 3, 4, 6$. Mean $= \frac{1+2+2+3+3+3+4+6}{8} = \frac{24}{8} = 3$. Median is the average of the $4^{th}$ and $5^{th}$ terms: $\frac{3+3}{2} = 3$. Mode is the value with the highest frequency,which is $3$. Since Mean $=$ Median $=$ Mode $= 3$,the mean deviations about these central tendencies will be identical. Therefore,$\alpha = \beta = \gamma$. Hence,option $(B)$ is correct.

Class	$1 - 10$	$11 - 20$	$21 - 30$	$31 - 40$	$41 - 50$
$f_i$	$5$	$7$	$8$	$6$	$4$

If $\alpha, \beta, \gamma$ are the mean deviations about the mean,median,and mode of the data $1, 2, 2, 3, 3, 3, 4, 6$ respectively,then

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Class $1 - 10$ $11 - 20$ $21 - 30$ $31 - 40$ $41 - 50$

$f_i$ $5$ $7$ $8$ $6$ $4$

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