(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

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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 $5^{th}$ 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 11 Mathematics Statistics Relation between mean, median and mode

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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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A
$\alpha < \beta < \gamma$
B
$\alpha = \beta = \gamma$
C
$\alpha \neq \beta = \gamma$
D
$\beta < \alpha < \gamma$

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Relation between mean, median and mode

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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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If in a frequency distribution,the mean and median are $21$ and $22$ respectively,then its mode is approximately

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Find the mode of the following frequency distribution.

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