(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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Karl Pearson's coefficient of skewness of a distribution is $0.32$. Its standard deviation $(S.D.)$ is $6.5$ and the mean is $39.6$. The median of the distribution is:

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Let $a_1, a_2, \dots, a_{101}$ be a group of real numbers such that $a_i > a_{i+1}$ for all values of $i$ and the mean square deviation of the group is minimum about the number $a_{51}$. Then the mode of the group will be:

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For a discrete series (where all values are not equal),what is the relationship between Mean Deviation $(M.D.)$,Mean $(Mean)$,and Standard Deviation $(S.D.)$?

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If $\text{Mean} = x(3 \times \text{Median} - \text{Mode})$,then what is the value of $x$?

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Find the mode of the following frequency distribution.

$x$ $4$ $5$ $6$ $7$ $8$

$f$ $6$ $7$ $10$ $8$ $3$

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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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Karl Pearson's coefficient of skewness of a distribution is $0.32$. Its standard deviation $(S.D.)$ is $6.5$ and the mean is $39.6$. The median of the distribution is:

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