The mean deviation from the mean 10 of the da | vedclass

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(A) Given,the mean $(\bar{x}) = 10$ and the number of observations $n = 8$. The sum of the observations is $6 + 7 + 11 + 12 + 13 + \alpha + 12 + 16 =

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$3.5$

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(A) Given,the mean $(\bar{x}) = 10$ and the number of observations $n = 8$. The sum of the observations is $6 + 7 + 11 + 12 + 13 + \alpha + 12 + 16 = 77 + \alpha$. Since $\bar{x} = \frac{\sum x_i}{n}$,we have $10 = \frac{77 + \alpha}{8}$. $80 = 77 + \alpha \Rightarrow \alpha = 3$. The data set is $6, 7, 11, 12, 13, 3, 12, 16$. The mean deviation about the mean is $\frac{1}{n} \sum |x_i - \bar{x}|$. $	ext{MD}(\bar{x}) = \frac{|6-10| + |7-10| + |11-10| + |12-10| + |13-10| + |3-10| + |12-10| + |16-10|}{8}$. $	ext{MD}(\bar{x}) = \frac{|-4| + |-3| + |1| + |2| + |3| + |-7| + |2| + |6|}{8}$. $	ext{MD}(\bar{x}) = \frac{4 + 3 + 1 + 2 + 3 + 7 + 2 + 6}{8} = \frac{28}{8} = 3.5$.

The mean deviation from the mean $10$ of the data $6, 7, 11, 12, 13, \alpha, 12, 16$ is

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