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Question
The mean deviation for n observations \[x_1 , x_2 , . . . , x_n\] from their mean \[\bar{X} \] is given by
Options
\[\sum^n_{i = 1} \left( x_i - X \right)\]
\[\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)\]
\[\sum^n_{i = 1} \left( x_i - X \right)^2\]
\[\frac{1}{n} \sum^n_{i = 1} \left( x_i - X \right)^2\]
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Solution
The mean deviation for n observations \[x_1 , x_2 , . . . , x_n\] from their mean \[\bar{ X } \] is \[\frac{1}{n} \sum^n_{i = 1} \left| x_i - X \right|\] .
\[\frac{1}{n} \sum^n_{i = 1} \left| x_i - X \right|\]
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