If the sum of the deviations of 50 observatio | vedclass

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(D) Let the $50$ observations be $x_1, x_2, ..., x_{50}$.Given that the sum of deviations from $30$ is $50$,we have:$\sum_{i=1}^{50} (x_i - 30) = 50$E

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(D) Let the $50$ observations be $x_1, x_2, ..., x_{50}$.Given that the sum of deviations from $30$ is $50$,we have:$\sum_{i=1}^{50} (x_i - 30) = 50$Expanding the summation:$\sum_{i=1}^{50} x_i - \sum_{i=1}^{50} 30 = 50$$\sum_{i=1}^{50} x_i - (50 	imes 30) = 50$$\sum_{i=1}^{50} x_i - 1500 = 50$$\sum_{i=1}^{50} x_i = 1550$The mean $\bar{x}$ is given by $\frac{\sum x_i}{n}$:$\bar{x} = \frac{1550}{50} = 31$Therefore,the mean of the observations is $31$.

If the sum of the deviations of $50$ observations from $30$ is $50$,then the mean of these observations is

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