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

Question

Aicharya · Accepted Answer

Let the observations be $x_1, x_{2,}, x_3, \ldots x_{50}$
Then, $\sum_{i=1}^{50}\left(x_i-30\right)=50$ $\begin{aligned} & \Rightarrow \sum_{\mathrm{i}=1}^{50} x_{\mathrm{i}}-(30 \times 50)=50 \\ & \Rightarrow \sum_{\mathrm{i}=1}^{50} x_{\mathrm{i}}=1550 \end{aligned}$ $\text {Mean }=\frac{\sum_{\mathrm{i}=1}^{50} x_{\mathrm{i}}}{\mathrm{n}}=\frac{1550}{50}=31$

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

Solution