3 While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25 . He obtained the mean and variance as 45 and 16, respectively. Find the correct mean and the variance.
$$\begin{aligned} & \text { Given, } \quad n=10, \bar{x}=45 \text { and } \sigma^2=16 \\ & \therefore \quad \bar{x}=45 \Rightarrow \frac{\Sigma x_i}{n}=45 \\ & \Rightarrow \quad \frac{\Sigma x_i}{10}=45 \Rightarrow \Sigma x_i=450 \\ & \text { Corrected } \Sigma x_i=450-52+25=423 \\ \therefore\quad & \bar{x}=\frac{423}{10}=42.3 \\ & \sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2 \\ & 16=\frac{\Sigma x_i^2}{10}-(45)^2 \\ & \Sigma x_i^2=10(2025+16) \\ & \Sigma x_i^2=20410 \\ & \therefore \quad \text { Corrected } \Sigma x_i^2=20410-(52)^2+(25)^2=18331 \\ & \text { and } \quad \text { corrected } \sigma^2=\frac{18331}{10}-(42.3)^2=43.81 \end{aligned} $$
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