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.

Solution:

Given that $n=10, \bar{x}=45$ and $\sigma^{2}=16$
$\begin{array}{l}
\bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\
\Rightarrow \quad \frac{\Sigma x_{i}}{10}=45 \Rightarrow \Sigma x_{i}=450
\end{array}$

$Corrected \Sigma x_{i}=450-52+25=423$

$\therefore \quad Corrected mean, \bar{x}=\frac{423}{10}=42.3$

$\Rightarrow \quad \sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}$

$\Rightarrow \quad 16=\frac{\Sigma x_{i}^{2}}{10}-(45)^{2}$

$\Rightarrow \quad \Sigma x_{i}^{2}=20410$

$\therefore \quad Corrected \Sigma x_{i}^{2}=20410-(53)^{2}+(25)^{2}=18331$

Corrected $\sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81$