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.
Given $n=10, \bar{x}=45$ and $\sigma^{2}=16$
$\begin{array}{c}\bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\\Rightarrow \quad \frac{\Sigma x_{i}}{10}=45 \Rightarrow \quad \Sigma x_{i}=450 \\\text { Corrected } \Sigma x_{i}=450-52+25=423\end{array}$
$\therefore \quad$ Corrected mean, $\bar{x}=\frac{423}{10}=42.3$
$\Rightarrow \quad \sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}$
$\begin{array}{ll}\Rightarrow & 16=\frac{\Sigma x_{i}^{2}}{10}-(45)^{2} \\ \Rightarrow & \Sigma x_{i}^{2}=20410\end{array}$
$\therefore \quad$ Corrected $\Sigma x_{i}^{2}=20410-(53)^{2}+(25)^{2}=18331$
And Corrected $\sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81$
The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......
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