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.
(N/A) Given $n=10, \bar{x}=45$ and $\sigma^{2}=16$.
$\bar{x} = \frac{\Sigma x_{i}}{n} = 45 \Rightarrow \Sigma x_{i} = 450$.
Corrected $\Sigma x_{i} = 450 - 52 + 25 = 423$.
Corrected mean $\bar{x} = \frac{423}{10} = 42.3$.
Using $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$:
$16 = \frac{\Sigma x_{i}^{2}}{10} - (45)^{2} \Rightarrow \Sigma x_{i}^{2} = 10(16 + 2025) = 20410$.
Corrected $\Sigma x_{i}^{2} = 20410 - (52)^{2} + (25)^{2} = 20410 - 2704 + 625 = 18331$.
Corrected $\sigma^{2} = \frac{18331}{10} - (42.3)^{2} = 1833.1 - 1789.29 = 43.81$.
$\bar{x} = \frac{\Sigma x_{i}}{n} = 45 \Rightarrow \Sigma x_{i} = 450$.
Corrected $\Sigma x_{i} = 450 - 52 + 25 = 423$.
Corrected mean $\bar{x} = \frac{423}{10} = 42.3$.
Using $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$:
$16 = \frac{\Sigma x_{i}^{2}}{10} - (45)^{2} \Rightarrow \Sigma x_{i}^{2} = 10(16 + 2025) = 20410$.
Corrected $\Sigma x_{i}^{2} = 20410 - (52)^{2} + (25)^{2} = 20410 - 2704 + 625 = 18331$.
Corrected $\sigma^{2} = \frac{18331}{10} - (42.3)^{2} = 1833.1 - 1789.29 = 43.81$.
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