$\begin{array}{1 1}(A)\; 42.3,43.81\\(B)\;67,45\\(C)\;16,10\\(D)\;6,45\end{array} $

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- The formulae used to calculate mean and variance are $Mean (\bar{x}) = \large\frac{\sum x}{n}$
- Variance $( \sigma^2)=\bigg[\large\frac{\sum x^2}{n}- \bigg(\frac{\sum x}{n}\bigg)^2\bigg]$

Mean $\bar{x}=45$

Variance $= (\sigma^2)=16$

$n=10$

Mean $= \large\frac{\sum x}{n}$

=> $45 = \large\frac{\sum x}{10}$

$\sum x=450$

Now replacing the wrong reading 52 by 25

$\therefore \sum x= 450-52+25=423$

The correct mean $\bar{x}= \large\frac{ \sum x}{n} =\frac{423}{10}$

$\qquad = 42.3$

Mean = 42.3

Given variance= 16

$\therefore \large\frac{\sum x^2}{n} - \bigg( \large\frac{\sum x}{n} \bigg)^2 $$=16$

$\large\frac{\sum x^2}{10} - \bigg( \large\frac{450}{10} \bigg)^2$$=16$

$\large\frac{ \sum x^2}{10 }$$= 16 +45^2$

$\large\frac{\sum x^2}{10} $$=16 +2025$

$\sum x^2=2041 \times 10$

$\qquad= 20410$

now replacing the wrong reading 52 by 25

$\sum x^2 =20410 -52^2 +25^2$

$\qquad= 20410-2704+625$

$\qquad= 18331$

Correct variance $\sigma ^2 =\large\frac{\sum x^2}{n} - \bigg( \large\frac{\sum x}{n}\bigg)^2$

$\qquad= \large\frac{18331}{10} - \bigg( \large\frac{423}{10}\bigg)^2$

$\qquad= 1833.1 -(42.3^2)$

$\qquad= 1833.1 -1789.29$

$\qquad= 43.81$

Hence A is the correct answer.

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