Let the mean and variance of 12 observations be (9)/(2) and 4...

Given,

The mean and variance of $12$ observations be $\dfrac{9}{2}$ and $4$ respectively,

So, mean will be,

$\overset{¯}{x} = \dfrac{{x}_{1} + {x}_{2} + \ldots \ldots + 9 + 10 + \ldots . . + {x}_{12}}{12} = \dfrac{9}{2}$

$⇒ {x}_{1} + {x}_{2} \ldots \ldots + {x}_{12} + 19 = 54$

Now removing the observation $9 \text{ and } 10$ and adding the observation $7 \text{ and } 14$,

So, the new total sum of the observation will be,

${x}_{1} + {x}_{2} + \ldots \ldots + {x}_{12} + 7 + 14 = 54 - 19 + 7 + 14$

Now new mean will be,

$\dfrac{{x}_{1} + {x}_{2} + \ldots . + {x}_{12} + 7 + 14}{12} = \dfrac{56}{12}$

$⇒ \overset{¯}{x}\text{new} = \dfrac{14}{3}$
Now using the formula of the variance we get,

$\dfrac{{x}_{1}^{2} + {x}_{2}^{2} + \ldots \ldots + {x}_{12}^{2} + {9}^{2} + {10}^{2}}{12} - {\left(\dfrac{9}{2}\right)}^{2} = 4$

$⇒ {x}_{1}^{2} + {x}_{2}^{2} + \ldots \ldots + {x}_{12}^{2} + {9}^{2} + {10}^{2} - 81 \times 3 = 4 \times 12$

$⇒ {x}_{1}^{2} + {x}_{2}^{2} + \ldots \ldots + {x}_{12}^{2} + {9}^{2} + {10}^{2} = 291$

Now removing ${9}^{2} \text{ and } {10}^{2}$ and adding ${7}^{2} \text{ and } {14}^{2}$ we get,

$⇒ {x}_{1}^{2} + {x}_{2}^{2} + \ldots \ldots + {x}_{12}^{2} + {7}^{2} + {14}^{2} = 355$
New variance $= \dfrac{∑ {x}_{i}^{2}}{N} - ( \overset{¯}{x}{)}^{2}$
$= \dfrac{355}{12} - {\left(\dfrac{14}{3}\right)}^{2}$
$= \dfrac{281}{36}$

Now on comparing with $\dfrac{m}{n} = \dfrac{281}{36}$ we get,

$⇒ m = 281$ and $n = 36$

$∴ m + n = 317$