JEE Main 2023 — Mathematics Probability & Statistics

Given,

The mean and variance of $12$12 observations be $\frac{9}{2}$29​ and $4$4 respectively,

So, mean will be,

$$\bar{x}=\frac{{x}_{1}+{x}_{2}+\ldots \ldots +9+10+\ldots ..+{x}_{12}}{12}=\frac{9}{2}$$xˉ=12x1​+x2​+……+9+10+…..+x12​​=29​

$$\Rightarrow {x}_{1}+{x}_{2}\ldots \ldots +{x}_{12}+19=54$$⇒x1​+x2​……+x12​+19=54

Now removing the observation 9&10 and adding the observation 7&14,

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

$${x}_{1}+{x}_{2}+\ldots \ldots +{x}_{12}+7+14=54-19+7+14$$x1​+x2​+……+x12​+7+14=54−19+7+14

Now new mean will be,

$$\frac{{x}_{1}+{x}_{2}+\ldots .+{x}_{12}+7+14}{12}=\frac{56}{12}$$12x1​+x2​+….+x12​+7+14​=1256​

$$\Rightarrow {\bar{x}}_{\text{new}}=\frac{14}{3}$$⇒xˉnew​=314​

Now using the formula of the variance we get,

$$\frac{{x}_{1}^{2}+{x}_{2}^{2}+\ldots \ldots +{x}_{12}^{2}+{9}^{2}+{10}^{2}}{12}-{(\frac{9}{2})}^{2}=4$$12x12​+x22​+……+x122​+92+102​−(29​)2=4

$$\Rightarrow {x}_{1}^{2}+{x}_{2}^{2}+\ldots \ldots +{x}_{12}^{2}+{9}^{2}+{10}^{2}-81\times 3=4\times 12$$⇒x12​+x22​+……+x122​+92+102−81×3=4×12

$$\Rightarrow {x}_{1}^{2}+{x}_{2}^{2}+\ldots \ldots +{x}_{12}^{2}+{9}^{2}+{10}^{2}=291$$⇒x12​+x22​+……+x122​+92+102=291

Now removing {9}^{2}&{10}^{2} and adding {7}^{2}&{14}^{2} we get,

$$\Rightarrow {x}_{1}^{2}+{x}_{2}^{2}+\ldots \ldots +{x}_{12}^{2}+{7}^{2}+{14}^{2}=355$$⇒x12​+x22​+……+x122​+72+142=355

New variance $\displaystyle =\frac{\sum {x}_{i}^{2}}{N}-(\bar{x}{)}^{2}$=N∑xi2​​−(xˉ)2

$$=\frac{355}{12}-{(\frac{14}{3})}^{2}$$=12355​−(314​)2

$$=\frac{281}{36}$$=36281​

Now on comparing with $\frac{m}{n}=\frac{281}{36}$nm​=36281​ we get,

$\Rightarrow m=281$⇒m=281 and $n=36$n=36

$$\therefore m+n=317$$∴m+n=317