JEE Main 2023 — Mathematics Probability & Statistics

Let the elements in set $A$A be ${{x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}}$x1​,x2​,x3​,x4​,x5​

Now given mean is$=5$=5

Now subtracting $3$3 from each term we get,

New mean as $({x}_{1}-3,{x}_{2}-3\ldots ,{x}_{5}-3)=5-3=2$(x1​−3,x2​−3…,x5​−3)=5−3=2

So, sum of elements will be $2\times 5=10$2×5=10

Also given variance,

$$Var(X)=12$$Var(X)=12

Now we know that subtracting $3$3 from each term will not change the variance,

So, $Var({x}_{1}-3,{x}_{2}-3,\ldots {x}_{5}-3)=12$Var(x1​−3,x2​−3,…x5​−3)=12

$$\Rightarrow \frac{\sum {({x}_{i}-3)}^{2}}{5}-4=12$$⇒5∑(xi​−3)2​−4=12

$$\Rightarrow \sum {({x}_{i}-3)}^{2}=80$$⇒∑(xi​−3)2=80

Now let elements in set $B$B be ${{y}_{1},{y}_{2}\ldots {y}_{5}}$y1​,y2​…y5​

Given mean $({y}_{1},{y}_{2}\ldots {y}_{5})=8$(y1​,y2​…y5​)=8

Now adding each element by $2$2 we get,

New mean $({y}_{1}+2,{y}_{2}+2,\ldots {y}_{5}+2)=10$(y1​+2,y2​+2,…y5​+2)=10

So, sum of elements will be $10\times 5=50$10×5=50

Also given $Var({y}_{1},{y}_{2}\ldots {y}_{5})=20$Var(y1​,y2​…y5​)=20

Similarly new variance,

$$Var({y}_{1}+2,{y}_{2}+2\ldots {y}_{5}+2)=20$$Var(y1​+2,y2​+2…y5​+2)=20

$$\Rightarrow \frac{\sum {({y}_{i}+2)}^{2}}{5}-100=20$$⇒5∑(yi​+2)2​−100=20

$$\Rightarrow \sum {({y}_{i}+2)}^{2}=120\times 5$$⇒∑(yi​+2)2=120×5

Now finding, combined mean we get,

$\displaystyle \frac{{\sum }_{i=1}^{5}({x}_{i}-3)+\sum ({y}_{i}+2)}{10}$10∑i=15​(xi​−3)+∑(yi​+2)​$=\frac{10+50}{10}=6$=1010+50​=6

And combined variance $\displaystyle =\frac{\sum {({x}_{i}-3)}^{2}+\sum {({y}_{i}+2)}^{2}}{10}-{6}^{2}$=10∑(xi​−3)2+∑(yi​+2)2​−62

$$=\frac{80+120\times 5}{10}-36=32$$=1080+120×5​−36=32

Hence, the sum of combined mean and variance will be $32+6=38$32+6=38