JEE Main 2023 — Statistics Question with Solution

Let the elements in set $A$ be $\left\{x_1,x_2,x_3,x_4,x_5\right\}$

Now given mean is$=5$

Now subtracting $3$ from each term we get, 

New mean as $\left(x_1-3,x_2-3\dots ,x_5-3\right)=5-3=2$

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

Also given variance,

$Var\left(X\right)=12$

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

So, $Var\left(x_1-3,x_2-3,\dots x_5-3\right)=12$

$\Rightarrow \frac{\sum {\left(x_i-3\right)}^2}{5}-4=12$

$\Rightarrow \sum {\left(x_i-3\right)}^2=80$

Now let elements in set $B$ be $\left\{y_1,y_2\dots y_5\right\}$

Given mean $\left(y_1,y_2\dots y_5\right)=8$

Now adding each element by $2$ we get,

New mean $\left(y_1+2,y_2+2,\dots y_5+2\right)=10$

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

Also given $Var\left(y_1,y_2\dots y_5\right)=20$

Similarly new variance, 

$Var\left(y_1+2,y_2+2\dots y_5+2\right)=20$

$\Rightarrow \frac{\sum {\left(y_i+2\right)}^2}{5}-100=20$

$\Rightarrow \sum {\left(y_i+2\right)}^2=120\times 5$

Now finding, combined mean we get,

$\frac{{\sum }_{i=1}^5\left(x_i-3\right)+\sum \left(y_i+2\right)}{10}$$=\frac{10+50}{10}=6$

And combined variance $=\frac{\sum {\left(x_i-3\right)}^2+\sum {\left(y_i+2\right)}^2}{10}-6^2$

$=\frac{80+120\times 5}{10}-36=32$

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