Let sets A and B have 5 elements each. Let the mean of the elements...

Let the elements in set $A$ be \left\{{x}_{1}, {x}_{2}, {x}_{3}, {x}_{4}, {x}_{5}\right\}$</p><p>Now given mean is$= 5$</p><p>Now subtracting$3$from each term we get, </p><p>New mean as \left({x}_{1} - 3, {x}_{2} - 3 \ldots , {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$</p><p>Now we know that subtracting$3$from each term will not change the variance,</p><p>So,$Var \left({x}_{1} - 3, {x}_{2} - 3, \ldots {x}_{5} - 3\right) = 12$</p><p>$⇒ \dfrac{∑ {\left({x}_{i} - 3\right)}^{2}}{5} - 4 = 12$</p><p>$⇒ ∑ {\left({x}_{i} - 3\right)}^{2} = 80$</p><p>Now let elements in set$B$be \left\{{y}_{1}, {y}_{2} \ldots {y}_{5}\right\}$

Given mean \left({y}_{1}, {y}_{2} \ldots {y}_{5}\right) = 8$</p><p>Now adding each element by$2$we get,</p><p>New mean \left({y}_{1} + 2, {y}_{2} + 2, \ldots {y}_{5} + 2\right) = 10$

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

Also given $Var \left({y}_{1}, {y}_{2} \ldots {y}_{5}\right) = 20$

Similarly new variance,

Var \left({y}_{1} + 2, {y}_{2} + 2 \ldots {y}_{5} + 2\right) = 20$</p><p>$⇒ \dfrac{∑ {\left({y}_{i} + 2\right)}^{2}}{5} - 100 = 20$</p><p>$⇒ ∑ {\left({y}_{i} + 2\right)}^{2} = 120 \times 5$</p><p>Now finding, combined mean we get,</p><p>$\dfrac{{∑}_{i = 1}^{5}\left({x}_{i} - 3\right) + ∑ \left({y}_{i} + 2\right)}{10}= \dfrac{10 + 50}{10} = 6$</p><p>And combined variance$= \dfrac{∑ {\left({x}_{i} - 3\right)}^{2} + ∑ {\left({y}_{i} + 2\right)}^{2}}{10} - {6}^{2}$</p><p>$= \dfrac{80 + 120 \times 5}{10} - 36 = 32$</p><p>Hence, the sum of combined mean and variance will be$32 + 6 = 38