Official solution
Given that,
\[{\sigma _1}^2 = 4\]
and \[{\sigma _2}^2 = 5\]
And also given,
\[\overline x = 2\,\,\] and \[\overline y = 4\,\,\]
So, \[\,\,\,\] \[{{\sum {{x_i}} } \over 5} = 2\]
\[ \Rightarrow \sum {{x_i}} = 10\] v
and \[{{\sum {{y_i}} } \over 5} = 4\]
\[ \Rightarrow \,\,\,\sum {{y_i}} = 20\]
\[\therefore\,\,\,\] \[{\sigma _1}^2 = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}\]
\[ \Rightarrow \,\,\,4 = {{\sum {x_i^2} } \over 5} - 4\]
\[ \Rightarrow \,\,\,\sum {x_i^2} = 40\]
and \[{\sigma _2}^2 = {{\sum {y_i^2} } \over 5} - {\left( {\overline y } \right)^2}\]
\[ \Rightarrow \,\,\,\,5 = {{\sum {y_i^2} } \over 5} - 16\]
\[ \Rightarrow \,\,\,\sum {y_i^2} = 105\]
Variance of combined data set
\[{\sigma ^2} = {1 \over {10}}\left( {\sum {x_i^2 + \sum {y_i^2} } } \right) - {\left( {{{\overline x + \overline y } \over 2}} \right)^2}\]
\[ = {1 \over {10}}\left( {40 + 105} \right) - 9\]
\[ = {{145 - 90} \over {10}}\]
\[ = {{55} \over {10}}\]
\[ = {{11} \over {2}}\]