A) \[\frac{5}{2}\]
B) \[\frac{11}{2}\]
C) \[6\]
D) \[\frac{13}{2}\]
Correct Answer: B
Solution :
\[{{n}_{1}}=5\] \[{{n}_{2}}=5\] \[\sigma _{1}^{2}=4\] \[\sigma _{2}^{2}=5\] \[\overline{{{x}_{1}}}=2\] \[\overline{{{x}_{2}}}=4\] sum of data = 10 sum of data = 20 \[4=\frac{1}{5}\](sum of squares) \[-4\] \[5=\frac{1}{5}\](sum of squares) − 16 (as variance\[=\frac{\sum{x_{i}^{2}}}{n}-{{\left( \frac{\sum{{{x}_{i}}}}{n} \right)}^{2}})\] sum of squares = 40 sum of squares = 105 \[\overline{x}=\frac{10+20}{10}=3\] new variance \[=\frac{1}{10}(145)-9=\frac{11}{2}\]
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