A) \[\frac{11}{2}\]
B) 6
C) \[\frac{13}{2}\]
D) \[\frac{5}{2}\]
Correct Answer: A
Solution :
\[\sigma _{x}^{2}=4,\,\sigma _{y}^{2}\,=5,\,\,\overline{x}=2,\,\overline{y}=4,\,n=5\] \[\therefore \,\frac{1}{5}\sum\limits_{{}}^{{}}{y_{i}^{2}-{{(4)}^{2}}=5\Rightarrow \,\sum\limits_{{}}^{{}}{y_{i}^{2}=105}}\] \[\frac{1}{5}\sum\limits_{{}}^{{}}{y_{i}^{2}\,-{{(4)}^{2}}\,=5\Rightarrow \,\sum\limits_{{}}^{{}}{y_{i}^{2}\,=105}}\] \[\therefore \,\sum\limits_{{}}^{{}}{(x_{i}^{2}\,+y_{i}^{2})\,=40\,+105\,=145}\] Also \[\sum\limits_{{}}^{{}}{({{x}_{i}}+{{y}_{i}})\,=5(2)+5(4)\,=30}\] So, variance of combined data \[=\frac{1}{10}\,\sum\limits_{{}}^{{}}{(x_{i}^{2}\,+y_{i}^{2})\,-{{\left( \frac{\sum\limits_{{}}^{{}}{({{x}_{i}}+{{y}_{i}})}}{10} \right)}^{2}}}\] \[=\frac{145}{10}-\,{{\left( \frac{30}{10} \right)}^{2}}=\frac{145}{10}-9\,=\frac{55}{10}\,=\frac{11}{2}\]
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