A) 40
B) 49
C) 48
D) 45
Correct Answer: C
Solution :
Let 7 observations be \[{{x}_{1}},\,\,{{x}_{2}},\,\,{{x}_{3}},\,\,{{x}_{4}},\,\,{{x}_{5}},\,\,{{x}_{6}},\,\,{{x}_{7}}\] \[\overline{x}=8\Rightarrow \sum\limits_{i=1}^{7}{{{x}_{i}}=56}\] ?..(a) Also\[{{\sigma }^{2}}=16\] \[\Rightarrow \]\[16=\frac{1}{7}\left( \sum\limits_{i=1}^{7}{x_{i}^{2}} \right)-{{(\overline{x})}^{2}}\] \[\Rightarrow \]\[16=\frac{1}{7}\left( \sum\limits_{i=1}^{7}{x_{i}^{2}} \right)-64\] \[\Rightarrow \]\[\left( \sum\limits_{i=1}^{7}{x_{i}^{2}} \right)=560\] ?.(b) Now, \[{{x}_{1}}=2,{{x}_{2}}=4,{{x}_{3}}=10,{{x}_{4}}=12,{{x}_{5}}=14\] \[\Rightarrow \]\[{{x}_{6}}+{{x}_{7}}=14\] (from (a)) & \[x_{6}^{2}+x_{7}^{2}=100\] (from (b)) \[\therefore \]\[x_{6}^{2}+x_{7}^{2}={{\left( {{x}_{6}}+{{x}_{7}} \right)}^{2}}-2{{x}_{6}}.{{x}_{7}}\]\[\Rightarrow \]\[{{x}_{6}}.{{x}_{7}}=48\]
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