A) 3 and 6
B) 2 and 6
C) 5 and 8
D) 1 and 7
Correct Answer: D
Solution :
[d] Let the other two observations be ?a? and ?b? \[\therefore \] mean \[=\frac{2+4+6+a+b}{5}\Rightarrow 4=\frac{12+a+b}{5}\] \[\Rightarrow a+b=8;\] Variance \[=\frac{1}{n}\sum{{{x}^{2}}}-{{x}^{-2}}=5.2\] \[\Rightarrow \frac{1}{5}\left( 4+16+36+{{a}^{2}}+{{b}^{2}} \right)-16=5.2\Rightarrow {{a}^{2}}+{{b}^{2}}\]\[=50\] From the options, it is clear that the two observations are 1 and 7.
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