A) 3
B) 1
C) 7
D) 5
Correct Answer: C
Solution :
Let the other two observations are \[{{x}_{1}}\]and \[{{x}_{2}}\] \[\therefore \]Mean\[(\overline{x})=\frac{3+4+4+{{x}_{1}}+{{x}_{2}}}{5}=4\] \[\Rightarrow \]\[{{x}_{1}}+{{x}_{2}}=9\] ?(i) And variance \[({{\sigma }^{2}})=\frac{\sum\limits_{{}}^{{}}{x_{i}^{2}}}{n}-{{(\overline{x})}^{2}}\] \[\Rightarrow \]\[5.20=\frac{9+16+16+x_{1}^{2}+x_{2}^{2}}{5}-16\] \[\Rightarrow \]\[\frac{41+x_{1}^{2}+x_{2}^{2}}{5}=21.2\Rightarrow x_{1}^{2}+x_{2}^{2}=106-41\] \[\Rightarrow \]\[x_{1}^{2}+x_{2}^{2}=65\] ...(ii) Using (i) and (ii), we get \[{{({{x}_{1}}+{{x}_{2}})}^{2}}=81\Rightarrow 65+2{{x}_{1}}{{x}_{2}}=81\]\[\Rightarrow \]\[2{{x}_{1}}{{x}_{2}}=16\] \[\therefore \]\[{{({{x}_{1}}-{{x}_{2}})}^{2}}=65-16=49\]\[\Rightarrow \]\[|{{x}_{1}}-{{x}_{2}}|=7\]
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