A) 582.5
B) 507.5
C) 586.5
D) 509.5
Correct Answer: B
Solution :
\[S.D.=3\] variance \[=\text{ }9\] \[9\,\,=\,\,\frac{1}{5}\,\sum\limits_{i\,=\,1}^{5}{{{x}^{2}}_{1}-{{(\overline{x})}^{2}}}\] \[{{x}^{2}}_{1}+{{x}^{2}}_{2}+{{x}^{2}}_{3}+{{x}^{2}}_{4}+{{x}^{2}}_{5}=5\times 109\] variance of \[{{x}_{1}},\,\,{{x}_{2}},\text{ }{{x}_{3}},\text{ }{{x}_{4}},\text{ }{{x}_{5}}\text{ }\And -\,50\] variance \[=\,\,\frac{1}{6}\sum\limits_{i\,=\,1}^{6}{{{x}^{2}}_{1}-{{(\overline{x})}^{2}}}\] \[=\frac{1}{6}({{x}^{2}}_{1}+{{x}^{2}}_{2}+{{x}^{2}}_{3}+{{x}^{2}}_{4}+{{x}^{2}}_{5}+2500)-{{(\overline{x})}^{2}}\] \[(\because \,\,\,\overline{x}\,=\,\,0\,\,of\,\,\,{{x}_{1}},\,\,{{x}_{2}},\text{ }{{x}_{3}},\text{ }{{x}_{4}},\text{ }{{x}_{5}}\text{, }-\,50)\] \[=\,\,\frac{1}{6}\left( 5\times 109+2500 \right)\] = 507.5
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