A) 20
B) 18
C) 16
D) 22
Correct Answer: A
Solution :
\[\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}}{5}\,\,=\,\,150\,cm\] \[\Rightarrow \,\,\,\,{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}\,\,=\,\,150\,\times 5\,=\,750\] \[\Rightarrow \,\,\,18=\frac{1}{5}-\left( {{x}_{1}}^{2}+{{x}^{2}}_{2}+{{x}^{2}}_{3}{{x}^{2}}_{4}{{x}^{2}}_{5} \right)-{{\left( 150 \right)}^{2}}\] \[\Rightarrow \,\,\,18+{{\left( 150 \right)}^{2}}\,\,=\,\,\frac{1}{5}\,\,\left( \sum\limits_{i\,=\,1}^{5}{{{x}_{i}}^{2}} \right)\] \[\Rightarrow \,\,\,112590\,\,=\,\,\,\left( \sum\limits_{i\,=\,1}^{5}{{{x}_{i}}^{2}} \right)\]. From (1) Now new student of height 156 cm \[\sigma \,=\,\frac{1}{6}\,\sum\limits_{i\,=\,1}^{6}{{{x}_{i}}^{2}}-{{(\overline{x})}^{2}}\] \[=\,\,\frac{1}{6}\,[{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}^{2}}_{3}+{{x}^{2}}_{5}+{{x}^{2}}_{6}]\,\,-\] \[\left( \frac{{{x}_{1}}\,+\,{{x}_{2}}\,\,+\,\,{{x}_{3}}\,\,+\,\,{{x}_{4}}\,\,+\,\,{{x}_{5}}\,\,+\,\,{{x}_{6}}}{6} \right)\] \[=\,\,\frac{1}{6}\,[{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}^{2}}_{3}+{{x}^{2}}_{5}+{{x}^{2}}_{6}]\,\,\] \[-\,\,{{\left( \frac{750\,\,+\,\,156}{6} \right)}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\,(2)\] Now (2) \[\frac{1}{6}\,[112590+{{156}^{2}}]-\,\,{{\left( \frac{750\,\,+\,\,156}{6} \right)}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\,(2)\] \[=\,\,22821\,-\,{{(151)}^{2}}\] \[=\text{ }22821\text{ }-\text{ }22801\] \[=\text{ }20\]
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