A) \[\sqrt{\frac{17}{12}}\]
B) \[\sqrt{\frac{{{47}^{2}}-1}{12}}\]
C) \[2\sqrt{6}\]
D) \[4\sqrt{3}\]
E) \[\frac{5}{12}\]
Correct Answer: C
Solution :
Given data is 31, 32, 33, ..., 46, 47 \[\therefore \]\[\overline{x}=\frac{31+32+33+.....+47}{17}\] \[=\frac{663}{17}=39\] \[\therefore \]\[\sum\limits_{i=1}^{17}{{{({{x}_{i}}-\overline{x})}^{2}}}={{(31-39)}^{2}}+{{(32-39)}^{2}}\] \[+{{(33-39)}^{2}}+{{(34-39)}^{2}}+{{(35-39)}^{2}}\] \[+{{(36-39)}^{2}}+{{(37-39)}^{2}}+{{(38-39)}^{2}}\] \[+{{(39-39)}^{2}}+{{(40-39)}^{2}}+{{(41-39)}^{2}}\] \[+{{(42-39)}^{2}}+{{(43-39)}^{2}}+{{(44-39)}^{2}}\] \[+{{(45-39)}^{2}}+{{(46-39)}^{2}}+{{(47-39)}^{2}}\] \[=64+49+36+25+16+9+4+1+0+1\] \[+4+9+16+25+36+49+64\] \[=408\] Standard deviation\[=\sqrt{\frac{\sum\limits_{i=1}^{17}{{{({{x}_{i}}-\overline{x})}^{2}}}}{n}}\] \[=\sqrt{\frac{408}{17}}=\sqrt{24}=2\sqrt{6}\]
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