Question

The standard deviation of the numbers $31, 32, 33, ..., 46, 47$ is

• A. $ \sqrt{\frac{17}{12}} $
• B. $ \sqrt{\frac{{{47}^{2}}-1}{12}} $
• C. $ 2\sqrt{6} $
• D. $ 4\sqrt{3} $
• E. $ \frac{5}{12} $

Answer 1

Answer: (C)

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} $