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
$\therefore$ $\overline{x}=\frac{31+32+33+.....+47}{17}$
$=\frac{663}{17}=39$
$\therefore$ $\sum _{i=1}^{17}{(x_i-\overline{x})}^2={(31-39)}^2+{(32-39)}^2$ $+{(33-39)}^2+{(34-39)}^2+{(35-39)}^2$ $+{(36-39)}^2+{(37-39)}^2+{(38-39)}^2$ $+{(39-39)}^2+{(40-39)}^2+{(41-39)}^2$ $+{(42-39)}^2+{(43-39)}^2+{(44-39)}^2$ $+{(45-39)}^2+{(46-39)}^2+{(47-39)}^2$
$=64+49+36+25+16+9+4+1+0+1$ $+4+9+16+25+36+49+64$
$=408$ Standard deviation
$=\sqrt{\frac{\sum _{i=1}^{17}{(x_i-\overline{x})}^2}{n}}$
$=\sqrt{\frac{408}{17}}=\sqrt{24}=2\sqrt{6}$