Question

The S.D. of the numbers 31, 32, 33, .... 47 is

• A. $2\sqrt{6}$
• B. $4\sqrt{3}$
• C. $\sqrt{\frac{{47}^2-1}{12}}$
• D. None of these

Answer 1

The correct option is A $2\sqrt{6}$

Given series $31,32,33,....47$
which is an AP.
Last term $47=a+(n-1)d$
$\Rightarrow 47=31+(n-1)$
$\Rightarrow n=17$
Sum of the series $\sum x_i=\frac{17}{2}(31+47)$
$\Rightarrow \sum x_i=663$
Mean $\overline{x}=\frac{\sum x_i}{n}$
$\overline{x}=\frac{663}{17}=39$
Deviations of $x_i$ from mean is $8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8$
$\sum (x_i-\overline{x}{)}^2=8^2+7^2+.....+6^2+7^2+8^2=408$
Now, $\sigma =\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{n}}$
$\sigma =\sqrt{\frac{408}{39}}=\sqrt{24}=2\sqrt{6}$