Question

If $x_1,x_2,...,x_{18}$ are observations such that $\sum _{j=1}^{18}\left(x_j-8\right)=9$ and $\sum _{j=1}^{18}{\left(x_j-8\right)}^2=45$, then the standard deviation of these observations is

• A. $\sqrt{\frac{81}{34}}$
• B. $5$
• C. $\sqrt{5}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D

$\frac{3}{2}$

Explanation for the correct option.

Find the standard deviation.

The standard deviation is given as:

$SD=\sqrt{\frac{{\displaystyle \sum _{j=1}^{18}}{\left(x_j-8\right)}^2}{n}-{\left(\frac{{\displaystyle \sum _{j=1}^{18}\left(x_j-8\right)}}{n}\right)}^2}$.

Here, the number of terms is $18$, $\sum _{j=1}^{18}\left(x_j-8\right)=9$ and $\sum _{j=1}^{18}{\left(x_j-8\right)}^2=45$, so the standard deviation is:

$\begin{array}{rcl}\mathrm{SD}& =& \sqrt{\frac{45}{18}-{\left(\frac{9}{18}\right)}^2}\\ & =& \sqrt{\frac{5}{2}-{\left(\frac{1}{2}\right)}^2}\\ & =& \sqrt{\frac{5}{2}-\frac{1}{4}}\\ & =& \sqrt{\frac{10-1}{4}}\\ & =& \sqrt{\frac{9}{4}}\\ & =& \frac{3}{2}\end{array}$

Thus the standard deviation of these observations is $\frac{3}{2}$.

Hence, the correct option is D.