Question

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

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

Answer 1

Answer: (D)

$\frac{3}{2}$

Answer 2

Standard deviation $=\sqrt{\frac{\underset{j=1}{\mathop{\overset{18}{\mathop{\Sigma }}\,}}\,{{({{x}_{j}}-8)}^{2}}}{n}-{{\left( \frac{\underset{j=1}{\mathop{\overset{18}{\mathop{\Sigma }}\,}}\,({{x}_{k}}-8)}{n} \right)}^{2}}}$
$=\sqrt{\frac{45}{18}-{{\left( \frac{9}{18} \right)}^{2}}}$
$=\sqrt{\frac{45}{18}-\frac{1}{4}}=\sqrt{\frac{81}{36}}=\frac{9}{6}=\frac{3}{2}$