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)

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} $