A) 12
B) 18
C) 24
D) 144
Correct Answer: A
Solution :
Let the observations be \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}}\]and \[{{x}_{6}},\] so their mean \[\overline{x}=\frac{\sum\limits_{i=l}^{6}{{{x}_{i}}}}{6}=8\] \[\Rightarrow \]\[\sum\limits_{i=l}^{6}{{{x}_{i}}}=8\times 6\]\[\Rightarrow \]\[\sum\limits_{i=l}^{6}{{{x}_{i}}}=48\] On multiplying each observation by 3, we get the new observations as \[3{{x}_{1}},3{{x}_{2}},3{{x}_{3}},3{{x}_{4}},3{{x}_{5}}\]and \[3{{x}_{6}}.\] Now, their mean \[\overline{x}=\frac{\sum\limits_{i=l}^{6}{3{{x}_{i}}}}{6}=\frac{3\sum\limits_{i=l}^{6}{{{x}_{i}}}}{6}\] \[\Rightarrow \]\[\overline{x}=\frac{3\times 48}{6}=24\]Variance of new observations\[=\frac{\sum\limits_{i=l}^{6}{{{(3{{x}_{i}}-24)}^{2}}}}{6}=\frac{{{3}^{2}}\sum\limits_{i=l}^{6}{{{({{x}_{i}}-8)}^{2}}}}{6}\] \[=\frac{9}{1}\times \]Variance of old observations = 9 x 42 = 144 Thus, standard deviation of new observations \[=\sqrt{\text{Variance}}=\sqrt{144}=12\]
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