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\[x=\frac{\sum\limits_{i=1}^{6}{{{x}_{1}}}}{6}=8\]. \[\Rightarrow \] \[\sum\limits_{i=1}^{6}{{{x}_{1}}}=8\times 6\Rightarrow \sum\limits_{i=1}^{6}{{{x}_{i}}}=48\] 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\[=x=\frac{\sum\limits_{i=1}^{6}{3{{x}_{1}}}}{6}=\frac{3\sum\limits_{i=1}^{6}{{{x}_{i}}}}{6}\] \[\Rightarrow \] \[\bar{x}=\frac{3\times 48}{6}=24\] Variance of new observations \[=\frac{\sum\limits_{i=1}^{6}{{{(3{{x}_{i}}-24)}^{2}}}}{6}=\frac{{{3}^{2}}\sum\limits_{i=1}^{6}{{{({{x}_{i}}-8)}^{2}}}}{6}\] \[=\frac{9}{1}\times \]Variance of old observations\[=9\times {{4}^{2}}=144\] Thus, standard deviation of new observations\[=\sqrt{Variance}=\sqrt{144}=12\]
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