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\[\bar{x}=\frac{\sum\limits_{i=1}^{6}{{{x}_{i}}}}{6}=8\] \[\Rightarrow \]\[\sum\limits_{i=1}^{6}{{{x}_{i}}=8\times 6\Rightarrow }\sum\limits_{i=1}^{6}{{{x}_{i}}}=48\] 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\[\vec{x}=\frac{\sum\limits_{i=1}^{6}{3{{x}_{i}}}}{6}\frac{3\sum\limits_{i=1}^{6}{{{x}_{i}}}}{6}\] \[\Rightarrow \]\[\bar{x}=\frac{3\times 48}{6}=24\]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 \[=\sqrt{Variance}=\sqrt{144}=12\]
You need to login to perform this action.
You will be redirected in
3 sec
