A) 5
B) 10
C) 20
D) 40
Correct Answer: C
Solution :
[c] Let \[{{x}_{1}},{{x}_{2}},...,{{x}_{20}}\] be the given observations. Given, \[\frac{1}{20}\sum\limits_{i=1}^{20}{{{({{x}_{i}}-\bar{x})}^{2}}=5}\] To find variance of \[2{{x}_{1}},2{{x}_{2}},2{{x}_{3}},...2{{x}_{20}},\] Let \[\bar{x}\] denotes the mean of new observation, Clearly, \[\bar{x}=\frac{\sum\limits_{i=1}^{20}{2{{x}_{i}}}}{20}=\frac{2\sum\limits_{i=1}^{20}{{{x}_{i}}}}{20}=2\bar{x}\] Now, variance of new observation \[=\frac{1}{20}\sum\limits_{i=1}^{20}{{{(2{{x}_{i}}-\bar{x})}^{2}}=\frac{1}{20}\sum\limits_{i=1}^{20}{{{(2{{x}_{i}}-2\bar{x})}^{2}}}}\] \[=\frac{1}{20}\sum\limits_{i=1}^{20}{4{{({{x}_{i}}-\bar{x})}^{2}}=4\left( \frac{1}{20}\sum\limits_{i=1}^{20}{{{({{x}_{i}}-\bar{x})}^{2}}} \right)}\] \[=4\times 5=20\]
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