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JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    Let \[{{x}_{1}},{{x}_{2}},\,....\,,{{x}_{n}}\] be n observations, and let \[\overline{x}\] be their arithmetic mean and \[{{\sigma }^{2}}\] be the variance Statement-1: Variance of \[2{{x}_{1}},2{{x}_{2}},\,......,\,2{{x}_{n}}\] is \[4{{\sigma }^{2}}\]. Statement-2: Arithmetic mean \[2{{x}_{1}},2{{x}_{2}},\,......,\,2{{x}_{n}}\] is \[4\overline{x}\].   AIEEE  Solved  Paper-2012

    A) Statement-1 is false, Statement-2 is true.

    B) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for  Statement-1.

    C) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

    D) Statement-1 is true, statement-2 is false.

    Correct Answer: D

    Solution :

                 A.M. of \[2{{x}_{1}},2{{x}_{2}}......\,2{{x}_{n}}\] is \[\frac{2{{x}_{1}}+2{{x}_{2}}+.....+2{{x}_{n}}}{n}\] \[=2\left( \frac{{{x}_{1}}+{{x}_{2}}+......+{{x}_{n}}}{n} \right)=2\overline{x}\] So statement-2 is false variance \[(2{{x}_{i}})={{2}^{2}}\] variance \[({{x}_{i}})=4{{\sigma }^{2}}\]so statement-1 is true.


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