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J & K CET Engineering J and K - CET Engineering Solved Paper-2013

  • question_answer
    The variance of the data      
    \[x:1\] \[a\] \[{{a}^{2}}\] ?.. \[{{a}^{n}}\]
    \[f{{:}^{n}}{{C}_{0}}\] \[^{n}{{C}_{1}}\] \[^{n}{{C}_{2}}\] ?.. \[^{n}{{C}_{n}}\]

    A)  \[{{\left( \frac{1+{{a}^{2}}}{2} \right)}^{n}}-{{\left( \frac{1+a}{2} \right)}^{2n}}\]

    B)  \[{{\left( \frac{1+{{a}^{2}}}{2} \right)}^{2n}}-{{\left( \frac{1+a}{2} \right)}^{n}}\]

    C)  \[{{\left( \frac{1+a}{2} \right)}^{2n}}-{{\left( \frac{1+{{a}^{2n}}}{2} \right)}^{n}}\]

    D)  \[{{\left( \frac{1+a}{2} \right)}^{n}}-{{\left( \frac{1+{{a}^{2n}}}{2} \right)}^{n}}\]            

    Correct Answer: A

    Solution :

    Given data is
    \[x:\] \[1\] \[a\] \[{{a}^{2}}.....{{a}^{n}}\]
    \[y:\] \[^{n}{{C}_{0}}\] \[^{n}{{C}_{1}}\] \[^{n}{{C}_{2}}{{....}^{n}}{{C}_{n}}\]
    \[\therefore \] Mean of given data, \[\bar{x}=\frac{^{n}{{C}_{0}}.1{{+}^{n}}{{C}_{1}}.a+....{{+}^{n}}{{C}_{n}}{{a}^{n}}}{^{n}{{C}_{0}}{{+}^{n}}{{C}_{1}}+.....{{+}^{n}}{{C}_{n}}}\] \[=\frac{{{(1+a)}^{n}}}{{{2}^{n}}}={{\left( \frac{1+a}{2} \right)}^{n}}\] \[\therefore \]  Variance \[=\frac{1}{N}\sum\limits_{i=1}^{n}{{{f}_{i}}{{({{x}_{i}}-\bar{x})}^{2}}}\] \[=\frac{1}{N}{{\sum\limits_{i=1}^{n}{{{f}_{i}}x_{i}^{2}-\left( \frac{1}{N}\sum{{{f}_{i}}{{x}_{i}}} \right)}}^{2}}\] \[=\frac{1}{{{2}^{n}}}\{\sum{{{(}^{n}}{{C}_{0}}{{.1}^{2}}}{{+}^{n}}{{C}_{1}}.\,\,{{a}^{2}}{{+}^{n}}{{C}_{2}}{{a}^{4}}\] \[+....{{+}^{n}}{{C}_{n}}\,{{a}^{2n}})\}\] \[-{{\left\{ {{\left( \frac{1+a}{2} \right)}^{n}} \right\}}^{2}}\] \[=\frac{1}{{{2}^{n}}}\,{{(1+{{a}^{2}})}^{n}}-{{\left( \frac{1+a}{2} \right)}^{2n}}\] \[=\left( \frac{1+{{a}^{2}}}{2} \right)-\left( \frac{1+a}{2} \right)h2n\]


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