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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    If\[m\]and\[{{\sigma }^{2}}\]are the mean and variance of the random variable, whose distribution is given by
    \[X=x\] 0 1 2 3
    \[P(X=x)\] \[\frac{1}{3}\] \[\frac{1}{2}\] 0 \[\frac{1}{6}\]
    Then

    A) \[m={{\sigma }^{2}}=2\]              

    B)        \[m=1,\,\,{{\sigma }^{2}}=2\]

    C) \[m={{\sigma }^{2}}=1\]              

    D)        \[m=2,\,\,{{\sigma }^{2}}=1\]

    Correct Answer: C

    Solution :

    Given, distribution is
    \[X=x\] 0 1 2 3
    \[P(X=x)\] \[\frac{1}{3}\] \[\frac{1}{2}\] 0 \[\frac{1}{6}\]
    \[\therefore \]Mean,\[m=\sum\limits_{i=1}^{4}{{{p}_{i}}{{x}_{i}}}\]                                 \[=0\times \frac{1}{3}+1\times \frac{1}{2}+2\times 0+3\times \frac{1}{6}\]                                 \[=0+\frac{1}{2}+0+\frac{1}{2}=1\] Variance,             \[{{\sigma }^{2}}=\underset{i=1}{\overset{4}{\mathop{\Sigma }}}\,{{p}_{i}}{{({{x}_{i}}-m)}^{2}}\]                 \[=\frac{1}{3}{{(0-1)}^{2}}+\frac{1}{2}{{(1-1)}^{2}}\]                                                 \[+0{{(2-1)}^{2}}+\frac{1}{6}{{(3-1)}^{2}}\]                 \[=\frac{1}{3}+0+0+\frac{2}{3}=1\] \[\therefore \]  \[m={{\sigma }^{2}}=1\]


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