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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    A random variable X has the following probability distribution
    \[X={{x}_{1}}\] 1 2 3 4
    \[P(X={{x}_{1}})\] 0.1 0.2 0.3 0.4
    The mean and the standard deviation are respectively

    A)  3 and 2                                

    B)  3 and 1

    C)  3 and\[\sqrt{3}\]            

    D)         2 and 1

    E)  3 and \[\sqrt{2}\]

    Correct Answer: B

    Solution :

    \[{{x}_{i}}\] \[{{p}_{i}}\] \[x_{i}^{2}\] \[{{p}_{i}}{{x}_{i}}\] \[{{p}_{i}}x_{i}^{2}\]
    1 0.1 1 0.1 0.1
    2 0.2 4 0.4 0.8
    3 0.3 9 0.9 2.7
    4 0.4 16 1.6 6.4
    \[\Sigma {{p}_{i}}{{x}_{i}}=3\] \[\Sigma {{p}_{i}}x_{i}^{2}=10\]
    \[\therefore \]  \[\overline{x}=\Sigma {{p}_{i}}{{x}_{i}}=3\] and standard deviation\[=\sqrt{\Sigma {{p}_{i}}x_{i}^{2}-{{(\Sigma {{p}_{i}}{{x}_{i}})}^{2}}}\]                                                 \[=\sqrt{10-9=1}\]


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