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J & K CET Engineering J and K - CET Engineering Solved Paper-2012

  • question_answer
    The following table gives the probability that certain computer will malfunction 0, 1, 2, 3, 4, 5 or 6 times on any day
    Number of malfunctions X 0 1 2 3 4 5 6
    Probability f(x) 0.17 0.29 0.27 0.16 0.07 0.03 0.01
    The mean of this probability distribution is

    A)  \[1.74\]             

    B)  \[1.80\]

    C)  \[0.74\]             

    D)  None of these

    Correct Answer: B

    Solution :

    Given,
    Number of Malfunctions \[({{x}_{i}})\] Probability \[({{p}_{i}})\] \[{{x}_{i}}{{p}_{i}}\]
    \[0\] \[0.17\] \[0\]
    \[1\] \[0.29\] \[0.29\]
    \[2\] \[0.27\] \[0.54\]
    \[3\] \[0.16\] \[0.48\]
    \[4\] \[0.07\] \[0.28\]
    \[5\] \[0.03\] \[0.15\]
    6 \[0.01\] \[0.06\]
    Total \[1.8\]
    \[\therefore \] Mean,  \[\bar{x}=\Sigma {{x}_{i}}{{p}_{i}}\] \[=1.8\]


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