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J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    The mean deviation about the mean for the values 18, 20, 12, 14, 19, 22, 26, 16, 19, 24 is

    A)  \[3.1\]             

    B)  \[3.4\]

    C)  \[3.2\]             

    D)  \[3.3\]

    Correct Answer: D

    Solution :

    \[{{x}_{i}}\] \[|{{x}_{i}}-\bar{x}|\]
    \[18\] \[1\]
    \[20\] \[2\]
    \[12\] \[7\]
    \[14\] \[5\]
    \[19\] \[0\]
    \[22\] \[3\]
    \[26\] \[7\]
    \[16\] \[3\]
    \[19\] \[0\]
    \[24\] \[5\]
    Total     \[190\] \[33\]
    \[\bar{x}=\frac{18+20+12+....+24}{10}\] \[=\frac{190}{10}=19\] Here,  \[=(n=10)=\] total number of observations. Mean deviation \[=\frac{\Sigma \,|{{x}_{i}}-\bar{x}|}{n}=\left( \frac{33}{10} \right)=3.3\]


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