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NDA - Mean, Median & Mode Test 948
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NDA - Mean, Median & Mode Test 948
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  • Question 1/5
    1 / -0.33
    Using given data points tabulated below, a straight line passing through the origin is fitted using least squares method. The slope of the line is

    Solutions

    Let the required line be  (passing through the origin) …….(i)

    Then the normal equation of equation (i) is given by

     

  • Question 2/5
    1 / -0.33
    The sum of square of deviation for 10 observation taken from mean 50 is 250. The coefficient of variation is
    Solutions

    Coefficient of variation

  • Question 3/5
    1 / -0.33
    The mean and variance of a binomial distribution are 4 and 3, respectively. Then the probability of getting exactly six successes in this distribution is
    Solutions

    Let  be a binomial variate with mean 4 and variance 3. Then,

    , and  

      and

     

     

  • Question 4/5
    1 / -0.33
    If  and 17.5 is the mode of the following frequency distribution.

    Then,

    Solutions

    We know that, which class has maximum frequency are called Modal class.

    So, here 15-20 is the modal class such that

    lower limit=15, Width  = 5, frequency = 6

    And Frequency of class preceding the modal class  = 3

    Frequency of class following the modal class  =

    We know,

    Mode =

     

     

  • Question 5/5
    1 / -0.33
    The median class in the following frequency distribution is

    Solutions

    Computation of Median

    We have,

    We know that cumulative frequency just greater than .

    i.e. Cumulative frequency in this distribution is 50 which is greater than 40 and the corresponding class is 20-30. Hence, 20-30 is the median class.

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