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Unit 8: Measurement of Central Tendency




          Solution:                                                                                Notes

                 CI             f      Mid point (X)        fX
                 40–44          3      42                   126
                 35–39          5      37                   185
                 30–34          10     32                   320            Mean = “fN

                 25–29          14     27                   378
                 20–24          8      22                   176
                 15–19          6      17                   102
                 10–14          4      12                    48
                                N = 50        “fX = 1335
          (C) Calculating the Mean through Assumed Mean Method (The short method)

          In the Assumed Mean Method we try to avoid lengthy calculations of multiplications of mid-
          points of class intervals with corresponding frequencies. First of all, we locate a class that lies
          almost at the middle of the distribution. Its mid-point is taken as the Assumed Mean (A.M.).
          Now the class intervals around this, i.e., the choosen class interval for (A.M.) would be 1,2 or
          3 class intervals above and below it. So deviations from this class interval would be + 1, + 2,
          + 3 etc. and – 1, – 2, – 3 etc. in the subsequent class intervals containing higher and lower
          scores respectively on the two sides. These figures are obtained by subtracting the A.M. from
          the mid point of the class interval and dividing by the size of the class interval. However, this
          calculation is not required in regular practice while solving the questions. This method is also
          known as ‘Step Deviation Method’.
          The steps involved may be summarized as below :
          •    Arrange the data in a tabular form i.e., making columns for class interval (CI), frequency (f),
               deviation (d), and frequency  x  deviation (fd).
          •    Locate the class interval which falls midway in the distribution. If you come across two
               class intervals, choose the one with greater frequency.
          •    Fill up the column of deviation : zero against the class interval containing A.M, and +1,
               +2, +3 etc. against class intervals with larger score limits and –1, –2, –3 etc. against class
               intervals with smaller score limits.

          •    Find out multiplications of frequency and corresponding deviation and place the obtained
               value in the column headed by fd.
          •    Find the sum of the column fd  i.e., Cfd.

          •    The following formula may then be applied for calculating the Mean
                                Mean = A.M.+ {(“f d/N)} x i

          Where, A.M. = Assumed Mean
                    f = frequency
                    d = deviation from Assumed Mean
                    i = size of the class interval
          Interpretation of Mean

          Mean reprresents the centre of gravity of the scores. Measurements in any sample are perfectly
          balanced about the mean. In other words the sum of deviations of scores from mean is always


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