• Q: Find the missing frequency in the following frequency distribution table,
    if N = 85 and median is 33 ,

    C 0-10 10-20 20-30 30-40 40-50 50-60 60-70  T
    f   6 22 f1 14 18 f2 3 85

    where:-
    C → class , f → frequency. T → Total. Share to Whatsapp



  • Given that data :
    Let f1 and f2 be the frequencies of class intervals 20-30 and 50-60 respectively. then,
    6 + 22 + f1 + 14 + 18 + f2 + 3 = 85
    f1 + f2 = 85 - 63

    f1 + f2 = 22
    f1 = 22 - f2 .................(1)

  • Median is 33 , which lies in 30-40. so, the median class is 30-40,
    since, L = 30, h = 10, f = 14, N = 85 and
      cf = 6 + 22 + f1 = 28 + f1

    Now,median = L + [h{(N/2)-cf}/f]
    =>33 = 30 + [ 10{(85/2)-(28 + f1)}/14 ]

    =>33 = 30 + [ 10{(42.5)-(28 + f1)}/14 ]
    =>33 = [30*14 + 10{(42.5)-(28 + f1)}]/(14)

    =>33 * 14 = [30*14 + 10{ 14.5 - f1}]
    => 462 = [420 + 145 - 10 *f1]

    => 462 = [565 - 10 *f1]
    => 10 *f1 = [565 - 462]

  • => 10 *f1 = 103
    => f1 = 10.3
    => f1 = 10
    From 1 , we get,
    f1 = 22 - f2
    => 10 = 22 - f2
    => f2 = 22 - 10
    => f2 = 12

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